Algebraic Method of Solving the Linear Harmonic Oscillator Lecture by Gable Rhodes

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

21)( qmqV

)(2

2

qVmpH


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

21

21 pqm

mqmp

mH

qppqimipqmipqmm

H 21

pqimipqmipqmm

H ,21


pqimipqmipqmm

H ,21


• We can now define the operator a

• Upon substitution

• And Simplifying

mpiqma

2

pqimaamm

H ,221 †

mpiqma

2†

pqiaaH ,

2†



• 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

††

pqiaapqiaaHH

0,†† pqiaaaa

pqiaa ,, †

pqiaaH ,

2†

pqiaaH ,

2†



• 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†

21†aaH

pqiaa ,, †

1, †

iiaa

21†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

qVmpH 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†

21

21 †† aaaaH

21†aaEH

21

21† aaE

21E


Raising & Lowering Operators -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

Eige

nval

ues


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


Raising & Lowering Operators -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†††

nnn 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)

21

21 nE

0† n

n a

21nEn

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

21)( qmqV

22

21 qmV

2*2

21 qmV


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>

mpiqma

2†

mpiqma

2

22 †aam

q

22 †aaimm

p

2†*22*2

221

21 aa

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

nEnV

21

212

21

)(22



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

obtained.

22 †aam

q

22 †aaimm

p

02

†**

aam

qq

02

†**

aaimm

pp

212 n

mq

2†*222*2

2aam

mpp

aaaaaaaamp ††††*2

2

21100

22 nmnnmp


Variance and Uncertainty?• The variance is therefore

• And this agrees with the uncertainty principle

21222 n

mqqq

21222 nmppp

221

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

1,12

1*

1†*†

nnnnnn BBBaa

2*† Baa nn nnnn nnB ,*2 11

1† 1 nn na


Normalization• Lets try ψ1.

• Next try ψ2.

• We can now write the general equation

0†

!1

nn a

n

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

mpiqma

2

02 00

mpiqma

00

mpiq

qip

00 qm

dqd


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

dqd

2

0 2exp qmconst

2

41

0 2exp qmm


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

nn a

n

01†

1 !11 a

01 2

mpiqm

01 22

qm


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

nn a

n

0!2

1

qmH

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

00400

300200

10

a



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

2020

010

100

00300

20010

010

221 a



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

• Adding the bra.

1† 1 nn na

03

002001

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,

22 †aam

q

03302

20110

222 †

maa

mq



• And the equivalent expression for momentum is,

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

22 †aaimm

p

03302

20110

222 †

miaaimm

p

Algebraic Method of Solving the Linear Harmonic Oscillator Lecture by Gable Rhodes

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Transcript of Algebraic Method of Solving the Linear Harmonic Oscillator Lecture by Gable Rhodes