nmayhall-vt.github.io · chem 6634: ˚antum chemistry and spectroscopy fall 2018 tr 11:00 am -...

CHEM 6634: antum Chemistry and SpectroscopyFall 2018TR 11:00 AM - 12:15 PMRoom: Davidson 325

Instructor: Prof. Nick Mayhall, Davidson Hall 421C, [email protected]

Oice Hours: 11:00 AM - 12:00 PM Wednesday.

Required Textbook: Molecular antum Mechanics: 5th edition, Atkins and Friedman,ISBN 978-0-19-954142-3

Helpful textbooks:• Modern antum Chemistry, Szabo and Ostlund;• Introduction to Computational Chemistry, Jensen;• antum Mechanics, Cohen-Tannoudji• Introduction to antum Mechanics in Chemistry, Ratner, Schatz

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Table of Contents

Contents1 Math Review 4

1.1 Wavefunction space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 State Space and Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Linear Operators with Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Representations in the state space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 Eigenvalue equations and observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4.1 Eigenvalue equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4.2 Hermitian operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.3 Sets of commuting Hermitian operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5 Two special representations, |r〉 and |p〉 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Postulates of antum Mechanics 242.1 Statement of the postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.2 Separation of variables: TDSE→ TISE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3 Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.4 Time-evolution of expectation values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Linear motion 313.1 Free particle wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.1.1 Wavepackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.1.2 Wavepackets at a given instant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Scaering across a finite barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Particle in a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.4 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 Rotational motion 474.1 Particle in a Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.2 Particle on a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.3 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5 Angular momentum 575.1 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.2 antization of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.3 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.4 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.5 Coupling of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.6 Dichromium Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

6 Techniques of Approximation 76

2

CONTENTS

6.1 Time-independent perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.1.1 Two-level system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.1.2 Many-level systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.2 Variational Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.3 Time-dependent perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.3.1 Two-level system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.3.2 Many-level systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.3.3 Transitions to several possible states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936.3.4 Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7 Many-electron systems 977.1 Hydrogen atom spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

7.1.1 Selection rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977.1.2 Spin Orbit Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.2 Helium Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1027.2.1 Slater Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1037.2.2 Statistical Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7.3 Hartree-Fock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.3.1 Energy of a Slater Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.3.2 Minimization of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1077.3.3 Canonical orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1097.3.4 Koopman’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1117.3.5 Solving the Hartree-Fock equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1127.3.6 One-electron wavefunction basis sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

7.4 Electron Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1167.4.1 DFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.4.2 Kohn-Sham DFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.5 Excited States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1277.5.1 Born-Oppenheimer Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

8 Appendix 1338.1 Properties of the Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

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1 Math Review1.1 Wavefunction space• Schrödinger Equation for one particle:

ihddt

Ψ(r, t) =− h2

2m∇2Ψ(r, t) + V (r, t)Ψ(r, t) (1)

=HΨ(r, t) (2)

• Without justification: Ψ(r, t) is a probability amplitude with |Ψ(r, t)|2d3r being the probability of findingthe particle at time t inside d3r about r.

• As a probability:∫

d3r|Ψ(r, t)|2 = 1

– thus, Ψ(r, t) is restricted in form

• Restrictions on functional form of Ψ(r, t):

Restrictions on functional form of Ψ(r, t), defining wavefunction space , F

Square integrable→ set “L2” which has the structure of a Hilbert space (complete,and has inner product)

Continuous

Infinitely-dierentiable: (any cusp would occur at a space scale which is beyondmeasurement, and thus cannot be physical, also, kinetic energy would blow up)

Too small to measureso must be unphysical

Bounded domain: system should be finite

• F forms a vector space

Ψ3 = λ1Ψ1 + λ2Ψ2 ∈ F (3)

– Familiar example: 3D space

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– Any linear combination of ~x, ~y, and~z is also a vector in that space.

~v =vx~x + vy~y + vz~z =

vx

vy

vz

xyz

(4)

subscript “xyz” indicates that the vector is in the cartesian coordi-nate system

• This vector space has a scalar product (or inner product or “overlap”)

(φ, ψ) =∫

d3rφ(r)∗ψ(r) < |∞| (5)

– Properties:

(φ, ψ) = (ψ, φ)∗ complex conjugate (6)

(φ, λ1ψ1 + λ2ψ2) =λ1 (φ, ψ1) + λ2 (φ, ψ2) linear wrt. |ψ〉 (7)

(λ1φ1 + λ2φ2, ψ) =λ∗1 (φ1, ψ) + λ∗2 (φ2, ψ) antilinear or conjugate linear wrt. |φ〉 (8)

– Scalar product generalizes dot product of 3D vectors

(~v, ~w) = ~v · ~w = ~v†~w =√~v ·~v√~w · ~w cos θ

Use cos θ to illustrate Schwartz inequality

– Schwarz inequality :

| (φ, ψ) | ≤√(φ, φ)

√(ψ, ψ) (9)

• linear operator

– operator: a transformation which maps one function onto another function

– for linear operator, A, and number, a:

Properties of linear operators

Aψ(r) = ψ(r)′ operator definition

A (ψ(r) + φ(r)) = Aψ(r) + Aφ(r) additivity

A (aψ(r)) = aAψ(r) homogeneity

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– Examples:

∗ Multiplication: Xψ = xψ

∗ Dierentiation: Dxψ = ddx ψ

∗ Exponentiation: eψ+φ 6= eψ + eφ

– Order maers! : AB 6= BAExample, consider an arbitrary function of x, f (x) :

xd

dxf (x) = x f (x)′

but reversing the order:d

dxx f (x) = f (x) + x f (x)′

so, in general:d

dxx 6= x

ddx

– commutator : [A, B] = AB− BA[d

dx, x]

f (x) =(

ddx

x− xd

dx

)f (x) = f (x) + x f (x)′ − f (x)′ = f (x)

such that [d

dx, x]= 1

reversing order inside commutator changes sign:

[A, B] = −[B, A]

• Bases in F : ui(r)

– Just like ~x, ~y, and~z form a basis for 3D space, we can use bases for F– For convieniance, ui(r) are typically orthonormal

(ui, uj

)=∫

ui(r)∗uj(r)d3r = δij =

1 if i = j0 else

(10)

Where, δij is the kronecker delta

– If ui(r) is a basis for F it can be used express any Ψ(r)

Ψ(r) = ∑i

ui(r)ci

– Coeicients are scalar products(uj(r), Ψ(r)

)=∑

i

(uj(r), ui(r)

)ci (11)

=∑i

δijci (12)

=cj (13)

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– Scalar products using a basis: let φ(r) = ∑j bjuj(r)

(φ, ψ) =∑i

∑j

(biui, cjuj

)(14)

= ∑i

∑j

b∗i cj(ui, uj

)(15)

= ∑i

∑j

b∗i cjδij (16)

= ∑i

b∗i ci (17)

(ψ, ψ) =∑i

c∗i ci = ∑i|ci|2 ≥ 0 (18)

– Completeness of ui(r) :

∗ ui must span the space of F

ψ(r) =∑i

ciui(r) = ∑i(ui, ψ) ui(r) (19)

=∑i

[∫u∗i (r

′)ψ(r′)d3r′]

ui(r) expand scalar product as integral (20)

=∫ [

∑i

u∗i (r′)ui(r)

]ψ(r′)d3r′ bring summation into integral (21)

=∫ [

f (r, r′)]

ψ(r′)d3r′ introduce some function, f (22)

∗ For this to equality to hold, the following must be true:

∑i

u∗i (r′)ui(r) = δ(r− r′) closure relation (23)

∗ Where we have introduced the Dirac delta function :

δ(r− r′) =

+∞ if r = r′

0 else(24)

which has the following property:∫ +∞

−∞δ(r− r′)dr′ =1 continuous analogy of δij (25)

• Generalization to continuous bases:

– Essentially, ∑i →∫

dα and ci → c(α)

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Relation Discrete: index i Continuous, index α

Orthonormalization(ui(r), uj(r)

)= δij (uα(r), uα′(r)) = δ(α− α′)

Closure ∑i ui(r)ui(r′)∗ = δ(r− r′)∫

α uα(r)uα(r′)∗dα = δ(r− r′)

Expansion ψ(r) = ∑i ciui(r) ψ(r) =∫

α c(α)uα(r)dα

Coeicients ci = (ui, ψ) c(α) = (uα, ψ)

Scalar product (φ, ψ) = ∑i b∗i ci (φ, ψ) =∫

α b(α)∗c(α)dα

– Example, 1D plane waves:

νp(x) =1√2πh

eipx/h (26)

– Apply the closure relation for a continuous basis:

ψ(x) =∫

dx′δ(x− x′)ψ(x′) (27)

=∫

dx′(∫

dpv∗p(x′)vp(x))

ψ(x′) (28)

=∫

dpvp(x)(∫

dx′v∗p(x′)ψ(x′))

(29)

=1√2πh

∫dpeipx/hψ(p) (30)

such that ψ(x) is wrien as the Fourier transform of ψ(p), which itself is the inverse Fourier trans-form of ψ(x),

ψ(p) =1√2πh

∫dx′e−ipx′/hψ(x′) (31)

Equivalantly, ψ(p) also happens to be the “expansion coeicients” in the plane-wave basis

ψ(p) =(νp(x), ψ(x)

)(32)

=∫ +∞

−∞dxνp(x)∗ψ(x) (33)

=1√2πh

∫ +∞

−∞dxψ(x)e−ipx/h (34)

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1.2 State Space and Dirac Notation

• Once a basis is chosen, Ψ(r) can be fully described by a vector of coeicients

Basis Components

ui(r) ci

uα(r) c(α)

: :

• the components of the Ψ(r) change depending on the basis

Assume ~x′,~y′,~z′ is rotated w.r.t. ~x,~y,~z

~v =

vx′

vy′

vz′

x′y′z′

=

vx

vy

vz

xyz

but

vx′ 6= vx

• In fact, Ψ(r) itself can be thought of as a coeicient of an abstract wavefunction (without any basis).Ψ(r) is thus the representation of the state “Ψ” in a continuous basis in real space, ur0(r) = δ(r− r0).

Ψ(r) =∫

dr0δ(r− r0)Ψ(r0) (35)

• To beer articulate this, let’s adopt a notation which is “basis agnostic”→ Dirac notation

• For every state, there exists a state vector called a ket, |〉

– |〉 fully specifies the state of the system, without referring to a basis

• |ψ〉 is a vector corresponding to ψ(r), which lives in a vector space, E , called state space

• F is isomorphic to Eψ(r) ∈ F ⇔ |ψ〉 ∈ E

• Dual space, E∗

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– Linear functional : χ (|ψ〉)→ k ∈ C

functionoperator−−−−→ function

functionf unctional−−−−−→ number

– The set of all linear functionals defined on |ψ〉 ∈ E constitutes a vector space, called dual space

– A vector in this dual space is called a bra , 〈|

χ (|ψ〉) = 〈χ|ψ〉 = k ∈ C 〈χ| ∈ E∗ (36)

– For finite dimensions, E∗ and E are isomorphic.

∗ each ket has an associated bra, but not all possible bra’s must have an associated ket∗ we won’t consider this further

– To every |ψ〉, there corresponds one 〈ψ|– Each linear functional, 〈φ|, transforms |ψ〉 into the scalar product of |φ〉 and |ψ〉

〈φ|ψ〉 = (|φ〉 , |ψ〉) = k ∈ C (37)

⇔∫

φ(r)∗ψ(r)d3r (38)

1.2.1 Linear Operators with Dirac Notation

• bra-ket = 〈φ|ψ〉 = k ∈ C

• ket-bra? : Consider action on arbitrary function

(|ψ〉〈φ|) |χ〉 = |ψ〉〈φ|χ〉 (39)

= |ψ〉 k (40)

• thus |ψ〉〈φ| is an operator

• Order of bras and kets is everything!

〈ψ|φ〉 = complex number

|ψ〉〈φ| = operator

• Analogy with 3-space vectors

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– basis vectors

~ei = ~x,~y,~z =

1

0

0

,

0

1

0

,

0

0

1

(41)

〈x|x〉 =(

1 0 0

)

1

0

0

= 1 (42)

〈x|y〉 =(

1 0 0

)

0

1

0

= 0 (43)

|x〉〈y| =

1

0

0

(

0 1 0

)=

0 1 0

0 0 0

0 0 0

(44)

In a given basis, kets are column vectors, and operators are matrices!

• Projector onto |ψ〉

– assume |ψ〉 is normalized, 〈ψ|ψ〉 = 1

– define Pψ = |ψ〉〈ψ|

Pψ |φ〉 = |ψ〉〈ψ|φ〉 ∝ |ψ〉 (45)

– proportionality constant = overlap

– this is called an orthogonal projection

• Projections are idempotent : i.e., PP = P

|ψ〉〈ψ| |ψ〉〈ψ| = |ψ〉 (〈ψ|ψ〉) 〈ψ| = |ψ〉〈ψ| since normalized (46)

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• Resolution of the Identity : ∑i|ψi〉〈ψi| = I

|x〉〈x|+ |y〉〈y|+ |z〉〈z| =

1 0 0

0 0 0

0 0 0

+

0 0 0

0 1 0

0 0 0

+

0 0 0

0 0 0

0 0 1

(47)

=

1 0 0

0 1 0

0 0 1

(48)

=I (49)

• subspace projections

Project cartesian 3D vector, |v〉, onto x, y plane.

Pxy = |x〉〈x|+ |y〉〈y| =

1 0 0

0 1 0

0 0 0

(50)

Pxy |v〉 = (|x〉〈x|+ |y〉〈y|)(vx |x〉+ vy |y〉+ vz |z〉

)(51)

=vx |x〉+ vy |y〉 (52)

=

1 0 0

0 1 0

0 0 0

vx

vy

vz

=

vx

vy

0

(53)

Hartree-Fock density matrix is a subspace projection operator. It isa projection onto the occupied space.

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• matrix representation of operators

A =IAI = ∑ij|ψi〉〈ψi| A

∣∣ψj⟩⟨

ψj∣∣ Use RI (54)

=∑ij|ψi〉 Aij

⟨ψj∣∣ matrix elements (55)

– Aij is the i, jth element of the matrix representation of A in the basis ψi

– since A is an operator, A |ψ〉 = |ψ′〉

〈ψi| A∣∣ψj⟩=⟨

ψi

∣∣∣ψ′j⟩ scalar product - we know what to do (56)

⇔∫

ψ∗i (r)Aψj(r)d3r if position space representation (57)

– What about AB?

AB =IAIBI (58)

=∑ijk|ui〉〈ui| A

∣∣uj⟩⟨

uj∣∣ B |uk〉〈uk| (59)

=∑ijk|ui〉 AijBjk 〈uk| (60)

〈ui| AB |uk〉 =(

AB)

ik = AijBjk (61)

• adjoint or Hermitian conjugate

|ψ〉 A−→ |ψ′〉 A |ψ〉 = |ψ′〉

⇓ ⇓ ⇓ Adjoint

〈ψ| A†−→ 〈ψ′| 〈ψ| A† = 〈ψ′|

Adjoint “†” is the operation which formally establishes the one-to-one relationship between bras and kets.

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Order changes(AB)†

=B† A† (62)

(|x〉〈y|)† = (〈y|)† (|x〉)† Example (63)

= |y〉〈x| (64)

=

0 1 0

0 0 0

0 0 0

†

=

0 0 0

1 0 0

0 0 0

If real, adjoint = transpose (65)

(A†)

ij=(

A)∗

ji (66)

To obtain Hermitian conjugate:

1. Replace:

– λ→ λ∗ for λ ∈ C

– |ψ〉 → 〈ψ|– 〈ψ| → |ψ〉– A→ A†

2. Reverse the order

• If A† = A, A is said to be Hermitian .

1.3 Representations in the state space

• Choosing an orthonormal basis in state space, is called choosing a representation .

• Introduction of a basis, |ui〉:

|ψ〉 =I |ψ〉 (67)

=∑i|ui〉〈ui| |ψ〉 (68)

=∑i

ci |ui〉 (69)

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• Change of basis, |ui〉 → ∣∣wj⟩:

|ψ〉 =∑i|ui〉 ci (70)

=∑i

I |ui〉 ci (71)

=∑i

∑j

∣∣wj⟩ ⟨

wj∣∣ui⟩

ci (72)

=∑i

∑j

∣∣wj⟩

Sjici (73)

get components in new basis, bk:

bk = 〈wk|ψ〉 (74)

=∑i〈wk|ui〉〈ui|ψ〉 (75)

=∑i

Skici (76)

• For change of basis:|ui〉 → ∣∣wj⟩:

Ski = 〈wk|ui〉 Overlap, or scalar product (77)

S∗ik = 〈ui|wk〉 adjoint (78)

∑k

S∗ikSkj =∑k〈ui|wk〉

⟨wk∣∣uj⟩

Matrix multiply of both (79)

=⟨ui∣∣uj⟩

(80)

=δij Orthogonalization (81)

thus,

SS† =S†S = I (82)

S† = S−1 S is thus a unitary matrix (83)

1.4 Eigenvalue equations and observables

1.4.1 Eigenvalue equations

• For λ ∈ C, if

A |ψ〉 = λ |ψ〉 (84)

Then |ψ〉 is said to be an eigenvector of operator A, with eigenvalueλ.

• An operator will (in general) have many eigenvalues:

A |ψs〉 = λs |ψs〉 (85)

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• If each λs corresponds to a single |ψs〉, we say that λs is non-degenerate

• The number of linearly-independant eigenvectors associated with λi is the order of degeneracy , gs

• Finding eigenvalues:

〈ui| A |ψs〉 = λs 〈ui|ψs〉 project onto some basis (86)

∑j〈ui| A

∣∣uj⟩⟨

uj∣∣ |ψs〉 = ∑

jλs 〈ui|

∣∣uj⟩⟨

uj∣∣ |ψs〉 RI (87)

∑j

Aijcjs = λscis (88)

Acs = csλs matrix form (89)

(A− Iλs) cs = 0 (90)

– obviously, cs = 0 is a trivial (yet unphysical) solution.

– if (A− Iλs)−1 exists, then cs is trivial

– thus we want values of λs for which (A− Iλs)−1 does not exist

– equivalent to solving

|A− Iλs| =0 called the secular equation (91)

Examples:

A =d

dx(92)

|ψ〉 =aebx (93)d

dx

(aebx

)=b(

aebx)

(94)

A =d2

dx2 (95)

|ψ〉 =a sin bx (96)

d2

dx2 (a sin bx) =− b2 (a sin bx) (97)

1.4.2 Hermitian operators

• Observables (or “measurable” quantities) are associated with a Hermitian operator

• Properties of Hermitian operators: A† = A

1. Hermitian operators have real eigenvalues

Mayhall 16

1 MATH REVIEW

A |λ〉 =λ |λ〉 given

〈λ| A† = 〈λ| λ∗ take adjoint

〈λ| A |λ〉 =λ le multiply by 〈ψ|〈λ| A† |λ〉 =λ∗ right multiply by |ψ〉

Since 〈λ| A |λ〉 = 〈λ| A† |λ〉 definition of Hermitian

then λ =λ∗

2. Eigenvectors with dierent eigenvalues are orthogonal


A∣∣λ′⟩ =λ′

∣∣λ′⟩ given

⟨λ′∣∣ A =λ′

⟨λ′∣∣ Since A is Hermitian

⟨λ′∣∣ A |λ〉 =λ

⟨λ′∣∣λ⟩ le multiply by

⟨λ′∣∣⟨

λ′∣∣ A |λ〉 =λ′

⟨λ′∣∣λ⟩ right multiply by |λ〉

(λ′ − λ

) ⟨λ′∣∣λ⟩ =0 subtracting the two equations

Thus, either λ = λ′ or |λ〉 and |λ′〉 are orthogonal

3. Linear combinations of e’vectors with same e’values can are also e’vectors, with same e’value.


A∣∣λ′⟩ =λ

∣∣λ′⟩ given

A(a |λ〉+ b

∣∣λ′⟩) =aA |λ〉+ bA∣∣λ′⟩ linear operator

=aλ |λ〉+ bλ∣∣λ′⟩ eigenvalues

=λ(a |λ〉+ b

∣∣λ′⟩)

Since they can be combined, they can be orthogonalized, and thusform a basis for the space - the eigenbasis

Mayhall 17

1 MATH REVIEW

– If any eigenvalues of A are degenerate, specifying the eigenvalue does not uniquely identify aneigenvector, but an eigensubspace

– A second commuting operator might be chosen which has distinct eigenvalues in that same eigen-subspace

Definition: Complete sets of commuting observables

a set of commuting operators whose eigenvalues completely specify thestate of a system

or a minimal set of commuting operators for which a list of eigen-values (one for each operator in the set) identifies a specific vector

0

1A

Original Basis

Eigenbasis, a

B

Original Basis

Eigenbasis, a0

Eigenbasis, (a,b)1 2 3

C

Original Basis

Eigenbasis, a0

Eigenbasis, (a,b)1 2 30 1 2 3

Eigenbasis, (a,b,c)

Done

Degenerate

Degenerate

No Degeneracy

1.5 Two special representations, |r〉 and |p〉• Let’s consider two possible bases for F :

χr0(r) =δ(r− r0) ⇔ |r0〉 (104)

νp0(r) =(

12πh

)3/2

eih p0·r ⇔ |p0〉 (105)

– Neither of these functions are members of F , but it’s ok

– Referred to as position and momentum functions

Mayhall 19

1 MATH REVIEW

Note: 3D plane waves are just product of 1D plane waves

νp(r) =(2πh)−3/2eih(px~x+py~y+pz~z)·(x~x+y~y+z~z)

=(2πh)−3/2eih pxxe

ih pyye

ih pzz

=(2πh)−1/2eih pxx(2πh)−1/2e

ih pyy(2πh)−1/2e

ih pzz

=νpx(x)νpy(y)νpz(z)

• Orthonormalization:⟨r0∣∣r′0⟩ = ∫ drδ(r− r0)δ(r− r′0) (106)

=δ(r0 − r′0) 3 (107)

(108)⟨p0∣∣p′0⟩ = ∫ dr

(1

2πh

)3

e−ih p0·re

ih p′0·r (109)

=∫

dr(

12πh

)3

eih(p0−p′0)·r (110)

=δ(p0 − p′0) trust me, or not! (111)

Use the following identity of the Fourier transform (typical forantum Mechanics):

δ(x− x0) =1

2πh

∫ ∞

−∞e

ih px(x−x0)dpx

• Closure relation: ∫dr0 |r0〉〈r0| =

∫dp0 |p0〉〈p0| = I

– consider action on arbitrary function, v(r)⇔ |v〉:[∫dr0 |r0〉〈r0|

]|v〉 =

∫dr0 |r0〉〈r0|v〉

=∫

dr0δ(r− r0)∫

dr′δ(r′ − r0)v(r′) closure def

=∫

dr0δ(r− r0)v(r0) use orthonormality

=v(r)⇔|v〉 function is le unchanged

– thus, identity!

Mayhall 20

1 MATH REVIEW

where,

〈r0|p0〉 =∫

dr χr0(r)νp0(r) scalar product

〈r0|p0〉 = (2πh)−3/2∫

dr δ(r− r0)eih p0·r function definitions

〈r0|p0〉 = (2πh)−3/2 eih p0·r0 delta kills r

Combining these last two results:

ψ(r0) =∫

(2πh)−3/2 eih p0·r0ψ(p0)dp0

Thus,

ψ(r0)Fourier transform←−−−−−−−→ ψ(p0)

• Position and momentum operators

– In position representation, the position operator is just multiplication

〈r0| X |ψ〉 =x 〈r0|ψ〉 = xψ(r0)

– In momentum representation, the momentum operator is just multiplication

〈p0| Px |ψ〉 =px 〈p0|ψ〉 = pxψ(p0)

– What about the position operator in the momentum basis?

ψ(x) =1√2πh

∫ ∞

−∞e

ih px·xψ(px)dpx Defined above (120)

xψ(x) =1√2πh

∫ ∞

−∞xe

ih px·xψ(px)dpx multiply by x (121)

uv−∫

udv =∫

vdu integration by parts (122)

u = −iheih px v = ψ(px)

du = xeih pxdp dv = ∂

∂pxψ(px)dpx

xψ(x) =1√2πh

[−ihe

ih pxψ(px) +

∫ihe

ih px ∂

∂pxψ(px)dpx

]∞

−∞subsitute (123)

xψ(x) =1√2πh

∫ ∞

−∞ihe

ih px ∂

∂pxψ(px)dpx wavefunction integrable (124)

Comparing the equations for xψ(x), one sees that x → ih ∂∂px

Mayhall 22

2 POSTULATES OF QUANTUM MECHANICS

2 Postulates of antum Mechanics2.1 Statement of the postulates

Definition: antum Postulate 1

At a fixed time t0, the state of a physical system is defined by specifying aket |ψ(t0)〉 belonging to the state space E .

[Corresponds to Postulate 1 in Atkins and Friedman]


Every measurable quantity A is described by a Hermitian operator A actingin E ; this operator is an observable, and can be expressed as a function ofoperators R and P.


• Example: Hamiltonian operator

– Consider the classical Hamiltonian for one particle (sum of kinetic and potential energies)

H(r, p) =p2

2m+ V(r)

– obtain quantum Hamiltonian operator by replacing dynamic variables with operators:

H =P2

2m+ V(R)

– Position representation:

H = − h2

2md2

dr2 + V(r)


The only possible result of the measurement of a physical quantity A is oneof the eigenvalues of the corresponding observable A.

[Corresponds to Postulate 3 and 3’ in Atkins and Friedman]

• Comments:

– Since Observables are Hermitian operators, measurements only give real values

Mayhall 24


– If the spectrum of an operator is discrete, this system must be “quantized”


(Case of a non-degenerate discrete spectrum)When the physical quantityA is measured on a system in a normalized state|ψ〉, the probability of observing the eigenvalue an of the corresponding oper-ator A is:

P(an) =| 〈un|ψ〉 |2

=|cn|2

where, |un〉 is the eigenvector of A corresponding to eigenvalue an.


• Comments:

– Assume, system is in state |ψ〉, and an operator has eigenvectors: A |λn〉 = λn |λn〉.Expectation value is the average value obtained aer making many measurements:⟨

A⟩= 〈ψ|A|ψ〉 definition

=∑nm〈ψ| |λm〉〈λm| A |λn〉〈λn| |ψ〉 RI

=∑nm

λn 〈ψ|λm〉〈λm|λn〉〈λn|ψ〉 eigenvalue

=∑n

λn 〈ψ|λn〉〈λn|ψ〉 orthogonality of m, n

=∑n

λn|cn|2 weighted average

– Each individual measurement returns one eigenvalue, weighted by the squared modulus of the co-eicient, as stated by the postulate.

– If |ψ〉 happens to be an eigenstate of A, then it will always give the corresponding eigenvalue

– Note:

|cn|2 =c∗ncn = 〈ψ|un〉〈un|ψ〉 (125)

= 〈ψ| Pn |ψ〉 (126)

Mayhall 25



If the measurement of the physical quantityA on the system in state |ψ〉 givesthe result an, the state of the system immediately aer the measurement isthe normalized projection, of |ψ〉 onto the eigensubspace associated with an:

|ψ〉 an−→ Pn |ψ〉√〈ψ| Pn |ψ〉

,

This is referred to as wavefunction collapse

[Corresponds to Postulate 3 and 3’ in Atkins and Friedman]

• Comments:

– Considering two sequential measurements of A justifies this

– What does this say about commuting observables, [A, B] = 0?

∗ What if we measure A, then immediately measure B? (assume non-degenerate eigenvalues)

– Uncertainty


The time evolution of the state vector |ψ(t)〉 is governed by the Schrödingerequation:

ihddt|ψ(t)〉 = H(t) |ψ(t)〉 ,

where H(t) is the Hamiltonian operator, or the Hermitian operator associ-ated with the total energy of the system.


2.2 Separation of variables: TDSE→ TISE

• For time-independent Potential V(x) 6= f (t):

ihddt|ψ(t)〉 =

[− h2

2md2

dx2 + V(x)

]|ψ(t)〉 TDSE

• Position and Time are separable, so we assume the following solution form:

|ψ(t)〉 =Ψ(x, t) = ψ(x)θ(t) Position representation

Mayhall 26


• Substitute into TDSE

ihddt

ψ(x)θ(t) =

[− h2

2md2

dx2 + V(x)

]ψ(x)θ(t)

ψ(x)ihddt

θ(t) =θ(t)

[− h2

2md2

dx2 + V(x)

]ψ(x) Rearrange

1θ(t)

ihddt

θ(t) =1

ψ(x)

[− h2

2md2

dx2 + V(x)

]ψ(x) divide byψ(x)θ(t)

• Since LHS depends only on t, and RHS depends only on x, the equality means that both must equal aconstant:

E =1

θ(t)ih

ddt

θ(t) constant E

=1

ψ(x)

[− h2

2md2

dx2 + V(x)

]ψ(x)

• Yields two equations:

Eθ(t) =ihddt

θ(t) time equation

Eψ(x) =

[− h2

2md2

dx2 + V(x)

]ψ(x) space equation

Eψ(x) =Hψ(x) TISE

• Time equation is easy:

θ(t) =e−ih Et time solution

• Thus:

Ψ(x, t) = ψ(x)e−ih Et

• Even for time-independent Hamiltonian, wavefunction evolves in time??

• Yes, but the wavefunction is not an observable. What about probability density?

Ψ(x, t)∗Ψ(x, t) =ψ(x)∗eih Etψ(x)e

−ih Et

=ψ(x)∗ψ(x)6= f (t) ψ(x) is a stationary state

Mayhall 27


2.3 Uncertainty Principle

• Assume for two observables, A and B, [A, B

]= iC

• Consider a state |φ〉 =(

A + iλB)|ψ〉

• For any λ the squared norm must be non-negative:

〈φ|φ〉 ≥0

〈φ|φ〉 = 〈ψ|(

A− iλB) (

A + iλB)|ψ〉 by definition

=⟨

A2⟩+ iλ 〈[A, B]〉+ λ2

⟨B2⟩

expand terms

=⟨

A2⟩− λ

⟨C⟩+ λ2

⟨B2⟩

〈φ|φ〉 =⟨

B2⟩(

λ−⟨C⟩

2⟨

B2⟩)2

+⟨

A2⟩−⟨C⟩2

4⟨

B2⟩ ≥ 0 completing the square

Remember how to complete the square?

ax2 + bx + c = a(

x +b

2a

)2+ c− b2

4a

– For this to always be true for any value of λ, even when it makes the red term zero:

⟨A2⟩−⟨C⟩2

4⟨

B2⟩ ≥0 since λ is arbitrary

⟨A2⟩ ⟨

B2⟩≥⟨C⟩2

4rearranging

• This is true for any two Hermitian operators

• Consider the following “spread” operator:

δA =A−⟨

A⟩[

δA, δB]=[A, B

]−[A,⟨

B⟩]−[⟨

A⟩

, B]+[⟨

A⟩

,⟨

B⟩]

commutator

=[A, B

]Expectation value = scalar

=iC definition above

δA =δA† also Hermitian

Mayhall 28


• Thus, this also must hold:⟨δA2

⟩ ⟨δB2⟩≥⟨C⟩2

4rearranging⟨

δA2⟩ ⟨

δB2⟩=⟨(

A−⟨

A⟩)2⟩ ⟨(

B−⟨

B⟩)2⟩

expanding

=(⟨

A2⟩−⟨

A⟩2) (⟨

B2⟩−⟨

B⟩2)

expanding

• Thus, taking square root we get the famous Heisenberg uncertainty principle :

∆A∆B ≥|⟨C⟩|

2square root

or

∆A∆B ≥|⟨[

A, B]⟩|

2

• Where we have introduced the root mean square deviation or RMSD :

∆A =

√⟨A2⟩−⟨

A⟩2

Example: position and momentum

∆x∆ px ≥h2

2.4 Time-evolution of expectation values• Take derivative

∂

∂t⟨

A⟩=

∂

∂t〈ψ|A|ψ〉 definition

=

⟨∂

∂tψ

∣∣∣∣A∣∣∣∣ψ⟩+ 〈ψ| ∂

∂tA|ψ〉+

⟨ψ

∣∣∣∣A∣∣∣∣ ∂

∂tψ

⟩• remember TDSE

ih∂

∂t|ψ〉 =H |ψ〉

−ih∂

∂t〈ψ| = 〈ψ| H

• Substitute

∂

∂t⟨

A⟩=−1ih〈ψ|HA|ψ〉+ 〈ψ| ∂

∂tA|ψ〉+ 1

ih〈ψ|AH|ψ〉

=ih〈ψ|[H, A]|ψ〉+ 〈ψ| ∂

∂tA|ψ〉

Mayhall 29


• If A 6= f (t) then,

∂

∂t⟨

A⟩=

ih〈ψ|[H, A]|ψ〉

– Anything that commutes with the Hamiltonian is a constant of motion!

Definition: constant of motion

is a quantity that is conserved throughout the motion, imposing ineect a constraint on the motion

• What if A = H 6= f (t)?

∂

∂t⟨

H⟩=

ih〈ψ|[H, H]|ψ〉

=0

– Energy is clearly a constant of motion!

– Is momentum a constant of motion? Yes, for a free particle (zero potential)

Mayhall 30

3 LINEAR MOTION

3 Linear motion3.1 Free particle wavefunctions

• Consider simplest case: constant potential (V(x) = 0) Hamiltonian for single particle in 1D

H =p2

x2m

position−−−−−−−→representation

− h2

2md2

dx2

such that we seek solutions of the TDSE:

ih∂

∂t|ψ〉 =− h2

2md2

dx2 |ψ〉

• Since H is independent of time,

|ψ(t)〉 = |ψ〉 φ(t) Sep. of vars

= |ψ〉 e−ih Et

• What is |ψ〉? Eigenvalue of d2

dx2 , thus Ae±ikx is a solution

− h2

2md2

dx2 |ψ〉 =h2

2mk2Ae±ikx

=k2h2

2m|ψ〉

=E |ψ〉

thus,

E =k2h2

2m

• what is k? Consider momentum operator in position representation

px |ψ〉 =− ih∂

∂xAeikx

=khAeikx

=kh |ψ〉

Thus, px = ±kh

• |ψ〉 is therefore an eigenvalue of both H and px, with ±kh eigenvalue being equal to the momentum inx-direction

Mayhall 31

3 LINEAR MOTION

|ψ〉 =Aeikx Forward momentum state

|ψ〉 =Ae−ikx Backward momentum state

• Total wavefunction: substitute for E

|ψ(t)〉 = |ψ〉 e−i hk22m t

= |ψ〉 e−iωt where, ω =hk2

2m=Aeikxe−iωt

=Aei(kx−ωt)

Definition: plane wave

A plane wave has the following form:

|ψ(t)〉 =Aei(kx−ωt)

Definition: dispersion relation

A free particle has the following dispersion relation, which relates theangular frequency ω = 2π f , to the wavenumber k:

ω(k) =hk2

2m

• Unlike px, the total energy operator H also has eigenvectors which are linear combinations of le andright states

– Does this make sense? Think of the properties of degenerate eigenvalues of Hermitian operators.

– therefore, the energy eigenstates have a more general form of being linear combinations of le andright states

– if |ψ〉 = Aeikx + Be−ikx,

− h2

2md2

dx2 |ψ〉 =−h2

2m

(−k2Aeikx +−k2Be−ikx

)=

k2h2

2m

(Aeikx + Be−ikx

)=E |ψ〉

Mayhall 32

3 LINEAR MOTION

3.1.1 Wavepackets

• A single plane wave is not square-integrable, and thus cannot be a description of a physical state∫ ∞

−∞dx(

A∗Ae−i(kx−ωt)ei(kx−ωt))= |A|2x

∣∣∣∞−∞

= ∞

• To get a good state, we use the fact that since the Schrödinger equation is a linear dierential equation,a linear superposition of solutions is also a solution.

|ψ(t)〉 = 1√2π

∫ ∞

−∞dk g(k)ei(kx−ωt) linear combination

3.1.2 Wavepackets at a given instant

• If we consider the wavepacket at only a given instant t = t0:

|ψ〉 = 1√2π

∫ ∞

−∞dk g(k)eikx at t0 = 0

we see that g(k) is the Fourier transform of wavefunction, and it is called the shape function .

Prob

ability

x

Prob

ability

px

Prob

ability

px

Prob

ability

x

Prob

ability

x

Prob

ability

px

• wavepackets spread out in time

– At t = 0 we have a localized wavefunction about x = x0, with momentum localized about k = k0

Mayhall 33

3 LINEAR MOTION

– At t = ∆t, the bulk of the wavepacket has translated according to momentum k, but some proba-bility of smaller and larger momentum exists

– Thus some of the wavepacket moves slowly, some moves faster, resulting in a spreading out of thewavepacket

– consider a perfectly localized in space function as an example

3.2 Scaering across a finite barrier

I II IIIV(x)

V

0

I II III0 L

V(x)

V

0

• Schrödinger equations:

– Zone 1:

− h2

2md2

dx2 ψ(x) =Eψ(x)

– Zone 2: (− h2

2md2

dx2 + V

)ψ(x) =Eψ(x)

− h2

2md2

dx2 ψ(x) = (E−V)ψ(x)

– Zone 3:

− h2

2md2

dx2 ψ(x) = Eψ(x)

• Solutions:

ψI(x) =Aeikx + Be−ikx with k =

√2mEh

ψI I(x) =A′eik′x + B′e−ik′x with k′ =√

2m(E−V)

hψI I I(x) =A′′eikx + B′′e−ikx

Mayhall 34

3 LINEAR MOTION

Case 1: E < V

• k′ =√

2m(E−V)/h = iκ

• At x = 0

ψI(0) =ψI I(0) continuityd

dxψI(0) =

ddx

ψI I(0) smooth

Aeikx + Be−ikx =A′e−κx + B′eκx∣∣∣x=0

ikAeikx − ikBe−ikx =− κA′e−κx + κB′eκx∣∣∣x=0

A + B =A′ + B′

ikA− ikB =− κA′ + κB′

• At x = L

ψI I(L) =ψI I I(L) continuityd

dxψI I(L) =

ddx

ψI I I(L) smooth

A′e−κL + B′eκL =A′′eikL + B′′e−ikL

−κA′e−κL + κB′eκL =ikA′′eikL − ikB′′e−ikL

• We’ve got 6 unknowns, but only 4 equations:

– Normalization

– Initial Condition, particle starts at −∞ with positive momentum: B′′ = 0

I II IIII II III0 L

• Important coeicients:

– B = Reflection

– A′′ = Transmission

• How “much” of the wavefunction is being reflected?

Mayhall 35

3 LINEAR MOTION

– Reflection is the ratio of reflected flux density to incident flux density

1 =Jout

Jin=

J← + J→Jin

= R + T

Definition: flux density

In one dimension, the flux density is given as:

Jx =1

2m(ψ∗ pψ + ψ p∗ψ∗)

=−ih2m

ψ∗(x)∂

∂xψ(x) + C.C.

– Calculate J←

J← =−ih2m

(Be−ikx

)∗ ddx

(Be−ikx

)+ C.C. definition

=−ih2m

(B∗eikx

) (Be−ikx

)(−ik) + C.C.

=−kh2m|B|2 + C.C.

=−kh

m|B|2

– Calculate Jin

Jin =−ih2m

(Aeikx

)∗ ddx

(Aeikx

)+ C.C. definition

=khm|A|2

– thus,

R =− |B|2

|A|2T =1− R

• Solving for R and T in terms of the wavefunction coeicients requires only basic (but tedious!) algebra -we will skip this and just post the results

T =1

1 + (eκL−e−κL)2

16(E/V)(1−E/V)

When E < V

• Likewise, we can repeat the procedure to find the case when E > V

T =1

1 + sin2(k′L)4(E/V)(E/V−1)

When E > V

Mayhall 36

3 LINEAR MOTION

• and

T =1

1 + L2mV2h2

When E = V

• Let’s plot these transmission coeicients

– Function of incident energy (remember, T1 (T2) is only defined when E < V (E > V))

1 2 3 4 5

- 0.5

0.5

1.0

1.5

E<V

E>V

– Function of barrier width

10 20 30 40

- 0.5

0.5

1.0

1.5

E<V

E>V

E=V

– Function of both

Mayhall 37

3 LINEAR MOTION

0.5 1.0 1.5 2.0 2.5 3.0

Incident Energy

2

4

6

8

Transm

issi

on P

robabili

ty

Incr

easi

ng B

arr

ier

Wid

th v

alu

es,

L

• When does T → 1 as a function of L? This happens when,

k′L =nπ from T where E > V

=

√2m(E−V)L

hso,

L =nπh√

2m(E−V)

• As L→ ∞, TE≤V → 0, TE>V → 1

• As L→ 0, T → 1

• As V → ∞, T → 0

• As m→ ∞, TE≤V → 0, TE>V → 1

3.3 Particle in a box

• So we have observed only free-particles, no aractive potentials.

• Consider now the following potential

Mayhall 38

3 LINEAR MOTION

I II IIIV(x)

V

0

I II III0 L

V(x)

V

0

∞ ∞

• Trial wavefunction

ψ(x) =Aeikx + B−ikx

• Boundary conditions:

ψ(0) =ψ(L) = 0A + B =0 ⇒ A = −B

AeikL + B−ikL =0

• Thus,

A(

eikL − e−ikL)=0

2iA sin(kL) =0 Euler relation

Thus,

kL =nπ for n = 0, 1, 2, . . .

Remember Euler relation :

eix = cos(x) + i sin(x)

Leading to,eix + e−ix

2= cos(x)

eix − e−ix

2i= sin(x)

• We get energy quantization!

kL =L√

2mEh

= nπ From above

E =n2π2h2

2mL2 orn2h2

8mL2

ψ(x) =

√2L

sin(nπ

Lx)

Mayhall 39

3 LINEAR MOTION

– What about n = 0?

ψ(x) =0 no particle!

– For a particle to exist, n = 1, 2, 3, . . .

– Thus the minimum energy is

E1 =h2π2

2mL2 zero-point energy

• distribution inside of box

00

L

Incr

easi

ng q

uant

um n

umbe

r

Cla

ssic

al L

imit

3.4 Harmonic Oscillator

-1.0 -0.5 0.0 0.5 1.0

0.1

0.2

0.3

0.4

0.5

V

Mayhall 40

3 LINEAR MOTION

• Schrödinger equation(P2

x2m

+12

mω2X2)|ψ〉 =E |ψ〉 ω is angular frequency

• To simplify things, let’s use dimensionless quantities

H′ =H

hω

=12

(1

mhωP2 +

mω

hX2)

=12

(P′2 + X′2

)E′ =

Ehω

– where we define the dimensionless operators:

X′ =√

mω

hX

P′ =

√1

mhωP

– How does the aect the commutator?

[X′, P′] =X′P′ − P′X′

=

√mω

hX

√1

mhωP−

√1

mhωP√

mω

hX

=1h[X, P]

[X′, P′] =i

• Let’s try solving the SE by factorization. If P′ and X′ were just numbers, we could easily factor this intoa binomial.

a2 + b2 = (a− ib)(a + ib)

– But this doesn’t work for operators,(X′ − iP′

) (X′ + iP′

)=X′2 + P′2 + i[X′, P′]

=X′2 + P′2 − 1

– But it does give us something useful:

H′ =12[(

X′ − iP′) (

X′ + iP′)+ 1]

Mayhall 41

3 LINEAR MOTION

– Let’s define the following non-Hermitian operator:

a =1√2

(X′ + iP′

)annihilation operator

a =1√2

(√mω

hX + i

√1

mhωP

)in original units

such that

H′ =a† a +12

and

H′ |λ〉 =(

a† a +12

)|λ〉 = E′λ |λ〉

• This operator has the following commutator:

[a, a†] =12[X′ + iP′, X′ − iP′

]=

12([

X′, X]− i[X′, P′

]+ i[P′, X′

]+[P′, P′

])=− i

[X′, P′

]=1

• Consider the operator a† a = N,

– N commutes with the Hamiltonian

a† a |λ〉 =(

E′λ −12

)|λ〉

=λ |λ〉

〈λ|a† a|λ〉 =(

E′λ −12

)Mult. by 〈λ|

=(

a |λ〉)† (

a |λ〉)≥ 0 norm squared

– Thus we see that the energy has a minimum value

E′ ≥12

norm squared

E ≥ hω

2original units

– This is minimum value is the zero-point energy

Mayhall 42

3 LINEAR MOTION

• What is the eect of the operator a?

aa† a |λ〉 =λa |λ〉 le mult. by a(a† a + 1

)a |λ〉 =λa |λ〉 use commutator(

a† a)

a |λ〉 = (λ− 1) a |λ〉 Eigenvalue shied down

– thus

a |λ〉 ∝ |λ− 1〉

– Can we continue to lower the energy indefinitely by application of a? No.Since,

〈λ|a† a|λ〉 ≥0

The only way for the equality to hold, is if

a |0〉 =0 a stops at ground state

• What is the eect of the operator a†?

a† a† a |λ〉 =λa† |λ〉 le mult. by a†

a†(

aa† − 1)|λ〉 =λa† |λ〉 use commutator(

a† a)

a† |λ〉 = (λ + 1) a† |λ〉 Eigenvalue shied up

thus

a† |λ〉 ∝ |λ + 1〉

• So a is a lowering operator , (also called annihilation operator ) and a† is a raising operator (A.K.A

creation operator )

• Proportionality constants

a |λ〉 =kλ−1 |λ− 1〉 ∝ |λ− 1〉〈λ|a† a|λ〉 =λ

=|kλ−1|2

kλ−1 =√

λ λ is ≥ 0

a |λ〉 =√

λ |λ− 1〉

Mayhall 43

3 LINEAR MOTION

a† |λ〉 =kλ+1 |λ + 1〉 ∝ |λ + 1〉〈λ|aa†|λ〉 = 〈λ|1 + a† a|λ〉

=λ + 1

=|kλ+1|2

kλ+1 =√

λ + 1

a† |λ〉 =√

λ + 1 |λ + 1〉

• What values are permied for λ?

– We know 〈λ|a† a|λ〉 = λ ≥ 0

– We know a |λ〉 =√

λ |λ− 1〉

λ=2λ=1λ=0

λ=1.2λ=0.2λ=-0.8

– Only non-negative integers of λ allow these conditions to be true, thus λ is quantized!

Harmonic Oscillator energies

E′λ =

(λ +

12

)Eλ =hω

(λ +

12

)∀ λ ∈ 0, 1, 2, . . .

– All states equally spaced

• What do the eigenstates look like?

– Use a |0〉 = 0

– Let’s use position representation

a |0〉 → 1√2

(X′ + iP′

)φ0(x) = 0 where, φ0(x) = 〈x|0〉

=

(√mω

hx +

√1

mhωh

ddx

)φ0(x)

=

(x +

hmω

ddx

)φ0(x)

Mayhall 44

3 LINEAR MOTION

– Solve the following simple dierential equation

−xφ0(x) =h

mω

ddx

φ0(x)

−xdx =h

mω

1φ0(x)

dφ0(x) multiply bydx

φ0(x)

−12

x2 + c =h

mωln φ0(x) integrate

−mω

2hx2 + c′ = ln φ0(x) rearrange

Ae−mω2h x2

=φ0(x) exponentiate(mω

hπ

) 14 e−

mω2h x2

=φ0(x) normalize( α

π

) 14 e−

α2 x2

=φ0(x) with α =mω

h

Integral of a Gaussian:

∫ ∞

−∞e−

12 ax2

dx =

√2π

a

– What about higher energy states?

a† |0〉 =√

1 |1〉

=1√2

(√mω

hX− i

√1

mhωP

)|0〉 expand

=1√2

(√mω

hx−

√h

mω

ddx

)( α

π

) 14 e−

α2 x2

position rep.

=1√2

(√mω

hx−

√h

mω(−αx)

)( α

π

) 14 e−

α2 x2

dierentiate

=1√2

(√αx− 1√

α(−αx)

)( α

π

) 14 e−

α2 x2

simplify

=√

2α( α

π

) 14 xe−

α2 x2

simplify

– Thus, φ1(x) is a gaussian times a polynomial

– Repeated application of a† will generate higher energy excited states of the form

〈x|n〉 = φn(x) =Nn√2nn!

Hn(√

αx)e−α2 x2

Mayhall 45

3 LINEAR MOTION

where, Hn are the Hermite polynomials , and Nn are the normalization constants

H0(z) =1H1(z) =2z

H2(z) =4z2 − 2...

Hn(z) =(−1)nex2 dn

dxn e−x2

-1.0 -0.5 0.5 1.0

5

10

15

20

V

Mayhall 46

4 ROTATIONAL MOTION

4 Rotational motion4.1 Particle in a Ring• Polar coordinates

r y

x

ϕ

– conversions

x =r cos φ

y =r sin φ

r =√

x2 + y2

H =− h2

2m

(∂2

∂x2 +∂2

∂y2

)= − h2

2m

(∂2

∂r2 +1r

∂

∂r+

1r2

∂2

∂φ2

)polar coordinates

– If the particle is trapped in a ring of fixed size, then ∂∂r → 0

H =− h2

2mr2∂2

∂φ2

=− h2

2I∂2

∂φ2 I is rotational inertia

Hψ(φ) =Eψ(φ)

– Trial wavefunction:

ψ(φ) =Aei√

2IEh2 φ

+ Be−i√

2IEh2 φ

=Aeimlφ + Be−imlφ with ml =

√2IEh2

– Periodic boundary condition:

ψ(φ) =ψ(φ + 2π)

Aeimlφ + Be−imlφ =Aeimlφeiml2π + Be−imlφe−iml2π

Mayhall 47

4 ROTATIONAL MOTION

∗ This can only be true i

e±iml2π =1cos ml2π ± i sin ml2π =1 Euler’s formula

∗ To make this happen, the sin part must be zero, and the cos part then must be 1

ml2π =2nπ for all integers nml =0,±1,±2, . . .

-10 -5 5 10

-1.0

-0.5

0.5

1.0

V

∗ Thus,

Eml =m2

l h2

2I∀ml ∈ 0,±1,±2, . . .

– Wavefunction

∗ Since ml = 0,±1,±2, . . . completely satisfies the boundary conditions, A and B can be any-thing such that ψ(φ) can be normalized

∗ Set B = 0

Note that any other choice would also be a valid decision, i.e.,

ψ(φ) = A′(

eimlφ + e−imlφ)= A′2 cos(mlφ)

∗ Normalize ∫ 2π

0ψ(φ)∗ψ(φ)dφ =1

=|A|2∫ 2π

0e−imlφeimlφ

=2π|A|2

A =1√2π

ψ(φ) =1√2π

eimlφ

– Observations

∗ Eml is discrete and energy levels become more spaced with energy∗ No zero-point energy, ml = 0 is fine! Why is this?

Mayhall 48

4 ROTATIONAL MOTION

– Angular momentum

∗ particle is spinning in a circle, ml is likely an angular momentum∗ classically, angular momentum is

l =r× p =

∣∣∣∣∣∣∣∣∣∣∣∣

~x ~y ~z

x y z

px py pz

∣∣∣∣∣∣∣∣∣∣∣∣=~x(ypz − zpy) +~y(zpx − xpz) +~z(xpy − ypx)

=~xlx +~yly +~zlz

∗ Remember quantum postulate 3? We can replace x and p variables with operators to get thecorresponding quantum operators

lz =− ih(

x∂

∂y− y

∂

∂x

)actually 〈x|lz|ψ〉

lz =− ih∂

∂φangular momentum operator

∗ Apply operator to PIR eigenstates

lzψ(φ) =− ih∂

∂φ

(1√2π

eimlφ

)=ml

(1√2π

eimlφ

)=ml

(1√2π

eimlφ

)ml is the angular momentum in z-direction

∗ Allowed angular momentum = ml h = 0,±h,±2h, . . .

If a dierent form was chosen

ψ(φ) = A′(

eimlφ + e−imlφ)= A′2 cos(mlφ)

these are not eigenstates of the lz operator

Mayhall 49

4 ROTATIONAL MOTION

4.2 Particle on a Sphere• Spherical coordinates

rθ

xy

zy

x

z

ϕ

– Conversions:

∇2 =∂2

∂x2 +∂2

∂y2 +∂2

∂z2

=∂2

∂r2 +1r

∂

∂r+

1r2 Λ2

where,

Λ2 =1

sin2 θ

∂2

∂φ2 +1

sin θ

∂

∂θsin θ

∂

∂θ

• A particle on a sphere has a fixed radius, too. Thus,

HΨ(θ, φ) =−h2

2IΛ2Ψ(θ, φ)

=−h2

2IΛ2Φ(φ)Θ(θ) Sep. of Var.

=EΦ(φ)Θ(θ)

sin2 θΛ2Φ(φ)Θ(θ) =

(∂2

∂φ2 + sin θ∂

∂θsin θ

∂

∂θ

)Φ(φ)Θ(θ) mult. by

sin2 θ

−h2

2I

• gives(∂2

∂φ2 + sin θ∂

∂θsin θ

∂

∂θ

)Φ(φ)Θ(θ) =− 2IE

h2 sin2 θΦ(φ)Θ(θ)

Θ∂2

∂φ2 Φ =−Φ sin θ∂

∂θsin θ

∂

∂θΘ− 2IE

h2 sin2 θΦΘ

1Φ

∂2

∂φ2 Φ =− 1Θ

sin θ∂

∂θsin θ

∂

∂θΘ− 2IE

h2 sin2 θ Divide by ΦΘ

• Φ(φ) is essentially particle in a ring:

∂2

∂φ2 Φ(φ) =cΦ(φ)

Φ(φ) =1√2π

eimlφ ∀ml ∈ 0,±1,±2, . . .

Mayhall 50

4 ROTATIONAL MOTION

• Plug this back into the Schrödinger equation:

HΦΘ =EΦΘ

−h2

2I

[1

sin2 θ

∂2

∂φ2 +1

sin θ

∂

∂θsin θ

∂

∂θ

]ΦΘ =EΦΘ

−h2

2I

[1

sin2 θ

∂2

∂φ2 +1

sin θ

∂

∂θsin θ

∂

∂θ

]eimlφΘ =EeimlφΘ substitute

−h2

2I

[−m2

l

sin2 θ+

1sin θ

∂

∂θsin θ

∂

∂θ

]eimlφΘ =EeimlφΘ dierentiate w.r.t φ[

−m2l

sin2 θ+

1sin θ

∂

∂θsin θ

∂

∂θ+

2IEh2

]Θ =0 Φ cancels, and rearrange

• One could then choose to solve this dierential equation for Θ(θ), we will not. The solutions to thisinvolve the associated Legendre polynomials, P:

Θ(θ) =

(2l + 1

2

)(l − |ml|)!(l + |ml|)!

P|ml |

l (cos θ)

• Taken together with the particle in a ring wavefunction, Φ, the final wavefunctions are called spherical harmonics ,Yl,ml

(φ, θ)

Yl,ml=Φ(φ)ml Θ(θ)l,ml

such that

Λ2Ylml=− l(l + 1)Ylml

∀ l ∈ 0, 1, 2, . . . ∀ ml ∈ −l,−l + 1, . . . , l − 1, l

• this leads to a restriction on the energy levels allowed:

E =l(l + 1)h2

2I

with, degeneracy

gl = 2l + 1

• If we convert the rigid rotor Hamiltonian into spherical coordinates, it is essentially identical to a particleon a sphere. Read about this in Atkins.

• We will come back to this soon...

Mayhall 51

4 ROTATIONAL MOTION

4.3 Hydrogen Atom• Center of mass coordinates

r

re

rN

R N

e

~r =~re −~rN

~R =me~re + mN~rN

Mwith M = me + mN

• we are going to drop the vector arrows...

• Because the coulombic interaction involves both nuclear and electronic degrees of freedom, we cannot useseparation of variables, and so we say that the electronic and nuclear degrees of freedom are correlated.In order to simplify our problem, we will find a dierent set of coordinates which will allow us to applythe separation of variables.

• Hamiltonian in particle coordinates:

H =P2

N2mN

+P2

e2me

+ V(r)

• Define new momentum operators: Think of classical mechanics:

~Pcm =M~V = Mddt~R “total” momentum

=Mddt

(mere + mNrN

M

)=~pe + ~pN

~p =µddt~r “internal” momentum

=mNme

Mddt

(~re −~rN)

=mNme

M

(~pe

me− ~pN

mN

)=

mN~pe −me ~pN

M

Mayhall 52

4 ROTATIONAL MOTION

P =PN + Pe just sum of the momenta

p =mN Pe −mePN

M

Definition: reduced mass

µ =m1m2

m1 + m2

• P and p commute

• We want to transform the Hamiltonian operator into center-of-mass coordinates

• let’s get inverse transformations:

p =mN P−mN PN −mePN

Msubstitute for Pe

=mN

MP− PN substitute for Pe

so

PN =mN

MP− p

Pe =me

MP + p

• Now substitute into the Hamiltonian

H =1

2mN

(mN

MP− p

)2− 1

2me

(me

MP + p

)2+ V(r)

=1

2mN

[(mN

MP)2

+ p2 − 2mN

MPp]+

12me

[(me

MP)2

+ p2 + 2me

MPp]+ V(r)

=1

2mN

[(mN

MP)2

+ p2]+

12me

[(me

MP)2

+ p2]+ V(r)

=12

[mN

M2 P2 +p2

mN+

me

M2 P2 +p2

me

]+ V(r)

=12

[mN + me

M2 P2 +(me + mN) p2

memN

]+ V(r)

=P2

2M+

p2

2µ+ V(r)

Mayhall 53

4 ROTATIONAL MOTION

• Thus, the Schrödinger equation is separable, and we can now focus only in the internal degrees of freedom(p2

2µ− Ze2

4πε0r

)ψ(r) =Eψ(r)

−(

h2

2µ

(∂2

∂x2 x+

∂2

∂y2 +∂2

∂z2

)+

Ze2

4πε0r

)ψ(r) =Eψ(r)(

1r

∂2

∂r2 r +1r2 Λ2 +

Zµe2

2h2πε0r

)ψ =−2µE

h2 ψ spherical coords

• Assume trial function ψ(r, φ, θ) = R(r)Ylml(φ, θ), with Y being the spherical harmonics(

1r

∂2

∂r2 r +1r2 Λ2 +

Zµe2

2h2πε0r

)RY =

−2µEh2 RY(

1r

∂2

∂r2 r− l(l + 1)r2 +

Zµe2

2h2πε0r

)RY =

−2µEh2 RY Λ2Ylml

= −l(l + 1)Ylml(1r

∂2

∂r2 r− br2 +

ar

)R =−2µE

h2 R simplify, and divide by Y(∂2

∂r2 −br2 +

ar

)u =−2µE

h2 u mult. by r, and let u = Rr

with,

a =2µZe2

h24πε0

b =l(l + 1)

• Notice here that the above equation amounts to solving the Schrödinger equation with an eective Hamil-tonian which depends on l(l + 1)

1/r1/r21/r + 1/r2

Mayhall 54

4 ROTATIONAL MOTION

• Consider long-distance first, limit as r → ∞:

→ ∂2

∂r2 u =−2µE

h2 u as r → ∞

u∞ =Aei√

2µEh2 r

+ Be−i√

2µEh2 r

however, we know that E < 0, so

u∞ =Ae−√

2µ|E|h2 r

+ Be

√2µ|E|

h2 r

and to remain square-integrable:

u∞ =Ae−√

2µ|E|h2 r

u∞ =Ae−λr where λ =

√2µ|E|

h2

• Now what about finite r?

u =u∞L(r) with L(r) = ∑n

cnrn

does this form make sense? Remember, the exponential will kill o any long-range polynomial tails, dueto the relative rates of decay

• Plug it into our original equation, and restricting ourselves again to bound states, such that −E = |E|:(∂2

∂r2 −br2 +

ar

)u =λ2u(

∂2

∂r2 −br2 +

ar

)Le−λr =λ2Le−λr

∂

∂r(L′e−λr − Lλe−λr) +

(− b

r2 +ar

)Le−λr =λ2Le−λr dierentiate

(L′′e−λr − L′λe−λr − L′λe−λr + Lλ2e−λr) +

(− b

r2 +ar

)Le−λr =λ2Le−λr dierentiate

L′′ − 2L′λ +

(− b

r2 +ar

)L =0 simplify

expand

∑n

cnn(n− 1)rn−2 − 2λ ∑n

cnnrn−1 − b ∑n

cnrn−2 + a ∑n

cnrn−1 =0

this means

. . . + cn

[n(n− 1)rn−2 − 2λnrn−1 − brn−2 + arn−1

]+ cn+1

[n(n + 1)rn−1 − 2λ(n + 1)rn − brn−1 + arn

]+ . . .

=0

Mayhall 55

4 ROTATIONAL MOTION

As a polynomial, for this to generally equal zero, each order of r must also go to zero:

cn

[−2λnrn−1 + arn−1

]+ cn+1

[n(n + 1)rn−1 − brn−1

]=0

cn [−2λn + a] + cn+1 [n(n + 1)− b] =0

cn+1 =2λn− a

n(n + 1)− bcn

• The polynomials defined by these coeicients are called Laguerre polynomials to ensure square-integrability,there must be some cn such that cn+1 = 0. Another reason, is that since at finite values of r the wavefunc-tion is primarily L, an arbitrarily large n will lead to an arbitrarily large kinetic energy. Our expectationof a finite energy leads to:

2λn =a

2

√2µ|E|

h2 n =2µ

h2Ze2

4πε0substitute

|E| = µ

n2h2Z2e4

32π2ε20

rearrange

Mayhall 56

5 ANGULAR MOMENTUM

5 Angular momentum5.1 Operators• Classical Angular momentum:

l =r× p =

∣∣∣∣∣∣∣∣∣∣∣∣

~x ~y ~z

x y z

px py pz

∣∣∣∣∣∣∣∣∣∣∣∣=~x(ypz − zpy) +~y(zpx − xpz) +~z(xpy − ypx)

=~xlx +~yly +~zlz• antum postulate: swap dynamical position and momentum variables with operators

lx =y pz − z py

ly =z px − x pz

lz =x py − y px

• Magnitude of angular momentum:

l2 =l2x + l2

y + l2z

• angular momentum commutation relations (remember, [q, pq′ ] = ihδq,q′ ):

[lx, ly] =[ypz − zpy, zpx − xpz]

=[ypz, zpx] + [zpy, xpz]

=ypx[pz, z] + xpy[z, pz]

=ihlz[ly, lz] =ihlx

[lz, lx] =ihly

[l2, lz] =[l2x + l2

y + l2z , lz] = 0

=[l2x + l2

y, lz]

=lx lx lz − lz lx lx + ly ly lz − lz ly ly

=lx

(lz lx − ihly

)− lz lx lx + ly

(lz ly + ihlx

)− lz ly ly

=[lx, lz]lx − ihlx ly + [ly, lz]ly + ihly lx

=− ihly lx − ihlx ly + ihlx ly + ihly lx

=0

Mayhall 57

5 ANGULAR MOMENTUM

and the same for lx and ly,

[l2, lx] =[l2x + l2

y + l2z , lx] = 0

[l2, ly] =[l2x + l2

y + l2z , ly] = 0

• We can only know the angular momentum in one direction at at time. And this can be simultaneouslyknown with the total magnitude, l2

• This suggests a vector model of angular momentum

• Since we can only know one of lx, ly, or lz, at a given time, we will choose one coordinate as the knowncoordinate, this will be lz.• angular momentum ladder operators

l+ =lx + ilyl− =lx − ily

– Note: If we would have picked a dierent coordinate (not lz) these operators would look a bit dif-ferent but have the same behavior.

– Consider their commutators on various operators

[lz, l+] =[lz, lx + ily]

=[lz, lx] + i[lz, ly]

=ihly + hlx

=hl+[lz, l−] =− hl−

– and

[l+, l−] =[lx + ily, lx − ily]

=− i[lx, ly] + i[ly, lx]

=− 2i[lx, ly]

=2hlz

– and

[l2, l+] =[l2, lx + ily]

=0 l2 commutes with both lx, ly[l2, l−] =0

Mayhall 58

5 ANGULAR MOMENTUM

– so,

c+(λ, ml)c−(λ, ml + 1) =λ−m2l −ml

– however,

〈λ, ml|l−|λ, ml + 1〉∗ = 〈λ, ml + 1|l+|λ, ml〉c−(λ, ml + 1)∗ =c+(λ, ml)

– now, substitute

c+(λ, ml)c+(λ, ml)∗ =|c+(λ, ml)|2

=λ−m2l −ml

c+(λ, ml) =√

λ−ml(ml + 1) implies choice of phase

c−(λ, ml) =√

λ−ml(ml − 1) by analogous manipulations

• Ladder operators lead to relate λ and ml– Let ml assume it’s maximal value, mmax

l , remembering that√

λ ≥ |ml|

l+ |λ, mmaxl 〉 =0 since this is maximal value of ml

l− l+ |λ, mmaxl 〉 =0 we can do this(

lx − ily) (

lx + ily)|λ, mmax

l 〉 =0 substitute(l2x + l2

y + i[lx, ly])|λ, mmax

l 〉 =0 expand(l2x + l2

y − hlz)|λ, mmax

l 〉 =0(l2 − l2

z − hlz)|λ, mmax

l 〉 =0 put in terms of good variables(h2λ− h2(mmax

l )2 − h2mmaxl

)|λ, mmax

l 〉 =0 eigenvalues

– For this to be true, regardless of the values of λ, mmaxl 6=

√λ, but rather

λ = mmaxl (mmax

l + 1) (127)

– let us give a label, l = mmaxl , to this quantum number such that:

l2 |λ, ml〉 = l(l + 1)h2 |λ, ml〉 (128)

– we can now replace λ with a value l indicating angular momentum, |λ, ml〉 → |l, ml〉

Mayhall 60

5 ANGULAR MOMENTUM

5.2 antization of angular momentum• Now, let us show that both l and ml are quantized

– Consider both l+ |l, ml〉 and l− |l, ml〉:

〈l, ml|l− l+|l, ml〉 =∣∣∣l+ |l, ml〉

∣∣∣2 ≥ 0 norm squared

〈l, ml|l+ l−|l, ml〉 =∣∣∣l− |l, ml〉

∣∣∣2 ≥ 0

l− l+ =(

lx − ily) (

lx + ily)

=l2 − l2z − hlz

l+ l− =(

lx + ily) (

lx − ily)

=l2 − l2z + hlz

– Requiring both l+ |l, ml〉 and l− |l, ml〉 to have non-negative squared norms, leads to two simulta-neous inequalities:

〈l, ml|l− l+|l, ml〉 = 〈l, ml|l2 − l2z − hlz|l, ml〉 ≥ 0

〈l, ml|l+ l−|l, ml〉 = 〈l, ml|l2 − l2z + hlz|l, ml〉 ≥ 0

h2(

l(l + 1)−m2l −ml

)≥0

h2(

l(l + 1)−m2l + ml

)≥0

– rearranging:

l(l + 1)−m2l −ml =l2 + l −m2

l −ml

=(l −ml)(l + ml) + l −ml factoring dierence of squares

=(l −ml)(l + ml + 1)

l(l + 1)−m2l + ml =(l −ml + 1)(l + ml) likewise

– since these must be term-wise non-negative:

−(l + 1) ≤ ml ≤l−l ≤ ml ≤l + 1

Mayhall 61

5 ANGULAR MOMENTUM

-l -l+1 ll-1

ml

1)

2)

– thus, −l ≤ ml ≤ l

– In order for all of these results to be true, the distance between −l and l (2l) must be an integer.The reason this must be so, is that l− acting on the minimum value must equal 0, and l+ acting onthe max value must equal 0. If 2l was not an integer, then repeated application of l+ |l,−l〉 wouldultimately lead to states with ml > l. Thus,

2l =n ∀ n ∈ I

l =0,12

, 1,32

, 2, . . .

5.3 Spherical Harmonics• To do this, we now move into the position representation (in spherical polar coordinates)

lx =ih(

sin φ∂

∂θ+ cot θ cos φ

∂

∂φ

)ly =− ih

(cos φ

∂

∂θ− cot θ sin φ

∂

∂φ

)lz =− ih

∂

∂φ

derive l+ and l−:

l+ =h(

i sin φ∂

∂θ+ i cot θ cos φ

∂

∂φ+ cos φ

∂

∂θ− cot θ sin φ

∂

∂φ

)=h((cos φ + i sin φ)

∂

∂θ+ i cot θ (cos φ + i sin φ)

∂

∂φ

)=heiφ

(∂

∂θ+ i cot θ

∂

∂φ

)

l− =− he−iφ(

∂

∂θ− i cot θ

∂

∂φ

)• PS. remember, these are really the position representations of the general operators

lz ⇒ 〈r|lz|l, ml〉 = −ih∂

∂φY(φ, θ)

Mayhall 62

5 ANGULAR MOMENTUM

• Consider l+ acting on maximum ml state:

l+ |l, l〉 =0

heiφ(

∂

∂θ+ i cot θ

∂

∂φ

)Yl,l =0

tan θ∂

∂θYl,l =− i

∂

∂φYl,l

• This equation is separable, so let Yll = ΦΘ

Φ tan θ∂

∂θΘ =− iΘ

∂

∂φΦ

1Θ

tan θ∂

∂θΘ =−iΦ

∂

∂φΦ = k

• Φ can immediately be wrien down:

Φ(φ) =Aeikφ

• To get Θ we just integrate:

1Θ

tan θ∂

∂θΘ =k

1Θ

∂Θ =k cot θ∂θ rearrange

1Θ

∂Θ =kcos θ

sin θ∂θ

1Θ

∂Θ =k1

sin θ∂(sin θ) rearrange

ln Θ + c = ln sink θ integrate

Θ =A sink θ

• What is k?

lzY =− ih∂

∂φΦΘ

=khΦΘk =ml

• So,

Yl,l(φ, θ) = Nl sinl θeilφ

Mayhall 63

5 ANGULAR MOMENTUM

• Normalization leads to:

Nl =(−1)l

2l l!

√(2l + 1)!

4π

• Examples:

Y00 =1

2√

π

Y11 =− 12

√3

2πsin θeiφ

Y22 =18

√30π

sin2 θei2φ

• What about when l 6= ml?

l− |l, l〉 =h√

l(l + 1)−ml(ml − 1) |l, l − 1〉

example: Find Y10

Y10 =−√

12

e−iφ(

∂

∂θ− i cot θ

∂

∂φ

)·(−1

2

√3

2πsin θeiφ

)

=e−iφ

√12

√3

8π

(cos θeiφ − i cos θ(i)eiφ

)=

√3

4πcos θ

5.4 Spin• It turns out, (through no derivation in this class) that elementary particles exhibit spatial quantization of

an intrinsic angular momentum called “spin”

– Fermions (including electrons) have an intrinsic angular momentum which has only half-integervalues

– Bosons (including photons) have an intrinsic angular momentum which has only integer values

– Stern-Gerlach experiment

• In general, dierent sources of angular momentum have dierent notations:

lz orbital angular momentum

Sz intrinsic spin angular momentum

Jz arbitrary angular momentum

Mayhall 64

5 ANGULAR MOMENTUM

• While it might appear like there are simply two kinds of particles, spin-up and spin-down particles, thisis not the case. In other words, if we split the particles into two channels with a magnetic field, and blocko one of these channels (say, spin-down), we do not observe a “filtering” of the kinds of particles. Wecan recover the spin-down signal by passing it through perpendicular magnetic fields first!

Fz = −∂U∂z

= µz∂Bz

∂z(129)

B B

Classical antum

S-G (z)Source S-G (z) S-G (x)z-

z+ x+

x-

S-G (z)z-

z+

• Choosing a basis

– we can choose as our basis, either eigenvectors of (S2, Sz) or (S2, Sx) or (S2, Sy),

– we will denote these dierent bases as:

Sx |±x〉 =±h2|±x〉

Sy |±y〉 =±h2|±y〉

Sz |±z〉 =±h2|±z〉

S2 |±•〉 =s(s + 1)h2 |±•〉

– We are able to assume this shorthand notation for spin, because fermions always have s = 12 , so

|s, ms〉 =∣∣∣∣12, ms

⟩(130)

contains the unnecessary specification of s.

– Let us again take z to be our preferred direction, and work the eigenvectors of (S2, Sz), with thepositive and negative states,

Mayhall 65

5 ANGULAR MOMENTUM

– Since Fermions are defined as having a single value for s = 12 , this restricts ms =

±h2 , and a particle’s

full state space can be wrien in a basis of one positive (α) and one negative (β) spin vector

∣∣∣∣12,12

⟩= |z+〉 = |α〉 =

1

0

∣∣∣∣12,−12

⟩= |z−〉 = |β〉 =

0

1

– Resolution of identity

I = |α〉〈α|+ |β〉〈β|

=

1 0

0 0

+

0 0

0 1

=

1 0

0 1

• Spin operators in the |α〉 , |β〉 basis

– What do the spin operators look like in this basis?

ISzI = (|α〉〈α|+ |β〉〈β|) Sz (|α〉〈α|+ |β〉〈β|)

=h2|α〉〈α| − h

2|β〉〈β|

=h2

1 0

0 −1

=

h2

σz

where σz is the Pauli matrix for Sz

– Likewise

S2 =h2s(s + 1) |α〉〈α|+ h2s(s + 1) |β〉〈β|

=3h2

4I

Mayhall 66

5 ANGULAR MOMENTUM

– Before trying to find Sx and Sy, let’s remember that

S+ |β〉 =h√

s(s + 1)−ms(ms + 1) |α〉

=h

√34−(−12

)12|α〉

=h |α〉S+ |α〉 =0

S− |α〉 =h |β〉

such that,

S+ = (|α〉〈α|+ |β〉〈β|) S+ (|α〉〈α|+ |β〉〈β|)= |α〉〈α| S+ |β〉〈β| all other terms = 0=h |α〉〈β|

=h

0 1

0 0

S− =h

0 0

1 0

– Recall that

S+ =Sx + iSy

S− =Sx − iSy

Sx =S+ + S−

2

Sy =S+ − S−

2i

– So simply plugging these in we get

Sx =h2

0 1

1 0

=h2

σx

Sy =h2

0 −i

i 0

=h2

σy

Mayhall 67

5 ANGULAR MOMENTUM

5.5 Coupling of Angular Momenta• The vector space spanned by two sources of angular momenta is the direct product of the spaces spanned

by each independently, V1 ⊗ V2 = V3, with operators acting on V3 as:

Jz = J1z ⊕ J2z = J1z ⊗ I + I ⊗ J2z

• It’s easily seen that these operators are also angular momenta:

[ Jx, Jy] =[ J1x ⊕ J2x, J1y ⊕ J2y] = ih Jz

• The operators that act on V3 are thus the sum of the operators

Jz = J1z + J2z

J2 =(

J1x + J2x)2

+(

J1y + J2y)2

+(

J1z + J2z)2

= J21x + J2

2x + J21y + J2

2y + J21z + J2

2z + 2 J1x J2x + 2 J1y J2y + 2 J1z J2z [ J1z, J2z] = 0

=J21 + J2

2 + 2 J1x J2x + 2 J1y J2y + 2 J1z J2z

=J21 + J2

2 + 2~J1 · ~J2

• A single source of angular momentum J1z operator doesn’t commute with total angular momentum:

[ J1z, J2] =[ J1z, J21 + J2

2 + 2 J1x J2x + 2 J1y J2y + 2 J1z J2z]

=[ J1z, 2 J1x J2x + 2 J1y J2y]

=2ih(

J1y J2x − J1x J2y)

6=0

• We can specify J21, J2

2, and J2 simultaneously:

[J2, J21] =[J2, J2

2] = 0 since, each component commutes

• the summed operators are still angular momentum operators, so all the results for we’ve recently dis-cussed still hold:

[J2, Jz] =0

[ Jx, Jy] =ih Jz

• Thus, we can add Jz to our list of simultaneous observables, allowing us to label states of angular mo-mentum as:

|j1, j2; J, M〉 coupled picture

where,

J2 |j1, j2; J, M〉 =J(J + 1)h2 |j1, j2; J, M〉Jz |j1, j2; J, M〉 =Mh |j1, j2; J, M〉J2

1 |j1, j2; J, M〉 =j1(j1 + 1)h2 |j1, j2; J, M〉J2

2 |j1, j2; J, M〉 =j2(j2 + 1)h2 |j1, j2; J, M〉

Mayhall 68

5 ANGULAR MOMENTUM

• However, since [J21, J1z] = 0, [J2

2, J2z] = 0, and [J21, J2

2] = 0, [ J1z, J2z] = 0

|j1, m1; j2; m2〉 uncoupled picture

also forms a good basis

• The quantum numbers of the total angular momenta are restricted to:

J =j1 + j2, j1 + j2 − 1, . . . , |j1 − j2|M =mj1 + mj2

this is known as the Clebsch-Gordan series

• To transform from the coupled basis to the uncoupled basis we simply resolve the identity

|j1, j2; J, M〉 = ∑m1,m2

|j1, m1; j2, m2〉〈j1, m1; j2, m2| |j1, j2; J, M〉

= ∑m1,m2

|j1, m1; j2, m2〉C JMm1m2

where, C JMm1m2 are the Clebsch-Gordan coeicients

• Example: for a d2 electronic configuration, find the Clebsch-Gordan coeicients for |2, 2; L = 4, Ml = 3〉

– We want to express |4, 3〉 in the basis of uncoupled states

– High ms state is easy, since Ml = ml1 + ml2, there is a one-to-one correspondence between thecoupled and uncoupled bases for the highest (and lowest) angular momentum state

|l1 = 2, l2 = 2; L = 4, Ml = 4〉 = |l1 = 2, ml1 = 2; l2 = 2, ml2 = 2〉

– The values for l1 and l2 never change (both electrons stay inside d orbitals), so we can simplify ournotation

|4, 4〉 = |2; 2〉

– To get |4, 3〉 we can use the ladder operators

l− |4, 4〉 =(

l−1 + l−2

)|2; 2〉

√8h |4, 3〉 =2h (|1; 2〉+ |2; 1〉)

|4, 3〉 = 1√2(|1; 2〉+ |2; 1〉)

– this can be continued,

|4, 2〉 =√

314

(|0; 2〉+ |2; 0〉) +√

814|1; 1〉

Mayhall 69

5 ANGULAR MOMENTUM

– What about lower angular momentum states like |3, 3〉?

|3, 3〉 =a |2; 1〉+ b |1; 2〉 only possibilities to sum to 3

but since states of dierent angular momentum must be orthogonal, 〈4, 3|3, 3〉 = 0

〈4, 3|3, 3〉 = 1√2(〈2; 1|+ 〈1; 2|) (a |2; 1〉+ b |1; 2〉)

=1√2(a + b) since 〈1; 2|2; 1〉 = 0

=0 so, a = −b

|3, 3〉 = 1√2(|2; 1〉 − |1; 2〉) normalized

• Basic process

Step Down operator

Orthogonalize

5.6 Dichromium Example• Consider a real-life example of a transition metal complex:

Cr Cr• Start by focusing on a single Cr atom

Mayhall 70

5 ANGULAR MOMENTUM

– First couple two electrons, then add the third: s1 = 12 and s2 = 1

2

S = 1, 0 (131)

Possible states for two electrons:

|s1, s2; S, Ms〉 Notation reminder (132)∣∣∣∣12,12

; 1, 1⟩

(133)∣∣∣∣12,12

; 1, 0⟩

(134)∣∣∣∣12,12

; 1,−1⟩

(135)∣∣∣∣12,12

; 0, 0⟩

(136)

– For the highest/lowest spin cases, there is only one possible choice for representing these coupledstates in the uncoupled basis: ∣∣∣∣12,

12

; 1, 1⟩

= |α〉 |α〉 (137)∣∣∣∣12,12

; 1,−1⟩

= |β〉 |β〉 (138)

– Use S− to get the lower spin states:

S−

∣∣∣∣12,12

; 1, 1⟩

= h√

2∣∣∣∣12,

12

; 1, 0⟩

(139)

=(S−1 + S−2

)|α〉 |α〉 (140)

= h (|β〉 |α〉+ |α〉 |β〉) (141)

revealing, ∣∣∣∣12,12

; 1, 0⟩

=1√2(|β〉 |α〉+ |α〉 |β〉) (142)

and to ensure orthgonality,⟨

12 , 1

2 ; 0, 0∣∣∣1

2 , 12 ; 1, 0

⟩= 0,∣∣∣∣12,

12

; 0, 0⟩

=1√2(|β〉 |α〉 − |α〉 |β〉) (143)

– Now couple both of these spin states to the third electron. How many states are possible?Case: s1 = 1 and s2 = 1

2 :

S =32

,12

Possible spin values (144)

Mayhall 71

5 ANGULAR MOMENTUM

We can easily enumerate all of these coupled states,∣∣∣∣1,12

;32

,32

⟩(145)∣∣∣∣1,

12

;32

,12

⟩(146)∣∣∣∣1,

12

;32

,−12

⟩(147)∣∣∣∣1,

12

;32

,−32

⟩(148)∣∣∣∣1,

12

;12

,12

⟩(149)∣∣∣∣1,

12

;12

,−12

⟩(150)

Case: s1 = 0 and s2 = 12 :

S =12

Possible spin values (151)

∣∣∣∣0,12

;12

,12

⟩(152)∣∣∣∣0,

12

;12

,−12

⟩(153)

– There are two distinct doublet states!∣∣∣1, 1

2 ; 12 , Ms

⟩and

∣∣∣0, 12 ; 1

2 , Ms

⟩– Notice that we have identified 8 unique states. Is this the full number of possible spin states? We

can check this by combinatorics. Each electron can be in one of two states, and there are threeelectrons, so the total number of states is 23 = 8 - we’ve got all of them!

– What do these states look like in the un-coupled basis?There is only one way to obtain Ms =

32 , that being ms1 = 1 and mss2 = 1

2 .So, for the highest spin state, there is a simple one-to-one mapping between the coupled and un-coupled representations. ∣∣∣∣1,

12

;32

,32

⟩= |1, 1〉

∣∣∣∣12,12

⟩(154)

– Once again, in order to obtain the non-highest angular momentum states, we will need the ladderoperators.

S− =Sx + iSy (155)

=(Sx1 + Sx2

)− i(Sy1 + Sy2

)(156)

=S−1 + S−2 (157)

Mayhall 72

5 ANGULAR MOMENTUM

S−

∣∣∣∣1,12

;32

,32

⟩=h√

3∣∣∣∣1,

12

;32

,12

⟩(158)

=(S−1 + S−2

)|1, 1〉

∣∣∣∣12,12

⟩(159)

=h(√

2 |1, 0〉∣∣∣∣12,

12

⟩+ |1, 1〉

∣∣∣∣12,−12

⟩)(160)∣∣∣∣1,

12

;32

,12

⟩=

√23|1, 0〉

∣∣∣∣12,12

⟩+

√13|1, 1〉

∣∣∣∣12,−12

⟩(161)

=1√3(|αβα〉+ |βαα〉+ |ααβ〉) (162)

(163)

– Now, to find the∣∣∣1, 1

2 ; 12 , 1

2

⟩state, we enforce orthogonality again,

∣∣∣∣1,12

;12

,12

⟩=

√13|1, 0〉

∣∣∣∣12,12

⟩−√

23|1, 1〉

∣∣∣∣12,−12

⟩(164)

=

√16(|αβα〉+ |βαα〉)−

√23|ααβ〉 (165)

• Now that we have the states on a single Cr atom enumerated, we are now in a position to couple the twoCr atoms together.

– Before we do this, how many states should we expect? A: 26 = 64. Instead of trying to do this, let’sfocus our aention only on the low-energy states first.

– These Cr atoms have a large energy separation between the∣∣∣1, 1

2 ; 32 , Ms

⟩states and the lower S

states. (Just believe me for now). So instead of coupling all states together, if we are interested inenumerating the low energy states, we can restrict our consideration to only the low-energy atomicstates (i.e., when S = 3

2 ).

– s1 = 32 and s2 = 3

2

S = 3, 2, 1, 0 (166)

We can easily enumerate all of these coupled states,∣∣∣∣32,32

; 3,−3 · · · 3⟩

7 states : Heptet (167)∣∣∣∣32,32

; 2,−2 · · · 2⟩

5 states : intet (168)∣∣∣∣32,32

; 1,−1 · · · 1⟩

3 states : Triplet (169)∣∣∣∣32,32

; 0, 0⟩

1 state : Singlet (170)

Mayhall 73

5 ANGULAR MOMENTUM

– So the low energy spectrum should be comprised of 16 total states among 4 distinct energy levels.

• Energy separation of spin states

– Heisenberg Hamiltonian for calculating the energy of two spins:

H =− 2J~S1 · ~S2 (171)

– Phenomenological model Hamiltonian which assumes that the low energy spectrum is determinedby the relative alignments of the two (can be generalized to multiple sites) spins. To see how itseparates the dierent S2 eigenstates let’s look at the S2 operator:

S2 =(~S1 + ~S2

)·(~S1 + ~S2

)(172)

=S21 + S2

2 + 2~S1 · ~S2 dierent sites commute (173)

so,

H =− J(

S2 − S21 − S2

2

)(174)

clearly depends on S2, S21, S2

2 but not S1z or S2z

– Heisenberg Hamiltonian is diagonal in the coupled representation:

H∣∣∣∣32,

32

; 3, Ms

⟩=− Jh2 (S(S + 1)− S1(S1 + 1)− S2(S2 + 1))

∣∣∣∣32,32

; 3, Ms

⟩(175)

=− 92

Jh2∣∣∣∣32,

32

; 3, Ms

⟩(176)

H∣∣∣∣32,

32

; 2, Ms

⟩=+

32

Jh2∣∣∣∣32,

32

; 3, Ms

⟩(177)

H∣∣∣∣32,

32

; 1, Ms

⟩=+

112

Jh2∣∣∣∣32,

32

; 3, Ms

⟩(178)

H∣∣∣∣32,

32

; 0, 0⟩

=+152

Jh2∣∣∣∣32,

32

; 3, Ms

⟩(179)

– So the energy diagram is:

S=0S=1

S=2

S=3

If J<0

S=0S=1

S=2

S=3If J>0

Antiferromagnetic Ferromagnetic

• What about the uncoupled-basis?

Mayhall 74

5 ANGULAR MOMENTUM

– What does Heisenberg Hamiltonian look like in un-coupled basis: start with operators with clearaction on un-coupled states:

S1+S2− + S1−S2+ =(S1x + iS1y

) (S2x − iS2y

)+(S1x − iS1y

) (S2x + iS2y

)(180)

=2S1xS2x + 2S1yS2y (181)

=2~S1 · ~S2 − 2S1zS2z (182)

leading to,

H =− 2J~S1 · ~S2 = −J(S1+S2− + S1−S2+ + 2S1zS2z

)(183)

– Act H on to uncoupled basis to determine matrix elements:

– The high and low spin cases are diagonal, since the only get acted upon by Sz

H |αα〉 = −2Jh2(

12

) (12

)|αα〉 = −1

2 Jh2 |αα〉 (184)

H |ββ〉 = −2Jh2(−1

2

) (−1

2

)|ββ〉 = −1

2 Jh2 |ββ〉 (185)

– The ms = 0 cases are not diagonal

H |αβ〉 = −J(S1+S2− |αβ〉+ S1−S2+ |αβ〉+ 2S1zS2z |αβ〉

)(186)

= −Jh2(|βα〉 − 1

2 |αβ〉)

(187)

– Thus, the Hamiltonian matrix in the basis: |αα〉 , |αβ〉 , |βα〉 , |ββ〉 is:

H = −Jh2

|αα〉 |αβ〉 |βα〉 |ββ〉

〈αα| 12 0 0 0

〈αβ| 0 −12 1 0

〈βα| 0 1 −12 0

〈ββ| 0 0 0 12

(188)

– Eigenvalues:

1. −12 Jh2 : |αα〉 , |ββ〉 , 1√

2(|αβ〉+ |βα〉)

2. 32 Jh2 : 1√

2(|αβ〉 − |βα〉)

Mayhall 75

6 TECHNIQUES OF APPROXIMATION

6 Techniques of Approximation6.1 Time-independent perturbation theory

6.1.1 Two-level system

• Consider a system with only 2 states, |1〉 , |2〉 with energies E1, E2

• Being eigenstates of H, the Hamiltonian in this basis is diagonal:

H =

E1 0

0 E2

• Now, let’s add a ”small” perturbation to the system, which potentially mixes the two states:

H =H(0) + H(1) =

E1 0

0 E2

+

0 ε

ε 0

=

E1 ε

ε E2

– Familiar examples:

∗ Harmonic Oscillator: adding αx3

∗ Heisenberg Hamiltonian: adding magnetic moment∗ Orbital mixing between atoms

– Kets |1〉 and |2〉 are no longer eigenstates

• We want to find the new energies and states of the new Hamiltonian such that |ψ〉 = c1 |1〉+ c2 |2〉

– We need to solve:

H |ψ〉 =E |ψ〉(H − IE

)|ψ〉 =0E1 − E ε

ε E2 − E

c1

c2

=0

– If(

H − IE)−1

exists, only trivial solution obtained

– To get a non-trivial solution,(

H − IE)−1

must not exist

– If det|H − IE| = 0, then the inverse doesn’t exist

Mayhall 76


The determinant of a matrix is equal to the product of the eigen-values. If at least one eigenvalue is zero, then the determinant iszero. If at least one eigenvalue is zero, then the inverse diverges.

– Thus, we seek solutions to the following secular equation , by solving for E:

0 =

∣∣∣∣∣∣∣∣E1 − E ε

ε E2 − E

∣∣∣∣∣∣∣∣= (E1 − E) (E2 − E)− ε2

=E2 + (−E1 − E2)E +(

E1E2 − ε2)

quadratic equation in E

Remember, quadratic equation :

x =−b±

√b2 − 4ac

2a

– Two solutions, E+ and E−

E± =E1 + E2

2±√

E21 + E2

2 + 2E1E2 − 4E1E2 + 4ε2/2

=E±√

∆E2 + 4ε2

2where E =

E1 + E2

2, ∆E = E2 − E1

– The perturbation shis the energies up and down, about the average energy

• Let’s inspect this a bit...

– If ∆E = 0, then E± = E± ε

– If ε = 0, then E± = E1,2

– Let’s plot this as a function of the energy gap of the original (unperturbed) eigenstates ∆E,

Mayhall 77


ε

E-

E+

E1

E2

– What about intermediate regions?

• Case 1: 4ε2

∆E2 1:

– First, rearrange

E± =E± 12

∆E(

1 +4ε2

∆E2

)1/2

rearranging

– let’s use the Taylor series form of (1 + x)1/2 = 1 + 12 x− 1

8 x2 + · · ·

Definition: Taylor series

f (x) =∞

∑n=0

1n!

f (n)(0)xn

– plugging this in,

E± =E± 12

∆E

[1 +

12

4ε2

∆E2 −18

(4ε2

∆E2

)2

− · · ·]

≈E± 12

(∆E + 2

ε2

∆E

)truncating at first order

– Ploing this...

Mayhall 78


E-

E+

– Good when ∆E is large, but blows up when ∆E goes to zero

• Case 2: 4ε2

∆E2 1:

– First, rearrange dierently, taking ∆E2

4ε2 = x

E± =E± 12

[2ε

(1 +

∆E2

4ε2

)1/2]

≈E± ε

(1 +

∆E2

8ε2

)first order

≈E±(

ε +∆E2

8ε

)– this is a quadratic equation in ∆E and can be ploed as such,

Mayhall 79


E-

E+

6.1.2 Many-level systems

• Many-level systems

– For a 2-level system, a small perturbation has a quadratic eect on the energies: ± ε2

∆E

– How do we approach a similar problem when we have a large number (greater than 2) of states?

– Perturbation theory: We will assume we have a good guess, and the use that guess to constructcorrections due to the perturbation

– We will also assume the energy eigenstates are non-degenerate.

• Start with the Time-independent Schrödinger equation

H |i〉 =E |i〉

• Generally this problem is very hard. Instead of trying to solve it directly, sometimes we can find a verysimple Hamiltonian, H(0), that is similar to the exact H, but whose form is simple enough to permit anexact solution.

• The remainder, H(1), is called a perturbation to our unperturbed H(0) Hamiltonian such that:

H =H(0) + λH(1)

where we have introduced an order-parameter, λ, which will ultimately be set to one. But for now thishelps keep track of the significance of terms

• We will also expand our wavefunction and energy in orders of the perturbation strength:

|i〉 =∣∣∣i(0)⟩+ λ

∣∣∣i(1)⟩+ λ2∣∣∣i(2)⟩+ · · ·

Ei =E(0)i + λE(1)

i + λ2E(2)i + · · ·

Mayhall 80


• Plugging this all into the Schrödinger equation:(H(0) + λH(1)

) (∣∣∣i(0)⟩+ λ∣∣∣i(1)⟩+ · · · ) =

(E(0)

i + λE(1)i + · · ·

) (∣∣∣i(0)⟩+ λ∣∣∣i(1)⟩+ · · · )

and collecting orders: (H(0) − E(0)

i

) ∣∣∣i(0)⟩ =0 : λ0(H(0) − E(0)

i

) ∣∣∣i(1)⟩+ (H(1) − E(1)i

) ∣∣∣i(0)⟩ =0 : λ1(H(0) − E(0)

i

) ∣∣∣i(2)⟩+ (H(1) − E(1)i

) ∣∣∣i(1)⟩− E(2)i

∣∣∣i(0)⟩ =0 : λ2

...

• Le multiply by⟨

i(0)∣∣∣, ⟨

i(0)∣∣∣ (H(0) − E(0)

i

) ∣∣∣i(0)⟩ =0 : λ0⟨i(0)∣∣∣ (H(0) − E(0)

i

) ∣∣∣i(1)⟩+ ⟨i(0)∣∣∣ (H(1) − E(1)

i

) ∣∣∣i(0)⟩ =0 : λ1⟨i(0)∣∣∣ (H(0) − E(0)

i

) ∣∣∣i(2)⟩+ ⟨i(0)∣∣∣ (H(1) − E(1)

i

) ∣∣∣i(1)⟩− E(2)i =0 : λ2

...

• Simplify,

0 =0 : λ0⟨i(0)∣∣∣ H(1)

∣∣∣i(0)⟩ =E(1)i : λ1⟨

i(0)∣∣∣ (H(1) − E(1)

i

) ∣∣∣i(1)⟩ =E(2)i : λ2

...

• Where does∣∣∣i(1)⟩ come from? We will use the zeroth-order states as a basis to represent the perturbed

wavefunctions: ∣∣∣i(1)⟩ =∑j

c(1)ij

∣∣∣j(0)⟩

Mayhall 81


• Substitute this back into expression for E(2)i ,

E(2)i =

⟨i(0)∣∣∣ (H(1) − E(1)

i

) ∣∣∣i(1)⟩=∑

jc(1)ij

⟨i(0)∣∣∣ (H(1) − E(1)

i

) ∣∣∣j(0)⟩=∑

jc(1)ij

⟨i(0)∣∣∣ H(1)

∣∣∣j(0)⟩− E(1)i

=∑j 6=i

c(1)ij

⟨i(0)∣∣∣ H(1)

∣∣∣j(0)⟩+ ⟨i(0)∣∣∣ H(1)

∣∣∣i(0)⟩− E(1)i

=∑j 6=i

c(1)ij

⟨i(0)∣∣∣ H(1)

∣∣∣j(0)⟩

• From where do we get the coeicients? Le multiply the Schrödinger equation for λ1 by⟨

j(0)∣∣∣, where

j 6= i, ⟨j(0)∣∣∣ (H(0) − E(0)

i

) ∣∣∣i(1)⟩+ ⟨j(0)∣∣∣ (H(1) − E(1)

i

) ∣∣∣i(0)⟩ =0 : λ1

∑k 6=i

⟨j(0)∣∣∣ (H(0) − E(0)

i

) ∣∣∣k(0)⟩ c(1)ik +⟨

j(0)∣∣∣ H(1)

∣∣∣i(0)⟩ =0

∑k 6=i

(E(0)

j − E(0)i

)δjkc(1)ik +

⟨j(0)∣∣∣ H(1)

∣∣∣i(0)⟩ =0(E(0)

j − E(0)i

)c(1)ij +

⟨j(0)∣∣∣ H(1)

∣∣∣i(0)⟩ =0

c(1)ij = −

⟨j(0)∣∣∣ H(1)

∣∣∣i(0)⟩E(0)

j − E(0)i

• We can plug this back into the expression for E(2)i ,

E(2)i =∑

j 6=i

⟨j(0)∣∣∣ H(1)

∣∣∣i(0)⟩ ⟨i(0)∣∣∣ H(1)

∣∣∣j(0)⟩E(0)

i − E(0)j

=∑j 6=i

∣∣∣H(1)ij

∣∣∣2E(0)

i − E(0)j

• Note the resemblance to the second-order expression for the two-level system,

E± ≈E±(

∆E2

+ε2

∆E

)

Mayhall 82


6.2 Variational Approximation• Perturbation theory does not provide actual eigenstates of the Hamiltonian. Also, it can blow up in the

presence of “accidental” near degeneracies. Think back to the behavior of a 2-level system. Further,it requires us being able to solve an appropriate zeroth order Hamiltonian, which may not always bepossible.

• Instead of the perturbative strategy that we just looked at, now consider the Rayleigh ratio :

ε =〈ψ′|H|ψ′〉〈ψ′|ψ′〉

• For any given “trial” wavefunction (one that is not necessarily an exact eigenstate of H), the expectationvalue of the energy is an upper limit to the exact energy of the ground state |ψ0〉:

ε ≥E0

• To prove this, let us first normalize the trial wavefunction⟨ψ′∣∣ψ′⟩ =1

and then insert the resolution of the identity for the basis of exact energy eigenstates:⟨ψ′∣∣H∣∣ψ′⟩ =∑

ij

⟨ψ′∣∣ψi⟩ ⟨

ψi∣∣H∣∣ψj

⟩ ⟨ψj∣∣ψ′⟩ RI

=∑i

⟨ψ′∣∣ψi⟩

Ei⟨ψi∣∣ψ′⟩ Eigenstates

=∑i

Ei|ci|2 ≥ E0

• ε is a weighted average of the exact eigenstates, and only equals the true ground state energy when thetrial wavefunction has unit overlap with the exact ground state

• Since ε is guaranteed to be higher than the exact ground state, we can obtain the best possible wavefunc-tion, by modifying it to yield lower and lower ε. We thus variationally optimize the wavefunction to yieldthe best approximation (for a given trial functional form)

• Choose a wavefunction ansatz, by assuming an explicit functional form of the wavefunction

• Linear optimization problems are easier than non-linear:→ Rayleigh-Ritz method

– Choose a fixed basis to represent your wavefunction∣∣ψ′⟩ =∑i|φi〉 ci

– Optimize the expansion coeicients, ci, to minimize the energy. UseLagrange’s method of undetermined multipliers , and minimize the following objective function:

L =⟨ψ′∣∣H∣∣ψ′⟩− λ

(⟨ψ′∣∣ψ′⟩− 1

)Mayhall 83


– This guarantees that the solution obtained will find the lowest energy, provided the wavefunctionremains orthonormal

L =∑ij

⟨φi∣∣H∣∣φj

⟩cicj − λ

(∑

ic2

i − 1

)

set the first-derivative of this to zero,

∂

∂ckL =0

=∑j

⟨φk∣∣H∣∣φj

⟩cj + ∑

j

⟨φj∣∣H∣∣φk

⟩cj − 2λck

=Hkjcj − λck = 0

in matrix form:

Hc =cλ

– Thus, when we use a linear parameterization of the wavefunction, we get the energies, λ, and states,c, by matrix diagonalization

• Hellmann-Feynman theorem

– How does the energy change when a perturbation is applied?

– Assume we have found a variational solution via the Rayleigh-Ritz method

H |ψ〉 =E |ψ〉

– Both the energy and states are functions of the perturbation

∂

∂λ〈ψ|H|ψ〉 =

⟨∂

∂λψ

∣∣∣∣H∣∣∣∣ψ⟩+ 〈ψ| ∂

∂λH|ψ〉+

⟨ψ

∣∣∣∣H∣∣∣∣ ∂

∂λψ

⟩=E

⟨∂

∂λψ

∣∣∣∣ψ⟩+ 〈ψ| ∂

∂λH|ψ〉+ E

⟨ψ

∣∣∣∣ ∂

∂λψ

⟩=E

(⟨∂

∂λψ

∣∣∣∣ψ⟩+

⟨ψ

∣∣∣∣ ∂

∂λψ

⟩)+ 〈ψ| ∂

∂λH|ψ〉

=E∂

∂λ〈ψ|ψ〉+ 〈ψ| ∂

∂λH|ψ〉 un-chain rule

= 〈ψ| ∂

∂λH|ψ〉 derivative of constant

Mayhall 84


6.3 Time-dependent perturbation theory• Overview:

1. Find eigenstates, |n〉, of the time-independent Hamiltonian first

2. Express the eigenfunctions of the TDSE as linear combinations of TISE eigenstates, with time-dependent coeicients

3. Solve TDSE by optimizing expansion coeicients

• TDSE:

H |Ψ(t)〉 =ihddt|Ψ(t)〉

– Remember, if H 6= f (t), then

|Ψ(t)〉 = |ψ〉 |φ(t)〉= |ψ〉 e−iEt/h

=∣∣∣Ψ(0)(t)

⟩where the "(0)", indicates solution of a time-independent zeroth-order Hamiltonian, and

H |ψ〉 =E |ψ〉 TISE

– However, when H = f (t), the above is not true

– Let’s write the Hamiltonian as a sum of a time-independent part, and a time-dependent operator:

H(t) =H(0) + V(1)(t)

and use the TISE solutions as a basis:

|Ψ(t)〉 =∑n

cn(t)∣∣∣Ψ(0)

n (t)⟩

=∑n

cn(t) |n〉 e−iEnt/h

– Notice, that these states have time dependence in both the coeicient, and the exponential phasepart

• Now, let’s use the expansion over stationary states for solving the TDSE

H |Ψ(t)〉 =ihddt|Ψ(t)〉(

H(0) + V(t))

∑n

cn(t)∣∣∣Ψ(0)

n (t)⟩=ih ∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩+ ih ∑

ncn(t)

ddt

∣∣∣Ψ(0)n (t)

⟩H(0) ∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩+ V(t)∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩=ih ∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩+ H(0) ∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩V(t)∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩=ih ∑

ncn(t)

∣∣∣Ψ(0)n (t)

⟩V(t)∑

ncn(t) |n〉 e−iEnt/h =ih ∑

ncn(t) |n〉 e−iEnt/h

Mayhall 85


Now, multiply by 〈m|

〈m|V(t)∑n

cn(t) |n〉 e−iEnt/h =ih ∑n

cn(t) 〈m|n〉 e−iEnt/h

1ih ∑

ncn(t) 〈m|V(t) |n〉 eiωmnt =cm(t) where, ωmn =

Em − En

h1ih ∑

ncn(t)Vmn(t)eiωmnt =cm(t)

6.3.1 Two-level system

• Since there are only two states, we can simply write out the explicit expressions for the coeicient deriva-tives:

c1 =1ih

c1V11 +1ih

c2V12eiω12t

c2 =1ih

c1V21eiω21t +1ih

c2V22

where it is understood that the coeicients and V depend on time.

• Oen the time dependent piece is an odd function, so the diagonal terms are zero. We will assume thisto be the case here.

c1 =1ih

c2V12eiω12t

c2 =1ih

c1V21eiω21t

• Initial Conditions: Let’s consider a situation in which a system starts out completely in state 1, c1(0) =1, and at t = 0, a constant potential is turned on until it is turned o at t = T. Thus, in this example, thetime-dependent potential is a step function.

– Start by geing an equation in terms of derivatives of only one variable.

c1 =ih

V21e−iω21t c2

– Now dierentiate the c2 equation again:

c2 =V21

iheiω21t c1 +

V21

ih(iω21) eiω21tc1

=V21

iheiω21t

(1ih

c2V12eiω12t)+

V21

ih(iω21) eiω21t

(ih

V21e−iω21t c2

)sub

=−∣∣∣∣V21

h

∣∣∣∣2 c2 + iω21c2

Mayhall 86


– Second-order dierential equation:

Ac2 + Bc2 + Cc2 =0

where,

A =1B =− iω21

C =

∣∣∣∣V21

h

∣∣∣∣2– Try a solution of the form: c2(t) = kert

kertr2 + kertrB + kertC =0

kert(

r2 + Br + C)=0 quadratic equation in r

using the quadratic equation,

r =− B2± 1

2

√B2 − 4C

=iω21

2± 1

2

√−ω2

21 − 4∣∣∣∣V21

h

∣∣∣∣2=

iω21

2± iΩ Ω =

12

√ω2

21 +4h2 |V21|2

so,

c2(t) =eiω21

2 t[k1eiΩt + k2e−iΩt

]– Use initial conditions, c2(t = 0) = 0:

k1 = −k2

so,

c2(t) =2ikeiω21

2 t sin Ωt

– Use above equation to solve for c1(t):

c1 =ih

V21e−iω21t c2

=ih

V21e−iω21t

[2ike

iω212 t sin Ωt

(iω21

2

)+ 2ike

iω212 t cos ΩtΩ

]=

khV21

e−iω21

2 t [−iω21 sin Ωt− 2Ω cos Ωt]

Mayhall 87


– Require normalization at boundary condition∫

dxΨ(0)∗Ψ(0) = 1

〈Ψ(0)|Ψ(0)〉 =1=c1(0)∗c1(0)

=

(−2khΩ

V21

)2

k =V21

2hΩ

– Now we can write down the full expressions,

c2(t) =iV21

hΩe

iω212 t sin Ωt

c1(t) =− e−iω21

2 t[

iω21

2Ωsin Ωt + cos Ωt

]– This is the exact solution, given the particular constant form of the time dependent operator V(t) =

V, and initial conditions (the state starts in level 1)

• How does state occupation evolve in time? -→ Rabi formula

– Probability of being in level 2:

P2(t) =|V|2

h2Ω2sin2 (Ωt)

=4|V|2

h2ω221 + 4|V|2

sin2

√ω2214

+|V|2

h2 t

– Case 1: Degenerate levels, ω21 = 0

P2(t) = sin2(|V|h

t)

here, the system goes back and forth between the states, with a frequency which depends on thestrength of the perturbation

Mayhall 88


– Case 2: Big Gap/Small perturbation, h2ω221 4|V|2

P2(t) ≈(

2|V|hω21

)2

sin2(ω21

2t)

here, the system goes partially back and forth between the states, with a frequency which dependson the gap between the states. Full population transfer does not occur, but depends on the strengthof the perturbation.

6.3.2 Many-level systems

• We have already discussed the general strategy for solving the TDSE: expand the TDSE solutions |Ψ(t)〉as a linear combination of stationary states,

∣∣∣Ψ(0)n (t)

⟩|Ψ(t)〉 =∑

ncn

∣∣∣Ψ(0)n (t)

⟩=∑

ncn |n〉 e−iEnt/h

• We then derived the general expression for the derivative of the coeicients

cm(t) =1ih ∑

ncn(t)Vmn(t)eiωmnt

• To solve this we will make some assumptions, and then some approximations.

1. Small perturbation: V(t) ωnk

2. System starts in ground state: c0(t = 0) = 1

• Since the perturbation is small, in the time immediately following t0, the system will largely remain inthe ground state, suggesting a perturbative expansion about t0

H =H(0) + λV(t)

ck(t) =ck(0) + λc(1)k (t) + λ2c(2)k (t) + · · · c(0)k (t) = ck(0)

Mayhall 89


• Substitute this expansion into the equation above:

cm(t) =1ih ∑

ncn(t)Vmn(t)eiωmnt

ddt

(cm(0) + λc(1)m (t) + · · ·

)=

1ih ∑

nλV(1)

mn (t)eiωmnt(

cn(0) + λc(1)n (t) + · · ·)

• Collect orders:

ddt

cm(0) =0 : λ0

ddt

c(1)m (t) =1ih ∑

nV(1)

mn (t)eiωmntcn(0) : λ1

However, at t = 0, only ground state is occupied c0(0) = 1, so,

ddt

c(1)m (t) =1ih

V(1)m0 (t)e

iωm0t

• We can directly integrate this to get the coeicients,

c(1)m (t) =cm(0) +1ih

∫ t′=t

t′=0V(1)

m0 (t′)eiωm0t′dt′

Notice that this is simply the Fourier Transform of the perturbation!

• In order to proceed, we must know something about the functional form of the perturbation so as tointegrate it!

• Case 1: Static potential

– If a constant perturbation is abruptly switched on at t = 0, and turned o at t = T (like we didwith the 2-level system),

V

t=0 t=T

– For this potential we can break up the integral,

c(1)k (t) = c(0)k −ih

[∫ T

0dt′ +

∫ t

Tdt′]

Vk0(t′)e−iωk0t′

Mayhall 90


– from this it’s clear that aer the potential is turned o, the coeicient remains unchanged fromc(1)k (T), so we will just focus on what happens leading up to that point, t < T.

c(1)k (t) =c(0)k −iVk0

h

∫ t

0dt′e−iωk0t′

=δ0k +Vk0

h1

ωk0

(e−iωk0t − 1

)– Let’s look at the probability for the system to be in an excited state, i.e., k 6= 0

P(1)k (t) =

V2k0

h2ω2k0

(eiωk0t − 1

) (e−iωk0t − 1

)=

V2k0

h2ω2k0

(1− eiωk0t − e−iωk0t + 1

)=

V2k0

h2ω2k0

(2− 2 cos (ωk0t)) Euler’s formula

=2V2

k0

h2ω2k0

(1−

[1− 2 sin2

(ωk0

2t)])

double angle

=4V2

k0

h2ω2k0

sin2(ωk0

2t)

Double angle :cos(x) = 1− 2 sin2( x

2 )

– This is exactly what we got for the 2-level system, when we had a big gap!

– Each excited state then accrues population in an identical (to first order) way as a 2-level system

• Case 2: Oscillating potential

– now let’s consider the case in which the time dependent part of the hamiltonian is an oscillatingelectric field,

H =H(0) + 2V cos(ωt) where, ω = 2πν

– Remembering our equation for expressing the TDSE in the basis of stationary states, we can substi-

Mayhall 91


tute the above potential:

c(1)m (t) =cm(0) +1ih

∫ t′=t

t′=0V(1)

m0 (t′)eiωm0t′dt′

=2Vm0

ih

∫ t′=t

t′=0cos(ωt′)eiωm0t′dt′

=Vm0

ih

∫ t′=t

t′=0

(eiωt′ + e−iωt′

)eiωm0t′dt′ Euler

=Vm0

ih

∫ t′=t

t′=0

(ei(ωmo+ω)t′ + ei(ωmo−ω)t′

)dt′

=Vm0

ih

∫ t′=t

t′=0

(eiω+t′ + eiω−t′

)dt′ ω± = ωmo ±ω

=− Vm0

hω+

(eiω+t − 1

)− Vm0

hω−

(eiω−t − 1

)

– Considering the typical case where our system starts in a low energy state and has the possibilityto transition to higher energy states: ωmo =

Em−E0h > 0, we can notice that if we add the frequency

of the oscillating potential ω+, then this value only gets larger.

– Thus, in the perturbative regime, Vmoωm0 1, the first term is small compared to the second term, and

becomes vanishingly small when ω approaches ωm0.

c(1)m (t) ≈− Vm0

hω−

(eiω−t − 1

)

– The probability of making a transition to state m is,

P(1)m (t) =

|Vm0|2

h2ω2−

(1− e−iω−t − eiω−t + 1

)=|Vm0|2

h2ω2−

(2− 2 cos(ω−t))

=|Vm0|2

h2ω2−

(2− 2(1− 2 sin2

(ω−t

2

))double angle

=4|Vm0|2

h2ω2−

sin2(

ω−t2

)

– If the frequency of the oscillating perturbation is zero, ω = 0, then the results are identical to theconstant field results (as one should expect).

– One can interpret this result as if the oscillating field changes the energy gaps of the system, pro-viding an “eective” energy gap.

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– If ω → ωm0, then the “eective energy gap” drops to zero, and the system should behave as adegenerate system, which can transfer all of it’s population.

limω→ωm0

P(1)m (t) = lim

ω→ωm0

4|Vm0|2

h2ω2−

sin2(

ω−t2

)

= limω→ωm0

4|Vm0|2

h2ω2−

(ω−t

2−

ω3−t3

23 · 3!+

ω5−t5

25 · 5!− · · ·

)2

Taylor series

= limω→ωm0

4|Vm0|2

h2

(t2− ω2

−t3

23 · 3!+

ω4−t5

25 · 5!− · · ·

)2

=|Vm0|2

h2 t2

sin(x) = x− x3

3!+

x5

5!− x7

7!+ · · ·

6.3.3 Transitions to several possible states

• Assuming we can truncate aer first order in the perturbation, the transition probability is given as:

Pm(t) =4|Vm0|2

h2ω2k0

sin2(

ωk0t2

)ωk0 either bare or shied energy gap

which (if Vk0 is slowly varying with k) is essentially just a squared “sinc” function.

Pm(t) =|Vm0|2

h2 sinc2(

ωk0t2

)sinc(x) = sin(x)/x

• It is easily seen that when the transition frequency ωk0 = 0 (or when ω − ωk0 = 0 in an oscillatingfield) the probability of transitioning to state k increases quadratically, also seen from the limit above.This results in a highly peaked function, which tells us that the only state to which a transition is likelyto occur is a transition where ωk0 = 0 (and Vm0 6= 0).

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t=2t=3

t=4

t=5

t=6

Pk

wk0

0

t=100

• That this can (unphysically) exceed unit probability is a consequence of first order perturbation theory.

6.3.4 Fermi’s Golden Rule

• Here, we will consider what happens if a system has the opportunity to transition from initial state |i〉(not necessarily the ground state) to a continuum of energy states, such that we must introduce a densityof states, ρ(Ek), (i.e., the number of states inside the region Ek + δk).

ρ(Ek) =dk

dEk=⇒ ρ(Ek)dEk = dk (189)

• Examples:

– Electron transfer kinetics: Marcus theory

– Förster energy transfer

– Internal conversion/non-radiative decay/intersystem crossing

– β-decay of a neutron

– many more!

• Start with a few assumptions:

1. assume “long” time compared to energy gaps: t 2πωk0

2. assume Vk0 is slowly varying with k: dVk0dk 1

3. assume ρ(Ek) is slowly varying with k: dρ(Ek)dk 1

• Each possible transition has a probability Pk(t), but since there are multiple states with a given energyEk (such that ωk0 = 0), then the total probability is a sum over all the total possible states with energy

Mayhall 94


Ek.

P(t) =∑k

Pk(t) add up all transitions (190)

⇒∫

dk Pk(t) continuous (191)

=∫

dEk ρ(Ek)Pk(t) substitute for dk (192)

Substitute in for Pk(t), and recognize that since the sinc function is sharply peaked around the initial statei, that only the behavior of both the density of states and the matrix element around k = i is needed.And since we have assumed these are slowly varying quantities, then they are eectively constant in theregion around i “plucked out” by the sinc function.

P(wki)

ρ(Ek)

wii

P(t) =∫

dEk ρ(Ek)4|Vmi|2

h2ω2ki

sin2(

ωkit2

)(193)

=|Vi|2ρ(Ei)

h2

∫dEk

sin2(

ωkit2

)(ωki

2

)2 (194)

This integral is easily evaluated, noting that∫ ∞−∞ dx sinc(x)2 = π. Let

x =ωkit

2=

(Ek − Ei)t2h

(195)

dx =dωkit

2=

dEkt2h

(196)

⇒ dEk =2ht

dx (197)

Inserting this into the above equation yields:

∫ ∞

−∞dx sinc(x)2 =

∫ ∞

−∞

dEkt2h

sin2(

ωkit2

)(

ωkit2

)2 (198)

= π (199)

We can rearrange this to see that:

∫dEk

sin2(

ωkit2

)(ωki

2

)2 = 2πht (200)

Mayhall 95


such that the probability is linear in time:

P(t) =2π|Vi|2ρ(Ei)

ht (201)

• The transition rate, is the change in the probability with respect to time:

k(t) =dP(t)

dt=

2π|Vi|2ρ(Ei)

h(202)

• Thus, the decay rate (i.e., the rate of population loss from state |i〉 to a continuum of states) is independentof time, and is completely determined by a characteristic coupling matrix element, and the continuum’sdensity of states near the initial state’s energy. This is known as Fermi’s Golden Rule !

Mayhall 96

7 MANY-ELECTRON SYSTEMS

7 Many-electron systems7.1 Hydrogen atom spectrum

7.1.1 Selection rules

• What happens when you hit a Hydrogen atom with a photon?

• An electromagnetic wave can be represented approximately as:

~E =2~E0 cos ωt

Relatively flat

• The interaction of an atom with this field gives the following interaction potential

V(1)(t) =− 2µ · ~E0 cos ωt,

where ~µ is the electic dipole moment operator:

µ =µx~ex + µy~ey + µz~ez

=− e(X~ex + Y~ey + Z~ez

)=− eR

• Thus, we know that the transition from |i〉 to | f 〉 occurs with the following probability (to first order):

P(1)i f (t) =

|Vi f |2

h2

(sin(

ω2 t)

ω2

)2

=| 〈i|µ| f 〉 |2 · ~E2

0

h2

(sin(

ω2 t)

ω2

)2

=e2| 〈i|R| f 〉 |2 · ~E2

0

h2

(sin(

ω2 t)

ω2

)2

∝| 〈i|µ| f 〉 |2

• Thus, 〈i|µ| f 〉 6= 0 is required to observe a transition

• Laporte Selection rule : Required (−1)li 6= (−1)l f , change in pairity (for systems with inversion sym-metry, u to g transitions allowed)

Mayhall 97


• ∆l = ±1Symmetry or conservation of the photon’s angular momentum

• ∆ml = 0,±1Suppose light is plane-polarized (z):

– 〈i|µ| f 〉 ⇒ 〈i|µz| f 〉– and, µz = −er cos θ, so

〈 f |µz|i〉 ∝∫ 2π

0e−iml f

φeimli

φdφ =∫

ei∆mlφdφ

which equals 0, unless ∆ml = 0

– Considering x and y polarized light leads to ∆ml = ±1 cases

7.1.2 Spin Orbit Coupling

• Charged particle + angular momentum = magnetic moment

m

IA=2πr2

• 2 sources of angular momentum: orbital and spin

• Orbital angular momentum has a classical analogue, so we consider this first. Classically, magnetic mo-mentum is given by:

~m =q2~r×~v

But remember that angular momentum looks very similar,

~l =~r× ~p=m~r×~v

so,

~m =q

2m~l

=γ~l

Mayhall 98


• The magnetic moment and the angular momentum are proportional. For a single electron moving abouta fixed radius (in an orbital), the proportionality constant, γe, is the magnetogyric ratio of the electron:

γe =−e2me

• We can multiply that by h to get the Bohr magneton , µB

µB =− γh

such that

mz =− µBml ml is quantum number

• What about spin angular momentum? No classical analogue. Need to solve the relativistic Dirac equation.

Mayhall 99


Mayhall 100


Mayhall 101


7.2 Helium Atom• Simplest example of many-electron system: He atom

• No analytic solution, due to electron-electron repulsion

H =−h2

2me

(∆2

1 + ∆22

)− 2e2

4πε0

(1r1

+1r2

)+

e2

4πε0

1r12

• Simplify this Hamiltonian - convert to atomic units (au)

– Start with Hydrogen atom Schödinger equation(− h2

2me∆2 − e2

4πεr

)ψ(r) =Eψ(r)

– Let us scale the coordinates such that r = λr′(− h2

2meλ2 ∆′2 − e2

4πελr′

)ψ(r′) =Eψ(r′)

– We’d like to be able to factor out the constants. We can only do that if:

h2

meλ2 =e2

4πε0λ= k

thus,

λ =4πε0h2

mee2

– Consider the units of λ: (F/m)(J s)2/(C2kg) = (C2/mJ)(J2 s2)/(C2kg) = (1/m)(Js2)/(kg)= (1/m)(kg m2s−2s2)/(kg) = (m) distance

– λ is a standard distance called a Bohr radius

– Consider k

k =me

(e2

4πε0h

)2

units of energy

– k is a unit of energy called a Hartree

– We can now rewrite the Schrödinger equation in these atomic units:(−1

2∆′2 − 1

r′

)ψ(r′) =

Ek

ψ(r′)

– and now the Helium atom Hamiltonian is,

H =−12

(∆2

1 + ∆22

)− 2

(1r1

+1r2

)+

1r12

dropping the ′

Mayhall 102


7.2.1 Slater Determinant

• Although, we have solved for the exact Hydrogen orbitals, the solutions to a Helium atom, is not possible.Perhaps we can use the Hydrogen atom solutions to help our eorts at solving the Helium atom.

• In this direction, if the interelectronic repulsion term were not present, we would have a separable equa-tion, such that a product of 1-particle solutions (hydrogenic “orbitals”) would also be a solution:(

H1 + H2)

ψ1(r1)ψ2(r2) = (E1 + E2)ψ1(r1)ψ2(r2)

• Remember, the only thing we get from the hydrogen atom is n, l, ml ; spin quantum numbers can only berigorously added aer solving the Dirac equation.

• Let us now refer to these purely spatial functions (defined by n, l, ml) as spatial orbitals , ψ(r)

• Defining functions which also involve spin coordinates, we further define spin orbitals , χ(x) = ψ(r)σ(ω),where σ(ω) could either be α or β spin

• An approximate many-electron wavefunction, which is obtained by taking a product of spin orbitals iscalled a Hartree product

• Unfortunately, a Hartree product does not satisfy one of the fundamental requirements of a Fermionicwavefunction:

Definition: Pauli exclusion principle

The total wavefunction for two Fermions is antisymmetric with respect toexchange of any two particles.

ψ(r1, r2) = −ψ(r2, r1)

Thus if we exchange the electron indices in a Hartree product:

P(1, 2)ψ1(r1)ψ2(r2) =ψ1(r2)ψ2(r1) 6= −ψ1(r1)ψ2(r2)

we don’t get the same wavefunction back plus a sign.

• Put in a dierent way, “No two electrons can have the same quantum address” (set of quantum numbers,n, l, ml, s, ms)

• A spatial orbital can have occupations, 0, 1, or 2, while a spin orbital can only have 0 or 1.

• Although a single Hartree product does not satisfy the Pauli exclusion principle, a linear combinationcan:

Ψ(r1, r2) =χ1(r1)χ2(r2)− χ2(r1)χ1(r2)

P(1, 2)Ψ(r1, r2) =χ1(r2)χ2(r1)− χ2(r2)χ1(r1)

=Ψ(r2, r1) = −Ψ(r1, r2)

Mayhall 103


• Notice that this linear combination can be wrien as the determinant of a matrix:

det

∣∣∣∣∣∣∣∣χ1(r1) χ1(r2)

χ2(r1) χ2(r2)

∣∣∣∣∣∣∣∣ =χ1(r1)χ2(r2)− χ2(r1)χ1(r2)

• This can be further generalized to N-electron systems. The normalized form is called a Slater determinantwavefunction.

ΨSD(x1, x2, . . . , xN) =1√N!

det

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

χ1(r1) χ1(r2) · · · χ1(rN)

χ2(r1) χ2(r2) · · · χ2(rN)

...

χN(r1) χN(r2) · · · χN(rN)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

• The determinantal form of the wavefunction, ensure antisymmetry and single occupation of spin orbitals.

For instance, suppose we tried to put two electrons inside of a single spin orbital:

ΨSD(r1, r2) =1√2

(χ1(r1)χ1(r2)− χ1(r2)χ1(r1)

)= 0

This is also a general property of a determinant of a matrix: if two columns are the same the determinantis zero, and if you swap to columns, the determinant picks up a negative sign

7.2.2 Statistical Correlation

• The Slater determinant introduces correlated motions of electrons of same spin

1 =∫

dx1dx2|Ψ(x1, x2)|2

=∫

dr1dr2dω1dω2 |Ψ(x1, x2)|2

=∫

dr1dr2P(r1, r2)

dr2

dr1

r1

r2

P(r1,r2)dr1dr2

Mayhall 104


P(r1, r2) =12

∫dω1dω2 |ψ1(r1)ψ2(r2)σ1(ω1)σ2(ω2)− ψ1(r2)ψ2(r1)σ1(ω2)σ2(ω1)|2

=12

∫dω1dω2[|ψ1(r1)|2|ψ2(r2)|2 + |ψ1(r2)|2|ψ2(r1)|2

− ψ∗1(r1)ψ2(r1) · ψ∗2(r2)ψ1(r2) · σ∗1 (ω1)σ2(ω1) · σ∗2 (ω2)σ1(ω2)− c.c.]

• CASE 1: The spins are not the same

– i.e., σ1 = α and σ2 = β then,

P(r1, r2) =12

(|ψ1(r1)|2|ψ2(r2)|2 + |ψ1(r2)|2|ψ2(r1)|2

)– if ψ1 = ψ2,

P(r1, r2) =|ψ1(r1)|2|ψ1(r2)|2

– the probability density becomes the product of the densities: uncorrelated

– They can exist in the same location P(r1, r1) 6= 0

• CASE 2: The spins are the same

– i.e., σ1 = α and σ2 = α then,

P(r1, r2) =12(|ψ1(r1)|2|ψ2(r2)|2 + |ψ1(r2)|2|ψ2(r1)|2

− ψ∗1(r1)ψ2(r1) · ψ∗2(r2)ψ1(r2)− ψ∗1(r2)ψ2(r2) · ψ∗2(r1)ψ1(r1))

– Particles are correlated, Fermi hole

P(r1, r1) =12(|ψ1(r1)|2|ψ2(r1)|2 + |ψ1(r1)|2|ψ2(r1)|2

− ψ∗1(r1)ψ2(r1) · ψ∗2(r1)ψ1(r1)− ψ∗1(r1)ψ2(r1) · ψ∗2(r1)ψ1(r1))

=0

7.3 Hartree-Fock

7.3.1 Energy of a Slater Determinant

• The Hamiltonian can be broken up into 1 and 2 electron terms

• For simplicity, let’s consider again, the Helium Hamiltonian

H =−12

(∆2

1 + ∆22

)− 2

(1r1

+1r2

)+

1r12

=O1 + O2

=h(x1) + h(x2) + v(x1, x2)

Mayhall 105


• Expectation value of a 1-electron operator with a Slater Determinant wavefunction⟨ΨSD

∣∣∣h(x1)∣∣∣ΨSD

⟩=

12

∫dx1dx2 [χ1(x1)χ2(x2)− χ1(x2)χ2(x1)]

∗

h(x1) [χ1(x1)χ2(x2)− χ1(x2)χ2(x1)]

=12

∫dx1

(χ1(x1)

∗h(x1)χ1(x1) + χ2(x1)∗h(x1)χ2(x1)

)=

12

(〈χ1|h(x1)|χ1〉+ 〈χ2|h(x1)|χ2〉

)=⟨

ΨSD∣∣∣h(x2)

∣∣∣ΨSD⟩

and so, ⟨ΨSD

∣∣∣O1

∣∣∣ΨSD⟩=⟨

ΨSD∣∣∣h(x1) + h(x2)

∣∣∣ΨSD⟩

= 〈χ1|h|χ1〉+ 〈χ2|h|χ2〉And for an N-electron wavefunction,⟨

ΨSD∣∣∣O1

∣∣∣ΨSD⟩=

N

∑i〈χi|h|χi〉

where i are the orbitals occupied in ΨSD

• Expectation value of a 2-electron operator with a Slater Determinant wavefunction

• Again, let’s start with Helium,⟨ΨSD

∣∣∣v(x1, x2)∣∣∣ΨSD

⟩=

12

∫dx1dx2 [χ1(x1)χ2(x2)− χ1(x2)χ2(x1)]

∗

v(x1, x2) [χ1(x1)χ2(x2)− χ1(x2)χ2(x1)]

=12

∫dx1dx2χ1(1)χ2(2)v(1, 2)χ1(1)χ2(2)

+12

∫dx1dx2χ1(2)χ2(1)v(1, 2)χ1(2)χ2(1)

− 12

∫dx1dx2χ1(1)χ2(2)v(1, 2)χ1(2)χ2(1)

− 12

∫dx1dx2χ1(2)χ2(1)v(1, 2)χ1(1)χ2(2)

In each case, we are integrating over both electron coordinates, so the first two, and last two are duplicates,⟨ΨSD

∣∣∣v(x1, x2)∣∣∣ΨSD

⟩=∫

dx1dx2χ1(1)χ2(2)v(1, 2)χ1(1)χ2(2)

−∫

dx1dx2χ1(1)χ2(2)v(1, 2)χ1(2)χ2(1)

= 〈χ1χ2|χ1χ2〉 − 〈χ1χ2|χ2χ1〉= 〈χ1χ2||χ1χ2〉

where 〈χ1χ2||χ1χ2〉 is an antisymmetrized two-electron repulsion integral, when v(1, 2) = r−112

Mayhall 106


• For an N-electron wavefunction, there is a single matrix element for each pair of spin-orbitals⟨ΨSD

∣∣∣r−112

∣∣∣ΨSD⟩=∑

i<j

⟨χiχj

∣∣r−112

∣∣χiχj⟩−⟨χiχj

∣∣r−112

∣∣χjχi⟩

=∑i<j〈ij||ij〉 = 1

2 ∑ij〈ij||ij〉

• The final energy expression is therefore,

ESD =EN + ∑i〈χi|h|χi〉+

12 ∑

ij

⟨χiχj

∣∣∣∣χiχj⟩

7.3.2 Minimization of Energy

• How do we get the molecular orbitals, χi?

• We will use the variational method. In ortherwords, the best molecular orbitals are those for which theSlater determinant is orthonormalized and a minimum in energy.

• This represents a constrained optimization, and as such we will useLagrange’s method of undetermined multipliers again. However, since we don’t know anything about

the form of the orbitals, we will use a more general notation, and describe the Lagrangian as a functionalof the orbitals.

L[χk] =EN + ∑i〈χi|h|χi〉+

12 ∑

ij

⟨χiχj

∣∣∣∣χiχj⟩−∑

ijεij(⟨

χi∣∣χj⟩− δij

)• Similar to what we’ve done before, we want to set the first “derivative” equal to zero. However, since this

is a functional, we should refer to this as seing, the first functional variation to zero, although this ismechanically identical to the derivative, i.e.,

L[χi]small change−−−−−−→ L[χi + δχi] =L[χi] + L[δχi]

=L[χi] + δL[χi]

then require,

δL[χi] =0

• Expand the first variation

δL[χk] =∑i

(〈δχi|h|χi〉+ 〈χi|h|δχi〉

)+

12 ∑

ij

( ⟨δχiχj

∣∣∣∣χiχj⟩+⟨χiδχj

∣∣∣∣χiχj⟩+⟨χiχj

∣∣∣∣δχiχj⟩+⟨χiχj

∣∣∣∣χiδχj⟩)

−∑ij

εij(⟨

δχi∣∣χj⟩+⟨χi∣∣δχj

⟩)= 0

Mayhall 107


– A couple of these terms can be combined,⟨δχiχj

∣∣∣∣χiχj⟩=∫

dx1dx2[δχi(1)χj(2)

]∗ 1r12

[χi(1)χj(2)− χj(1)χi(2)

]=∫

dx2dx1[δχi(2)χj(1)

]∗ 1r12

[χi(2)χj(1)− χj(2)χi(1)

]flip 1, 2

=∫

dx1dx2[χj(1)δχi(2)

]∗ 1r12

[χj(1)χi(2)− χi(1)χj(2)

]reorder terms

=⟨χjδχi

∣∣∣∣χjχi⟩

– Now, combine those terms,

δL[χk] =∑i

(〈δχi|h|χi〉+ 〈χi|h|δχi〉

)+ ∑

ij

( ⟨δχiχj

∣∣∣∣χiχj⟩+⟨χiχj

∣∣∣∣δχiχj⟩)−∑

ijεij(⟨

δχi∣∣χj⟩+⟨χi∣∣δχj

⟩)= 0

– and seperate out the complex conjugate,

δL[χk] =∑i〈δχi|h|χi〉+ ∑

ij

⟨δχiχj

∣∣∣∣χiχj⟩−∑

ijεij⟨δχi∣∣χj⟩+ C.C. = 0

=∑i

∫dx1δχ∗i (1)hχi(1) + ∑

ij

∫ ∫dx1dx2δχ∗i (1)χ

∗j (2)

1r12

(χi(1)χj(2)− χj(1)χi(2)

)−∑

ijεij

∫dx1δχ∗i (1)χj(1) + C.C. = 0

– Now pull out the integration over electron 1,

=∑i

∫dx1δχ∗i (1)

×[

hχi(1) + ∑j

∫dx2χj(2)

1r12

χj(2)χi(1)−∑j

∫dx2χj(2)

1r12

χi(2)χj(1)−∑j

χj(1)εij

]+ C.C. = 0

– The non-trivial solution requires the following expression to hold:

hχi(1) + ∑j

∫dx2χj(2)

1r12

χj(2)χi(1)−∑j

∫dx2χj(2)

1r12

χi(2)χj(1) = ∑j

χj(1)εij

=

[h + ∑

j

∫dx2χj(2)

1r12

χj(2)−∑j

∫dx2χj(2)

1r12

P(1, 2)χj(2)

]χi(1) = ∑

jχj(1)εij

• Fock operator

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– The second term, can be rewrien as an eective 1-electron coulomb operator ,

∑j

∫dx2χj(2)

1r12

χj(2)χi(1) =∑j

Jj(1)χi(1)

where,

Jj(1) =∫

dx2χj(2)1

r12χj(2)

– The third term, can be rewrien as an eective 1-electron exchange operator ,

∑j

∫dx2χj(2)

1r12

χi(2)χj(1) =∑j

Kj(1)χi(1)

where,

Kj(1) =∫

dx2χj(2)1

r12P(1, 2)χj(2)

and P(1, 2) is the permutation operator, P(1, 2)χi(1)χj(2) = χi(2)χj(1)

– The exchange operator, K, is a purely quantum mechanical quantity with no classical analogue. Be-cause the integral over x2 takes place when electron 2 is in two dierent orbitals, this is commonlycalled non-local exchange

– We can now use these definitions to define a Fock operator, f ,

f =h(1) + ∑j

(Jj(1)− Kj(1)

)such that,

f χi(1) =∑j

εijχj(1)

this constitutes the Hartree-Fock equations

7.3.3 Canonical orbitals

• In order to transform the Hartree-Fock equations into the pseudo-eigenvalue equation, we need to con-sider how a Slater determinant behaves under unitary transformation.

• Consider the eect of combining all of the occupied molecular orbitals via multiplication with a unitary matrix

(i.e., one for which U† = U−1),

χ′i =∑j

χjUji

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• Now, recall that the Slater determinant is the determinant of a matrix of orbitals with varying electroncoordinates:

ΨSD(x1, x2, . . . , xN) =1√N!

det

χ1(1) χ1(2) · · · χ1(N)

χ2(1) χ2(2) · · · χ2(N)

...

χN(1) χN(2) · · · χN(N)

=

1√N!

det(A)

• If we were to multiply this matrix A by U, we would get,

AU =

(∑

jχi(1)Uij|∑

jχi(2)Uij| · · · |∑

jχi(N)Uij

)=(

χ′i(1)|χ′i(2)| · · · |χ′i(N))

=A′

• Taking the determinant of this equation, and using the fact that det(AU) = det(A)det(U),

det(A)det(U) =det(A′)∣∣∣ΨSD

⟩det(U) =

∣∣∣Ψ′SD⟩

∣∣∣ΨSD⟩=∣∣∣Ψ′SD

⟩• since, det(U) = 1 via,

det(

U†U)=det(I) = 1

=det(U)† det(U)

=|n|2 = 1

• As a result, Unitary transformation of the occupied orbitals leaves the wavefunction unchanged! Thismeans, that if χi are solutions to the Hartree-Fock equations, then so are χ′i

• Now let’s choose a specific unitary transformation which makes the matrix of Lagrange multipliers (εij)diagonal,

f χi =∑j

εijχj

=∑jk

UikεkUkjχj

=∑k

Uikεkχ′k

Mayhall 110


now, multiply both sides by Uil and sum,

∑i

f χiUil =∑i

∑k

UilUikεkχ′k

f χ′l =∑k

δlkεkχ′k

f χ′l =εlχ′l

• We have now arrived at the canonical orbital form of the Hartree-Fock equation. These are the typicalorbitals that one associates with Hartree-Fock, and are delocalized throughout the molecule. It is in thisform that the Lagrange multiplier has a physical significance, the energy of the one-electron orbital!

7.3.4 Koopman’s Theorem

• What do the orbital “energies” mean?

• First, the sum of orbital energies does not correspond to the HF energy,

εi = 〈χi| f |χi〉= 〈χi|h + ∑

j

(Jj − Kj

)|χi〉

= 〈χi|h|χi〉+ ∑j

∫dx1

∫dx2χi(1)∗χ∗j (2)r

−112(1− P(1, 2)

)χj(2)χi(1)

= 〈χi|h|χi〉+ ∑j

⟨χiχj


∑i

ε =∑i〈χi|h|χi〉+ ∑

ij

⟨χiχj


so,

EHF =∑i

εi −12 ∑

ij

⟨χiχj


• Thus, we see that the sum of orbital energies double counts the electron-electron repulsions

e-

e- e-

Definition: Koopman’s theorem

Occupied orbital energies, are approximations to the ionization energies.

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– To illustrate this, consider the kth ionization potential, IPk = EN−1k − EN

EN =∑i〈χi|h|χi〉+

12 ∑

ij

⟨χiχj


– If we make the approximation that the optimal orbitals are unchanged during the ionization process,then,

EN−1k = ∑

i 6=k〈χi|h|χi〉+

12 ∑

i 6=k,j 6=k

⟨χiχj


– Then,

EN−1k − EN =− 〈χk|h|χk〉 −

12 ∑

j 6=k

⟨χkχj

∣∣∣∣χkχj⟩− 1

2 ∑i 6=k〈χiχk||χiχk〉

=− 〈χk|h|χk〉 −∑j

⟨χkχj

∣∣∣∣χkχj⟩

allowing j = k

=− εk

– Thus, the ionization potential associated with removing an electron from the kth occupied orbital isjust the negative of that orbital energy.

– A similar relationship holds for the Electron ainity, although this is oen much less accurate

EA

IP

KoopmanΔSCF

Exact

– This is because if orbital relaxation is included, both |N − 1〉 and |N + 1〉 states will lower in energy(∆SCF values). However, when electron correlation is added, all states will decrease in energy. Theamount of correlation depends on the number of electrons. As such, the |N + 1〉 energy will decreasethe most. This ultimately leads to Koopman IP’s being more accurate than Koopman EA’s.

7.3.5 Solving the Hartree-Fock equations

• Notice that although the canonical HF equations look like a typical eigenvalue equation, it is not. TheFock operator depends on the solution:

f [χi]χi =εχi

• As a consequence, we have a non-linear optimization problem, which we solve iteratively.

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1. Guess initial form of MO’s, χ(it)i

2. Build Fock operator with initial MO’s, f [χ(it)i ]

3. Find eigenvalues/vectors of f [χ(it)i ] to yield update MO’s, χ

(it+1)i for the next iteration

4. Did the MO’s change? Compare χ(it)i and χ

(it+1)i

– If Yes: Go back to step 2.– If No: done.

• What should the form of our molecular orbitals be? The answer to this is not obvious. The most successfulstrategy has been to assume that our MO’s can be wrien as a linear combination of fixed functions. Thefinal MO’s are defined by the expansion coeicients, Cµi, and thus these are the variables we want tosolve for.

χi(x1) =∑µ

φµ(x1)Cµi

or, in Dirac notation,

|χi〉 =∑µ

∣∣φµ

⟩Cµi

• This allows us to replace the complex integro-dierential equation with a series of linear equations.

f |χi〉 = |χi〉 εi

∑µ

f∣∣φµ

⟩Cµi =∑

ν

|φν〉Cνiεi

Multiply by 〈φλ|

∑µ

⟨φλ

∣∣ f ∣∣φµ

⟩Cµi =∑

ν

〈φλ|φν〉Cνiεi

∑µ

FλµCµi =∑ν

SλνCνiεi

FC =SCe

We can use symmetric orthogonalization to get rid of the overlap matrix. Right multiply by S−12

S−12 FS−

12 S

12 C =S

12 Ce

FC =Ce

Now, it’s just a simple eigenvalue problem.

Mayhall 113


• What does this Fock matrix look like? Let’s just use the definition of the Fock operator that we’veobtained above.⟨

φλ

∣∣ f ∣∣φµ

⟩=∫

dx1φ∗λ(1)h(1)φµ(1)

+12

∫dx1φ∗λ(1)

(∑

j

∫dx2χ∗j (2)

1r12

(1− P(1, 2)

)χj(2)

)φµ(1)

=hλµ +12 ∑

νσ∑

jC∗νjCσj

∫ ∫dx1dx2φ∗λ(1)φ

∗ν(2)

1r12

(1− P(1, 2)

)φσ(2)φµ(1)

=hλµ +12 ∑

νσ

Pνσ

⟨φλφν

∣∣∣∣φµφσ

⟩where we have introduced the density matrix,

Pµν =∑i

CµiCνi

• So the Fock matrix depends on the density matrix, which ultimately comes from the MO coeicients. Wethus solve this iteratively until a self consistent field is achieved.

• Although there are N4 individual two-electron integrals (⟨φµφν

∣∣∣∣φλφσ

⟩), for large systems, it turns out

that only a linear (instead of quadratic) number of Pµν matrix elements are asymptotically non-zero. Thus,the actual number of integrals that need to be computed is much less, and this leads to HF requiring onlyabout ≈ N3 computational operations.

7.3.6 One-electron wavefunction basis sets

• What types of functions should we use for our basis set?

• Slater-type orbitals

ψSTOnζ (r) =NYl,ml

(φ, θ)rn−1e−ζr

– Similar to the analytical solutions of Hydrogen, except STO’s don’t have radial nodes

– Become extremely complicated (and computationally demanding) when computing the 2-electronrepulsion integrals,

⟨φµφν

∣∣φλφσ

⟩• Gaussian-type orbitals

ψGTOnζ (r) =NYl,ml

(φ, θ)e−αr2

– Not as accurate, but eicient

Mayhall 114


n=1

n=2n=3 n=4

GTO

STOSTO's

• Gaussian product rule

– Consider a product of two s-type gaussian functions located on separate atoms, A and B.

AB

e

rArB

RAB

r

R

– We will use the following relationships:

~RAB =~rB −~rA

~rB =~RAB +~rA

~R =~RAB/2

~r =~rA + ~R

– Consider the following product of gaussian functions:

φµA(r1)φνB(r1) ∝e−α~r2A e−α~r2

B

∝e−α~r2A e−α(2~R+~rA)

2

– now, let’s focus on the exponents,

−α~r2A − α

(2~R +~rA

)2=− α~r2

A − α(

4~R2 + 4~R ·~rA +~r2A

)=− 2α

(2~R2 + 2~R ·~rA +~r2

A

)=− 2α

(~R2 +

(~R +~rA

)2)

=− 2α~R2 − 2α~r2

Mayhall 115


– so we see that the product of two gaussians, is itself a Gaussian, referenced to the midpoint of thetwo original atoms,

φµA(r1)φνB(r1) ∝e−2α~R2e−2α~r2

∝Ae−2α~r2

– This significantly reduces the computational cost of evaluating the numerous two-electron repulsionintegrals

7.4 Electron Correlation• Configuration Interaction

– HF was obtained by applying the variational method to an approximate wavefunction with the formof a Slater determinant. Consequently, the HF wavefunction is the “best” single Slater determinantwavefunction for a given system.

– As we have discussed, HF includes only the averaged interactions between electrons, with same spinelectrons being statistically correlated to enforce adherence to the Pauli antisymmetry principle.

– The goal of most quantum chemistry research is focused on trying to obtain a beer wavefunction(beyond HF) which includes electron correlation.

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7.4.1 DFT

• Although we know that the exact energy (or any observable) can be obtained if we have found the exactwavefunction, is there a simpler way? Density functional theory (DFT) provides a simpler approachto solving the Schrödinger equation using the electron density as the central quantity instead of thewavefunction.

• Start with the normalization criterion,

1 =∫

dx1dx2 · · · dxNψ∗(x1x2 · · · xN)ψ(x1x2 · · · xN)

and then carry out the integrations over all coordinates except x1,

1 =1N

∫dx1ρ(x1)

where the one-electron reduced density ρ(x1) is,

ρ(x1) =∫

dx2 · · · dxNψ∗(x1x2 · · · xN)ψ(x1x2 · · · xN)

• Although ρ(x1) depends only on a single electrons coordinates, it can tell us a lot:

1. The integration of the density yields number of electrons

2. The positions of the cusps tell us where the nuclei are located

3. the height of the cusps tell us the nuclear charge of the atom

O C O

• In terms of complexity, the density (a function of only 3 coordinates) is a much simpler quantity than thewavefunction (a function of 3N coordinates)

• DFT asserts that the energy is a functional of the density, E[ρ(x1)], and that we should search for theform of the functional, then find the density The following theorem makes this possible.

•Definition: Hohenberg-Kohn Theorem 1

The ground-state density ρ uniquely determines the ground state en-ergy/properties

• To prove that E↔ ρ (or equivalently that H ↔ ρ) we must show that

Mayhall 123


1. the Hamiltonian has a one-to-one correspondence with the ground state wavefunction

2. the wavefunction has a one-to-one correspondence with the density

– Assume we have two Hamiltonians

H =− 12 ∑

i∇2

i −∑i

∑A

ZA

riA+ ∑

i<j

1rij

=− 12 ∑

i∇2

i + ∑i

vext(ri) + ∑i<j

1rij

H′ =− 12 ∑

i∇2

i + ∑i

v′ext(ri) + ∑i<j

1rij

where the external potentials dier by more than a constant,

vext 6= v′ext + c

such that

H |ψ〉 =E0 |ψ〉H′∣∣ψ′⟩ =E′0

∣∣ψ′⟩– By the variational principle we know that,

E0 <⟨ψ′∣∣H∣∣ψ′⟩

<⟨ψ′∣∣H′∣∣ψ′⟩+ ⟨

ψ′∣∣H − H′

∣∣ψ′⟩<E′0 + ∑

i

⟨ψ′∣∣[vext(ri)− v′ext(ri)

]∣∣ψ′⟩<E′0 + ∑

i

∫dx1dx2 · · · dxNψ′∗(x1x2 . . . )

[vext(ri)− v′ext(ri)

]ψ′(x1x2 · · · )

<E′0 + ∑i

∫dxi

1N ρ′(xi)

[vext(ri)− v′ext(ri)

]<E′0 +

∫dr1ρ′(r1)

[vext(r1)− v′ext(r1)

]sum

– Likewise,

E′0 <E0 −∫

dr1ρ(r1)[vext(r1)− v′ext(r1)

]– Now, suppose vext 6= v′ext but ρ = ρ′

E′0 <E0 +∫

dr1ρ(r1)∆(r1)

E0 <E′0 −∫

dr1ρ(r1)∆(r1)

Mayhall 124


leading to,

E′0 − E0 <E′0 − E0

which tells us that our supposition was not possible.

– Consequently,

E[ρ] =T[ρ] + Vee[ρ] +∫

ρ(r1)vext(r1)dr1

where we now define the Hohenberg-Kohn universal functional

FHK[ρ] =T[ρ] + Vee[ρ]

• Now that we know that the energy is indeed a functional of the density, how do we find the density?

•Definition: Hohenberg-Kohn Theorem 2

Any trial density ρ′(r), the energy of that density is an upper bound theexact ground state density

E[ρ] = 〈ψ|H|ψ〉 ≤⟨ψ′∣∣H∣∣ψ′⟩ = E[ρ′]

E[ρ] ≤E[ρ′]

7.4.2 Kohn-Sham DFT

• What should the kinetic energy functional look like? This is a hard question. Solve it and you will probablyget a Nobel prize. A simple (and very inaccurate) approximation that is based on the homogeneouselectron gas is called the Thomas-Fermi functional,

ETF =3

10

(3π2

) 23∫

[ρ (r)]53 d3r.

• Kohn-Sham DFT provides a much more accurate approach for evaluating the kinetic energy.

• What if we assumed that the system was non-interacting, such that the exact wavefunction happenedto be a single Slater determinant?

• In other words, assume we know how to modify the external potential such that we can delete theelectron-electron repulsion, and still get the same density

Hλ =T + λVee + Vext(λ)

=∑i

hKS(ri) when λ→ 0

=∑i−1

2∇2

i + Vext(0)

where Vext(0) corresponds to some fictitious external potential energy, and where Vext(1) = VeN

Mayhall 125


• As a one-particle operator, a single Slater determinant provides an exact solution. The spin-orbitals thatform this determinant are called Kohn-Sham orbitals , χKS

i (x1)

N

N

N

N

N

e

e

e

e

ee

N

N

N

N

N

e

e

e

e

ee

• The kinetic energy would simply be the expectation value of the kinetic energy operator:

Ts[ρ] =− 〈ψ|∑i

12∇2|ψ〉

=− 12 ∑

i

⟨χKS

i∣∣∇2∣∣χKS

i⟩

• the s subscript indicates that this is only the form of the functional when the state is a single Slaterdeterminant.

• Unfortunately, the system is not non-interacting, and thus T[ρ] 6= Ts[ρ]

• However, Ts[ρ] gets “most” of the kinetic energy, and it becomes helpful to write the kinetic energy asTs[ρ] plus a small correction:

T[ρ] =Ts[ρ] + (T[ρ]− Ts[ρ])

• Further, we can easily write down the Vee[ρ] functional, as a sum of the classical electron-electron repul-sion functional and a correction:

J[ρ] =∫

dr1dr2ρ(r1)ρ(r2)

|r1 − r2|Vee[ρ] =J[ρ] + (Vee[ρ]− J[ρ])

• Finally, returning to the original energy functional, we can substitue these values in:

E[ρ] =Ts[ρ] + (T[ρ]− Ts[ρ]) + J[ρ] + (Vee[ρ]− J[ρ]) +∫

ρ(r1)vext(r1)dr1

=Ts[ρ] + J[ρ] +∫

ρ(r1)vext(r1)dr1 + Exc[ρ]

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7.5 Excited States

7.5.1 Born-Oppenheimer Approximation

• In all of the previous examples, we were talking about calculating the ground states. What about excitedelectronic states? This language already assumes that we can talk about the electronic degrees of freedom,and the nuclear degrees of freedom separately. Let’s develop this idea more carefully.

• The mass of proton is ≈ 2000 x as large as an electron. As such, for a given amount of kinetic energy, theelectron moves about 45x faster than a proton. Perhaps we should expect dierent timescales.

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8 APPENDIX

8 Appendix8.1 Properties of the Fourier transform

Definition: Fourier transform

The form of the Fourier transform used can vary slightly. Here are a few for1D (3D is just the same but with (2πh)−3/2):

F [ f (x)] = f (p) =1√2πh

∫f (x)e−ipx/hdx

F [ f (p)] = f (x) =1√2πh

∫f (p)eipx/hdp

other defs.

F [ f (x)] = f (k) =∫

f (x)e−2πikxdx

F [ f (x)] = f (k) =1√2πh

∫f (x)e−ikx/hdx with Planck’s const

The square root of 2π is the chosen form for symmetry purposes. Usingthe time-scaling property, one could expect to see the constant withoutthe√. The forward and backward transforms would then look a bit dif-ferent, but still just as good.

The Fourier transform between variables x and k has the following properties (using the first definition):

1. Linearity

if h(x) =a f (x) + bg(x) ∀ a, b ∈ C

then h(k) =a f (k) + ag(k)

2. Translation/Time-shiing

if h(x) = f (x− x0) ∀ x0 ∈ R

then h(k) =e−ikx0 f (k)

3. Modulation/Frequency-shiing

if h(x) =eik0x f (x) ∀ k0 ∈ R

then h(k) = f (k− k0)

4. Time scaling

if h(x) = f (ax) ∀ a > 0 ∈ R

then h(k) =1|a| f (

ka)

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8 APPENDIX

Alphabetical Index

Aadjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13angular frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32angular momentum commutation relations . . . . .57angular momentum ladder operators . . . . . . . . . . . 58angular momentum operator . . . . . . . . . . . . . . . . . . . 49annihilation operator . . . . . . . . . . . . . . . . . . . . . . . 42, 43atomic units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

BBohr magneton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99Bohr radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102bra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Ccanonical orbital . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Clebsch-Gordan coeicients . . . . . . . . . . . . . . . . . . . . 69Clebsch-Gordan series . . . . . . . . . . . . . . . . . . . . . . . . . . 69closure relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7commutator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Complete sets of commuting observables . . . 18, 19constant of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30coulomb operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109coupled picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68creation operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

DDensity functional theory . . . . . . . . . . . . . . . . . . . . . 123Dirac delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Dirac notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9dispersion relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Double angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91dual space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Eeigenvalue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15eigenvector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Euler relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39exchange operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109Expectation value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

FFermi hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96flux density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Fock matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114Fock operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .108Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . 8, 20, 133functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107, 123

GGaussian product rule . . . . . . . . . . . . . . . . . . . . . . . . . 115Gaussian-type orbitals . . . . . . . . . . . . . . . . . . . . . . . . . 114

HHamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Harmonic Oscillator energies . . . . . . . . . . . . . . . . . . . 44Hartree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Hartree product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Hartree-Fock equations . . . . . . . . . . . . . . . . . . . . . . . . 109Heisenberg Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . 74Heisenberg uncertainty principle . . . . . . . . . . . . . . . 29Hellmann-Feynman theorem . . . . . . . . . . . . . . . . . . . 84Hermite polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 46Hermitian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14, 16Hermitian conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Hohenberg-Kohn Theorem 1 . . . . . . . . . . . . . . . . . . 123Hohenberg-Kohn Theorem 2 . . . . . . . . . . . . . . . . . . 125Hohenberg-Kohn universal functional . . . . . . . . . 125

Iidempotent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

KKohn-Sham DFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125Kohn-Sham orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . 126Koopman’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 111kronecker delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

LLagrange’s method of undetermined multipliers

83, 107Laguerre polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . .56Laporte Selection rule . . . . . . . . . . . . . . . . . . . . . . . . . . 97Linear functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

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linear operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5lowering operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Mmagnetogyric ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Nnon-degenerate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16non-local . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Oone-electron reduced density . . . . . . . . . . . . . . . . . . 123order of degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16orthogonal projection . . . . . . . . . . . . . . . . . . . . . . . . . . 11orthonormal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

PPauli exclusion principle . . . . . . . . . . . . . . . . . . . . . . .103Pauli matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66plane wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Qquadratic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .77antum Postulate 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 24antum Postulate 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 24antum Postulate 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 24antum Postulate 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 25antum Postulate 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 25antum Postulate 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

RRabi formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .88raising operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43Rayleigh ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83Rayleigh-Ritz method . . . . . . . . . . . . . . . . . . . . . . . . . . .83reduced mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Resolution of the Identity . . . . . . . . . . . . . . . . . . . . . . . 12

RMSD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29root mean square deviation . . . . . . . . . . . . . . . . . . . . . 29

Sscalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Schwarz inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5secular equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16, 77self consistent field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114shape function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Slater determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104Slater-type orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7spatial orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103spherical harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51spin orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103state space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9state vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

TTaylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78TISE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27

Uuncoupled picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69unitary matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15, 109

Vvector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9vector model of angular momentum . . . . . . . . . . . . 58vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Wwavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21wavefunction collapse . . . . . . . . . . . . . . . . . . . . . . . . . . 26wavefunction space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Zzero-point energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 40, 42

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