Midterm 1 Review

Wei Zhang

September 20, 2006

1 History

1.1 Photoelectric Effect

Einstein found that electrons were released from metal when shining light onthe metal. The K.E. of an electron freed is given by

K.E. = hνlight − E0

where E0 is the ionization potential, or work function of the metal.This illustrates that light was behaving as particles!

1.2 Hydrogen Atomic Spectrum

Discrete series of lines were observed in the hydrogen atomic spectrum. TheRydberg Formula accounts for the positions of the lines:

ν =1

λ= 109 680

(1

n21

− 1

n22

)cm−1

where the Rydberg constant is RH = 109 677.57cm−1.Lyman series n1 = 1, Balmer series n1 = 2, Paschen series n1 = 3, etc.

1.3 Wave-particle Duality

de Broglie relation:

λ =h

p

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2 Bra-ket Notation

Notation Analog|ψ〉 ψ(x) ket, function〈ψ| ψ∗(x) bra, complex conjugate of function|Ψ(t)〉 Ψ(x, t) ket, function〈Ψ(t)| Ψ∗(x, t) bra, complex conjugate of function

〈f |g〉∫

dxf ∗(x)g(x) inner product〈ψ|ψ〉

∫dxψ∗(x)ψ(x) norm squared

〈ψ|A|ψ〉∫

dxψ∗(x)Aψ(x) 〈A〉, expectation of A; or diagonal matrixelement

〈f |A|g〉∫

dxf ∗(x)Ag(x) off-diagonal matrix element of A

Table 1: Introduction to bra-ket notation.

3 The Schrodinger Equation

Time Dependent Schrodinger Equation (1-dimensional case) In Quan-tum Mechanics, we want to solve for the wavefunction Ψ(x, t), which de-scribes a particle, and we get it by solving the Schrodinger Equation

ih∂Ψ(x, t)

∂t= HΨ(x, t)

where H is the Hamiltonian Operator :

ih∂Ψ(x, t)

∂t= − h2

2m

∂2

∂2xΨ(x, t) + V (x)Ψ(x, t)

This is a partial differential equation, and so we solve it by separation ofvariables, which means to look for solutions of the form

Ψ(x, t) = ψ(x)f(t)

Separation of variables results in a separation constant E, energy. Thetime dependence comes out as

f(t) = e−iEt/h

and it remains to solve the time independent Schrodinger Equation.

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Time Independent Schrodinger Equation

Hψ(x) = Eψ(x)

To solve this we first need to know V (x), and we shall further discuss thiswith Various Potentials in a later section.

Note that the time independent Schrodinger Equation is an eigenvalueequation. In essence, we are looking for eigenstates of H, ψn(x), with energyeigenvalue En.

Hψn(x) = Enψn(x)

The eigenstates of the Hamiltonian, ψn(x), are also called stationarystates, see later section Wavefunctions for further discussion.

Solutions to the Schrodinger Equation Once we have found an eigen-value and an eigenstate, we can immediately spot a particular solution tothe time dependent Schrodinger Equation,

Ψn(x, t) = ψn(x)e−iEnt/h

but since H is a linear operator, the most general solution to the SchrodingerEquation would be a linear combination of these

Ψmostgeneral(x, t) =∑n

cnΨn(x, t) =∑n

cnψn(x)e−iEnt/h

We also showed in class that the expansion coefficients, cn (which may becomplex in value), can be found from the initial condition Ψn(x, 0) by takingthe inner product with the nth eigenstate ψn(x).

cn = 〈ψn|Ψ(0)〉

This results from the fact that the initial state Ψ(x, 0) can be written asan expansion of the eigenstates:

|Ψ(0)〉 =∑n

cn|ψn〉

and cn was found by taking the inner product with 〈ψm|, and then using theorthogonality relation (due to the fact that H is Hermitian):

〈ψm|ψn〉 = δmn =

{0 for n 6= m1 for n = m

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4 Operators

An operator acts on a function, and outputs another function, for example,

Af(x) = g(x)

4.1 Linear Operators

An operator A is said to be linear if

A (af(x) + bg(x)) = aAf(x) + bAg(x)

where a and b are scalars.Below are some examples of linear operators that you need to know:

Multiplicative Operator .In one dimension:

x position operator

V (x) potential energy operator

In three dimensions:

V (r) potential energy operator

Differential operators .In one dimension:

px =h

i

∂

∂xmomentum operator

Kx =p2

x

2m= − h2

2m

∂2

∂2xkinetic energy operator

In three dimensions:

p =h

i

(∂

∂x+

∂

∂y+

∂

∂z

)momentum operator

K =p2

2m= − h2

2m

(∂2

∂2x+

∂2

∂2y+

∂2

∂2z

)kinetic energy operator

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Other operators

H = − h2

2m

∂2

∂2x+ V (x) Hamiltonian operator (1D)

H = − h2

2m∇2 + V (r) Hamiltonian operator (3D)

4.2 Hermitian Operators

Definition Operator A is hermitian if 〈ψ|A|ψ〉 is real for all functions ψ(x).This is the same as saying that the expectation value 〈A〉t for any wave-

function Ψ(x, t) is real.

Turnaround Rule If Operator A is hermitian, then it’s off-diagonal ma-trix elements satisfy the turnaround rule:

〈ψ2|A|ψ1〉 = 〈ψ1|A|ψ2〉∗

i.e. the matrix representation of A would be equal to its transpose complexconjugate (also called Hermitian conjugate).

A = (AT )∗

Consequences of Hermiticity

• Diagonal matrix elements 〈ψ|A|ψ〉 are real.

• Off-diagonal matrix elements 〈ψ2|A|ψ1〉 satisfy the turnaround rule.

• All eigenvalues of A are real.

• The eigenfunctions of A are orthogonal.

〈ψm|ψn〉 = δmn =

{0 for n 6= m1 for n = m

• The eigenfunctions ψn(x) of Hermitian operator A form a complete setof functions. This is equivalent to saying that any arbitrary functionχ(x) defined on the same space can be written as a linear combinationof the eigenfunctions (basis functions)

|χ〉 =∑n

cn|ψn〉

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Examples of Hermitian Operators : x, p, p2, H, etc.

4.3 Eigenfunctions and Eigenvalues

Definition |ψn〉 is called an eigenstate of linear operator A, with eigenvaluean, if

A|ψn〉 = an|ψn〉

In this case |ψn〉 is called a pure state of A, because the uncertainty in Ais zero

∆A =√〈ψn|A2|ψn〉 − 〈ψn|A|ψn〉2 = 0

If A is an observable (i.e. A is Hermitian), then measuring A on the stateψn would always give back the eigenvalue an.

4.4 Commutators

See Shervin’s commutator worksheet in the Extra Study Aids section of thecourse website.

Heisenberg’s equation of motion For an operator A that has no explicitdependence on time, the time evolution of 〈A〉t follows the Heisenberg’s equa-tion of motion:

ihd

dt〈A〉t = −〈Ψ(t)| [H, A] |Ψ(t)〉

This means that if [H, A] = 0, then ddt〈A〉t = 0 for any state Ψ(x, t). In

this case A is called a constant of motion.

5 Wavefunctions

The total wavefunction for a 1-dimensional system is Ψ(x, t). More generally,the wavefunction is Ψ(q, t), where q is the generalized position coordinate.

The wavefunction is a complex-valued function.

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5.1 Probability Amplitude

The probability that the position of particle lies between x and (x + dx) attime t is given by

P (x, t)dx = |Ψ(x, t)|2dx = Ψ∗(x, t)Ψ(x, t)dx

Probability amplitude P (x, t) is real-valued.

5.2 Normalization

The wavefunction must be normalized so that the probability for the particleto be somewhere at any time t is 1.∫ +∞

−∞|Ψ(x, t)|2dx = 1

5.3 Stationary States

Stationary states areΨn(x, t) = ψn(x)e−iEnt/h

such that ψn(x) are eigenfunctions of the Hamiltonian operator H.Notice that stationary states DO have time dependence, i.e. in the phase

factor e−iEnt/h. However, the probability does not depend on time

P (x, t) = |Ψn(x, t)|2 = |ψn(x)|2

Furthermore, the expectation value 〈A〉t, for any observable A is alsotime-invariant for a stationary state.

d

dt〈A〉t =

d

dt〈Ψn(t)|A|Ψn(t)〉 = 0

6 Postulates of Quantum Mechanics

1. The state of a physical system s completely specified by its normalizedwavefunction Ψ(x, t) (in 1D) or Ψ(q, t) (in general).

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2. To every observable ACL(x, p) in classical mechanics, there correspondsa linear Hermitian operator

A = ACL(x, p)

The expectation value of A at time t in the state Ψ(x, t) is given by

〈A〉t = 〈Ψ(t)|A|Ψ(t)〉 =∫ +∞

−∞dxΨ∗(x, t)AΨ(x, t)

and this would be real.

3. In any measurement of the observable associated with the operator Aon a state |Ψ〉, you are certain to get one of the eigenvalues an of A,where an satisfies

Aψn = anψn

Since A is Hermitian, the measured value an would be real, and theeigenfunctions are all orthogonal.

6.1 Classical Mechanics vs. Quantum Mechanics

In classical mechanics, the state of the system is completely specified by itstrajectory {−→p (t),−→q (t)}, where −→p is momentum and −→q is position.

In Quantum mechanics, it is not possible to simultaneously specify theposition and momentum of a particle because this violates the uncertaintyprinciple.

7 Uncertainty Principle

This is a quantum mechanical result. It says that the more precisely deter-mined a particle’s position is, the less precisely its momentum is determined.Quantitatively,

∆x∆p ≥ h

2where ∆x is the standard deviation in x, ∆p is the standard deviation in p,given by

∆x =√〈(x− 〈x〉)2〉 =

√〈x2〉 − 〈x〉2

∆p =√〈(p− 〈p〉)2〉 =

√〈p2〉 − 〈p〉2

This is the Heisenberg Uncertainty Principle.

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7.1 Generalized Uncertainty Principle

For any two Hermitian operators A, B,

∆A∆B ≥ 1

2

∣∣∣〈ψ|[A, B]|ψ〉∣∣∣

where ∆A and ∆B are the standard deviations in A and B; langleψ|[A, B]|ψ〉is the expectation of the commutator.

The above uncertainty relation in x and p therefore results from the com-mutator relation

[p, x] =h

i

8 Various Potentials

So far we have studied several potentials V (x), and we have solved the time-independent Schrodinger Equation

Hψ(x) = Eψ(x)

for some of them, either in lecture or in your homework.In each case we’ve found the eigenstates ψn(x) and the discrete energy

levels En (for bound states) through some quantization condition. For scat-tering states energy levels are continuous, and we’ve also found the relation-ship between the wavefunctions ψ(x) and energy E.

8.1 Particle in a 1D box

V (x) =

{0 for 0 < x < a∞ otherwise

You should be able to solve the time-independent Schrodinger Equationfor a particle in a 1D box and find the following:

Eigenstates:

ψn(x) =

√2

asin(

nπx

a)

Energy levels

En =n2π2h2

2ma2for n = 1, 2, 3, . . .

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Figure 1: Particle in a 1D box, energy spacings increase with n.

8.2 Particle in a 3D box

V (x, y, z) =

{0 for 0 < x < a, 0 < y < b, 0 < z < c∞ otherwise

You should be able to solve the time-independent Shrodinger Equationfor a particle in a 3D box using separation of variables in x, y and z.

Eigenstates:

ψnx,ny ,nz(x, y, z) =

√2

asin

(nπx

a

)√2

bsin

(nπy

b

)√2

csin

(nπz

c

)Energy levels

Enx,ny ,nz =π2h2

2m

(n2

x

a2+n2

y

b2+n2

z

c2

)for nx, ny, nz = 1, 2, 3, . . .

Degeneracies Degeneracy comes about through symmetry in the system.If we introduce symmetry and make our box cubic, such that a = b = c,then degeneracies in energy levels would arise. An energy level is said tohave degeneracy = 2 if two and only twodifferent eigenstates ψnx,ny ,nz andψn′

x,n′y ,n′

zshare the same energy.

Enx,ny ,nz = En′x,n′

y ,n′z

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8.3 Step Potential

See PS2 Problem 2.

Figure 2: The Step Potential, with E > V0.

V (x) =

{0 for x < 0V0 for x > 0

We have considered the scattering states such that E > V0, and we havefound that for a unit incident flux from −∞ there is non-zero reflectionprobability R = |B|2/|A|2 > 0, which is surprising because classically aparticle with enough energy would always go over the barrier. We have alsofound that the momentum of the particle in the R region, p2, is less than itsmomentum p1 in the L region, as expected in classical mechanics.

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8.4 Tunneling through a Barrier

V (x) =

0 for x < 0V0 for 0 < x < a0 for x > a

Figure 3: Tunneling through a barrier, with E < V0.

We have considered the scattering states such that 0 < E < V0, andwe have found that for a unit incident flux from −∞ there is a non-zerotransmission probability T = |c6|2/|c1|2 > 0, which is surprising becauseclassically a particle with not enough energy would always be reflected andwould go back to −∞. Here we’ve demonstrated that the wavefunction isnon-zero in the classically forbidden region, region II, where it has negativekinetic energy, and imaginary momentum. This process is called tunnelling,and is a quantum effect.

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8.5 Other Potentials

Below are some more potentials V (x) that we have merely touched upon inclass.

8.5.1 Harmonic Oscillator

V (x) =1

2kx2

Energy levels

En = hω(n+1

2). for n = 0, 1, 2, . . .

Figure 4: The Harmonic Oscillator. Energy levels are equally spaced.

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8.5.2 Hydrogen Atom

V (r) = −1

r

Energy levels

En = −mee4

8ε20h

2

1

n2. for n = 1, 2, 3, . . .

Figure 5: The Hydrogen Atom. Energy spacings decrease with n.

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