Algebraic Method of Solving the
Linear Harmonic Oscillator
Lecture by Gable Rhodes
PHYS 773: Quantum Mechanics
February 6th, 2012
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
V (q) 
1
m 2 q 2
2
• The Hamiltonian is formulated in the usual way for
canonical variables q and p.
p2
H
 V (q)
2m
Gable Rhodes, February 6th, 2012
2
Raising & Lowering Operators - Defined
• After substituting in our potential, we can rearrange
the constants slightly to give the Hamiltonian a
more suggestive appearance
H



1
1
p 2  m 2 2 q 2 
m 2 2 q 2  p 2
2m
2m

• If we “factor” the operators, we get
H
1
mq  ip mq  ip   im  pq  qp 
2m
• Recognizing that the last term is the commutator of
the canonical variables, we get
H
1
mq  ip mq  ip   imq, p 
2m
Gable Rhodes, February 6th, 2012
3
Raising & Lowering Operators - Defined
1
mq  ip mq  ip   imq, p 
2m
H
• We can now define the operator a
a
m 
p 
q  i

2 
m 
a† 
m 
p 
q i

2 
m 
• Upon substitution


1
H
2maa †  im q, p 
2m
• And Simplifying
 † i

H   aa  q, p 
2


Gable Rhodes, February 6th, 2012
4
Raising & Lowering Operators - Defined
• If we reverse the order of the operators, a similar
expression is obtained
 † i

H   a a q, p 
2


 † i

H   aa  q, p 
2


• Subtracting the two forms yields the commutator
i
i




H  H   aa †  q, p    a † a  q, p   0
2
2




• Simplifying to


i
aa  a a  q, p   0

†
†
Gable Rhodes, February 6th, 2012
a, a    i q, p
†
5
Raising & Lowering Operators - 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.
a, a    i q, p
†
a, a    i i  1
†
• And the Hamiltonian can be written as a linear
function of a, a†
1
 †
H   a a  
2

Gable Rhodes, February 6th, 2012
 † 1
H   aa  
2

6
Properties of a, a†
• If a wave function has the property that it is an
eigenfunction
of the Hamiltonian
2
H
p
 V (q)
2m
H  E
• We can introduce the equivalent statement for the
operator aa† with generic eigenvalue, λ
a † a  
• We then apply the a† operator to a†ψ and test if the
result is the same eigenvector.
 




a † a a †  a † aa †  a † 1  a † a   a † 1     1   a †
Use commutator
Gable Rhodes, February 6th, 2012
7
Properties of a, a†
• And the equivalent method for a



a † a  

a † aa   aa †  1 a   a a † a  1   a  1    1a
• Using the relationship of Hamiltonian, we can then
relate eigenvalues
H  E
a † a  
1
 † 1
 †
H    aa      a a  
2
2


1
 †
H  E    a a  
2

1
1
 †

E    a a        
2
2


1

E      
2

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Raising & Lowering Operators -Properties
• a† is called the raising
(or creation) operator.
a† a† ψ
• And a is the lowering (or
annihilating) operator.
• The rungs of the ladder
are all evenly
separated.
• No degeneracy.
Gable Rhodes, February 6th, 2012
λ+2
λ+ 1
a† ψ
ψ
aψ
aaψ
λ
λ- 1
λ-2
9
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
a  0
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10
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 (nondegenerate) eigenvalues.
 n  a

n  0,1,2...
† n
0
• And introducing an arbitrary starting point λ0.
 
a an  a a a
†
†
† n
0
 
 0  n  a
† n
0
• But for the special case of the ground state
a † a  0  0  0   0
• Must be zero on the right side (by our definition), so
λ0 is exactly 0.
0  0  0
0
Gable Rhodes, February 6th, 2012
  0  n   n
11
Determine the Energy Levels
• Using the previous equation relating the Hamiltonian and
aa†, we can relate energy to λ (n).
1
1


E          n  
2
2


• And since we started at the ground state, we can relate
the energy level to the eigenvectors
1

† n
E



n



 n  a   0
n
2


• If we use a normalization constant
 n  An a

† n
0
• Where An can be found by direct integration at each
step or by algebraic tricks (later)
Gable Rhodes, February 6th, 2012
12
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.
1
V (q) 
1
m 2 q 2
2
V 
2
m 2 q 2
• Can this be done without knowing the
wavefunction?
1
2
* 2
V 
Gable Rhodes, February 6th, 2012
2
m  q 
13
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
a
m 
p 
q  i

2 
m 
2  a  a † 


q
m  2 
a† 
m 
p 
q i

2 
m 
 a  a† 
2

mi 
p
m
 2 
• Now these can be substituted into <V>


1
1

2
* 2
2
*
† 2
V  m  q   m
 aa 
2
2
2m 
1 
*
† †
†
†
V 

(
aa

a
a

aa

a
a )
2 2 
Gable Rhodes, February 6th, 2012
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Expectation value of Potential
V 
1 
*
† †
†
†

(
aa

a
a

aa

a
a )

2 2
• Of the four terms in the integral, we see that two of
them vanish due to the orthogonality of the
wavevectors.
*
*

aa



n

  n 2   n,n 2
*
* † †

a
a



n

  n  2   n,n  2  0
• And the other two are known to us from the
previous work.
*
*
†

aa



n

 n  1 n  n  1 n,n  n  1
*
* †

a
a



n

 n n  n n,n  n
Gable Rhodes, February 6th, 2012
15
Expectation value of Potential
• This makes the result pretty straightforward
V 
1 
 * (aa  a † a †  aa †  a † a)

2 2
V 
1 
0  0  n  n  1
2 2
V 
1  2n  1  1
 
  En
2  2  2
• 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.
Gable Rhodes, February 6th, 2012
16
<q>, <p>, <q2>, <p2>
• Other key expectation values can be easily
obtained.
 a  a† 
2
2  a  a † 
q
q
2

m 
2


p

m
mi 


*
†

a

a
 0

2m
p   * p  

mi  * a  a †   0
2m

p 2   * p 2 
Gable Rhodes, February 6th, 2012


 

2

 m 2 2  * a  a † 
2m
m
*
† †
†
†

aa

a
a

aa

a
a

2
m
1

0  0  n  n  1  m  n  

2
2

p2  
p2
2

q   *q  
 
1

n  
m 
2




17
Variance and Uncertainty?
• The variance is therefore
 q2  q 2  q 
2
 
1
n



m 
2


1
2
 p2  p 2  p  m  n  
2
• And this agrees with the uncertainty principle
Gable Rhodes, February 6th, 2012
18
What is left then?
• The Normalization constant, which can also be
determined algebraically.
a †  n  B  n 1
• Square both sides and integrate
 a
†
n
a
*
†
 n   B  n 1
 B
*
 B  n 1,n 1
2
n 1
• After integration by parts and throwing out the
boundary terms.
 aa
†
n

*
n
 B
2
B   n  1  n
2

*
n
 n  1 n ,n
• So that gives us our wavefunction in terms of the
ground state and raising operator
a †  n  n  1  n 1
Gable Rhodes, February 6th, 2012
19
Normalization
a †  n  n  1  n 1
• Lets try ψ1.
0  1  01  a †  0
• Next try ψ2.
1  1  11  a †  1
 1  a†  0
2 
 
1 †2
a 0
2!
• We can now write the general equation
n 
Gable Rhodes, February 6th, 2012
 
1 †n
a 0
n!
20
And what is ψ0?
• Starting with our condition for the ground state
a0  0
• And using the definition of the operator
m 
p 
a
q  i

2 
m 
p
 
i q
• We get a first order ODE
a0 
m 
p 
q  i
0  0
2 
m 
p 

q

i

0  0
m 

Gable Rhodes, February 6th, 2012
d 0
m

q 0
dq

21
And what is ψ0?
• The equation is separable and easily solved.
d 0
m

q 0
dq

 m 2 
q 
 2 
 0  constexp 
• After normalization we get
1
 m  4
 m 2 
0  
exp
q 


  
 2 
• Which is consistent with the solutions in chapter 5.
Gable Rhodes, February 6th, 2012
22
Finding ψ1.
• Here we can use the raising operator to generate
further solutions
n
1
1
 
1 †1

a 0
1!
n 
a  
†
n!
0
• Substitute in the operator
 m 
p 
q

i

   0
m  
 2 
 1  
• After rearranging, we get the desired result.
 m

2q   0
 2

 1  
Gable Rhodes, February 6th, 2012
23
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
n 
n
 
1 †n
a 0
n!
 m 

H n 
q   0
n
2 n!   
1
Gable Rhodes, February 6th, 2012
24
Matrix representation of a, a†
• We can show that the lowering operator relates
neighboring wave vectors with the normalization
factor.
a  n  n  n 1
• Adding the bra.
 n 1 a  n   n 1 n  n 1  n
• This gives the matrix elements of a as shown.
0

0
a




1
0
0
Gable Rhodes, February 6th, 2012
2
0
0
3
0
0





4
0 
25
Matrix representation of a, a†
• As an example, lowering n=2,
1 a  2  2
0
0  
1  0
 
0  
 
 
 
1
0
0

2
0
0

Gable Rhodes, February 6th, 2012
3
0

 n 1 a  n   n 1 n  n 1  n

 0  0  
    
0 1 






 1 0  
    
   


0 
2 
 2

0

 
26
Matrix representation of a, a†
• The equivalent representation of the raising
operator is derived from the expression
a †  n  n  1  n 1
• Adding the bra.
 n 1 a †  n   n 1 n  1  n 1  n  1
0

 1
a†  




0
0
2
Gable Rhodes, February 6th, 2012
0
0
3
0
0







 
27
Matrix representation of a, a†
• 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  a  a † 


q
m  2 
0

1
2  a  a † 
 

 
q
m  2 
2m 



Gable Rhodes, February 6th, 2012
1
0
2
2
0
3
3
0







 
28
Matrix representation of a, a†
• And the equivalent expression for momentum is,
 a  a† 
2

mi 
p
m
 2 
 0

 1

†


2
mi  a  a   i m 
p
m
2 
 2 



1
0
 2
2
0
 3
3
0






 
 
• The complex constant insures that the antisymmetric matrix is Hermitian.
Gable Rhodes, February 6th, 2012
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