algebraic method of solving the linear harmonic oscillator lecture by gable rhodes phys 773: quantum...

Algebraic Method of Solving the Linear Harmonic

Oscillator

Lecture by Gable Rhodes

PHYS 773: Quantum MechanicsFebruary 6th, 2012

Gable Rhodes, February 6th, 2012 2

Simple Harmonic Oscillator• Many physical problems can be modeled as small

oscillations around a stable equilibrium• Potential is described as parabolic around the

minimum energy

• The Hamiltonian is formulated in the usual way for canonical variables q and p.

22

2

1)( qmqV

)(2

2

qVm

pH


Raising & Lowering Operators -

Defined• After substituting in our potential, we can

rearrange the constants slightly to give the Hamiltonian a more suggestive appearance

• If we “factor” the operators, we get

• Recognizing that the last term is the commutator of the canonical variables, we get

22222222

2

1

2

1pqm

mqmp

mH

qppqimipqmipqmm

H 2

1

pqimipqmipqmm

H ,2

1


pqimipqmipqmm

H ,2

1


Defined

• We can now define the operator a

• Upon substitution

• And Simplifying

m

piq

ma

2

pqimaamm

H ,22

1 †

m

piq

ma

2†

pq

iaaH ,

2†



Defined

• If we reverse the order of the operators, a similar expression is obtained

• Subtracting the two forms yields the commutator

• Simplifying to

0,2

,2

††

pq

iaapq

iaaHH

0,†† pqi

aaaa

pqiaa ,, †

pq

iaaH ,

2†

pq

iaaH ,2

†



Defined

• It is important to note that the a, a† operators are defined in such a way that as long as the state variables follow the canonical commutation relation, the a, a† commutator will be 1.

• And the Hamiltonian can be written as a linear function of a, a†

2

1†aaH

pqiaa ,, †

1, †

ii

aa

2

1†aaH


Properties of a, a†

• If a wave function has the property that it is an eigenfunction of the Hamiltonian

• We can introduce the equivalent statement for the operator aa† with generic eigenvalue, λ

• We then apply the a† operator to a†ψ and test if the result is the same eigenvector.

)(2

2

qVm

pH EH

aa†

†††††††† 111 aaaaaaaaaaa

Use commutator


Properties of a, a†

• And the equivalent method for a

• Using the relationship of Hamiltonian, we can then relate eigenvalues

aa†

aaaaaaaaaaa 1111 †††

EH aa†

2

1

2

1 †† aaaaH

2

1†aaEH

2

1

2

1† aaE

2

1E



Properties• a† is called the raising

(or creation) operator.

• And a is the lowering (or annihilating) operator.

• The rungs of the ladder are all evenly separated.

• No degeneracy.

ψ

a† ψ

aψ

aaψ

a† a† ψ

λ

λ+ 1

λ+2

λ-2

λ- 1W

ave

vect

ors

Eig

enva

lues


What is the Significance?• If we have any solution, we can find infinitely more

solutions by repeated application of the raising and lowering operators

• But, importantly, although there are infinite solutions we know that the are are no solutions with negative energy (both kinetic and potential components for the Hamiltonian are always positive)

• Therefore, a ground state must exist. (this is also a property of Sturm-Liouville PDE)

• Applying the lowering operator to the ground state will result in a null vector (trivial state).

• Eq.10-77 0a



Eigenvectors• Once we have a ground state, repeated

application of the raising operator will result in an infinite set of eigenvectors with distinct (non-degenerate) eigenvalues.

• And introducing an arbitrary starting point λ0.

• But for the special case of the ground state

• Must be zero on the right side (by our definition), so λ0 is exactly 0.

0† n

n a

0†

00††† nn

n anaaaaa

000† 0 aa

00 00 nn 0

...2,1,0n


Determine the Energy Levels• Using the previous equation relating the Hamiltonian

and aa†, we can relate energy to λ (n).

• And since we started at the ground state, we can relate the energy level to the eigenvectors

• If we use a normalization constant

• Where An can be found by direct integration at each step or by algebraic tricks (later)

2

1

2

1nE

0† n

n a

2

1nEn

0† n

nn aA


What Are These Operators Good For

Anyway?• We found Energy exactly and wavevectors in

abstract form.• What else can we do with them?• What about expectation values? In chapter 5,

problem 2, we were asked to find <V>. This required direct integration with the (explicitly known) wavefunction.

• Can this be done without knowing the wavefunction?

22

2

1)( qmqV

22

2

1qmV

2*2

2

1qmV


Expectation value of Potential• If we use the definition of a†, a and rewrite q and

p operators in terms of a and a† we get

• Now these can be substituted into <V>

m

piq

ma

2†

m

piq

ma

2

2

2 †aa

mq

2

2 †aaim

mp

2†*22*2

22

1

2

1aa

mmqmV

)(

22

1 ††††* aaaaaaaaV


Expectation value of Potential

• Of the four terms in the integral, we see that two of them vanish due to the orthogonality of the wavevectors.

• And the other two are known to us from the previous work.

2,2**

nnnnaa 02,2*††* nnnnaa

111 ,*†* nnnaa nnnn

nnnaa nnnn ,*†*

)(

22



Expectation value of Potential• This makes the result pretty straightforward

• And the this result agrees with problem 5.2, and the virial theorem

• But, we did not need to know the explicit form of the wavefunction.

10022

1 nnV

nEn

V2

1

2

12

2

1

)(

22



<q>, <p>, <q2>, <p2>• Other key expectation values can be easily

obtained.

2

2 †aa

mq

2

2 †aaim

mp

02

†**

aam

qq

02

†**

aaimm

pp

2

12 nm

q

2†*222*2

2aam

mpp

aaaaaaaa

mp ††††*2

2

2

1100

22 nmnn

mp


Variance and Uncertainty?• The variance is therefore

• And this agrees with the uncertainty principle

2

1222 nm

qqq

2

1222 nmppp

22

1

npq


What is left then?• The Normalization constant, which can also be

determined algebraically.

• Square both sides and integrate

• After integration by parts and throwing out the boundary terms.

• So that gives us our wavefunction in terms of the ground state and raising operator

1†

nn Ba

1,1

2

1

*

1†*†

nnnnnn BBBaa

2*† Baa nn nnnn nnB ,

*211

1† 1 nn na


Normalization• Lets try ψ1.

• Next try ψ2.

• We can now write the general equation

0†

!

1 n

n an

1† 1 nn na

0†

1010 a 0†

1 a

1†

1111 a 0

2†2

!2

1 a


And what is ψ0? • Starting with our condition for the ground state

• And using the definition of the operator

• We get a first order ODE

00 a

m

piq

ma

2

02 00

m

piq

ma

00

mp

iq

qip

00

qm

dq

d


And what is ψ0?• The equation is separable and easily solved.

• After normalization we get

• Which is consistent with the solutions in chapter 5.

00

qm

dq

d

2

0 2exp q

mconst

2

41

0 2exp q

mm


Finding ψ1. • Here we can use the raising operator to generate

further solutions

• Substitute in the operator

• After rearranging, we get the desired result.

0†

!

1 n

n an

0

1†1

!1

1 a

01 2

m

piq

m

01 22

q

m


Finding ψn. • We find that continuing up the ladder is in fact a

generating algorithm for the Hermite polynomials, and the general equation then identical to eq. 5.39 0

†

!

1 n

n an

0!2

1

q

mH

nnnn


Matrix representation of a, a†

• We can show that the lowering operator relates neighboring wave vectors with the normalization factor.

• Adding the bra.

• This gives the matrix elements of a as shown.

1 nn na

nna nnnn 111

00

400

300

200

10

a



• As an example, lowering n=2,nna nnnn 111

20

2

0

0

1

0

1

0

0

00

300

200

10

0

1

0

221 a



• The equivalent representation of the raising operator is derived from the expression

• Adding the bra.

1† 1 nn na

03

002

001

00

†a

11 11†

1 nna nnnn



• With the lowering and raising operators in matrix form, we can then solve for the q and p operators in terms of a, a†.

• For position we get,

2

2 †aa

mq

03

302

201

10

22

2 †

m

aa

mq



• And the equivalent expression for momentum is,

• The complex constant insures that the anti-symmetric matrix is Hermitian.

2

2 †aaim

mp

03

302

201

10

22

2 †

mi

aaim

mp

algebraic method of solving the linear harmonic oscillator lecture by gable rhodes phys 773: quantum...

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