5.

61 Fall, 2013 Lecture #9 Page 1

Lecture #9: Harmonic Oscillator:

Creation and Annihilation Operators
Last time

Simplified Schrödinger equation: ξ = α1/2 x, α = (kμ)1/2 n

⎡ ∂2 2E ⎤
⎢ − ∂ξ 2 + ξ − nω ⎥ ψ = 0
2
(dimensionless)
⎣ ⎦
reduced to Hermite differential equation by factoring out asymptotic form of ψ. The asymptotic
ψ is valid as ξ2 → ∞. The exact ψv is

Hermite polynomials
2
ψ v (x) = N v H v (ξ)e− ξ 2
v = 0, 1, 2, … ∞

orthonormal set of basis functions

Ev = nω(v + ½), v = 0, 1, 2, …
even v, even function
odd v, odd function
v = # of internal nodes
what do you expect about Tv ? Vv ? (from classical mechanics)
pictures
* zero-point energy
* tails in non-classical regions
* nodes more closely spaced near x = 0 where classical velocity is largest
* envelope (what is this? maxima of all oscillations)
* semiclassical: good for pictures, insight, estimates of ψ *vOpψ ∫
ˆ v′ integrals without
solving Schrödinger equation
⎣ μ( E − V (x)) ⎦⎤
1/2
pE (x) = pclassical (x) = ⎡2
envelope of ψ(x) in classical region (classical mechanics)
⎛ 1 1/2 ⎡ 2k / π
2
⎤
1/4
⎞
⎜ ψ * ψdx ∝ , ψ(x) envelope
= 2 ⎢ E − V (x) ⎥ for H. O. ⎟
⎜⎝ vv ⎣ ⎦ ⎟⎠
velocity

spacing of nodes (quantum mechanics): # nodes between x1 and x2 is

2 x2
h ∫x1
pE (x) dx (because λ(x) = h/p(x) and node spacing is λ/2)
2 x+ (E )
# of levels below E: ∫ pE (x) dx “Semi-classical quantization rule”
h x− (E )
“Action (h) integral.”

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 2

Non-Lecture
Intensities of Vibrational fundamentals and overtones from
1
μ(x) = μ0 + μ1 x + μ2 x 2 + …
2

∫ dx ψ *v x n ψ v+ m “selection rules”
m = n, n − 2, … −n

Today some amazing results from av† ,â (creation and annihilation operators)

* dimensionless xpˆ, ppˆ → exploit universal aspects of problem — separate universal from
specific → a,
ˆ aˆ † annihilation/creation or “ladder” or “step-up” operators
* integral- and wavefunction-free Quantum Mechanics
* all Ev and ψv for Harmonic Oscillator using â,â †
* values of integrals involving all integer powers of x̂ and/or p̂
* “selection rules”
* integrals evaluated on sight rather than by using integral tables.

1. Create dimensionless x̂p and p̂p operators from x̂ and p̂

1/2
⎡ n ⎤
1/2
⎡ mf2t −1 ⎤ ⎛ ⎡ kμ ⎤ x ⎞
1/4
x̂ = ⎢ ⎥ xp̂, units = ⎢ =f ξ=α =
1/2
⎝⎜
recall x ⎟
⎣ μω ⎦ ⎣ mt −1 ⎦⎥ ⎣⎢ n 2 ⎥⎦ ⎠
p̂ = [ nμω ] p,
p̂ units = [ mf2t −1mt −1 ] = mft −1 = p
1/2 1/2

replace x̂ and p̂ by dimensionless operators

v p̂ 2 1 2 nμω 2 k n 2
H= + kx̂ = pp̂ + xp̂
2μ 2 2μ 2 mω
� �
� � nω
nω
2 2

nω 2
= ⎡ pp̂ + xp̂ 2 ⎤⎦
⎣ factor this?
2
nω
= ⎡( ipp̂ + xp̂ )( −ipp̂ + xp̂ ) ⎤⎦ ?
2 ⎣ does this work? No, this attempt at factorization
↓ ↓ generates a term i ⎣⎡ p̂p , x̂p ⎤⎦ , which must be subtracted
21/2 aˆ 21/2 aˆ †
v = nω ⎛ 2ââ − i ⎣⎡ p̂p , x̂p ⎦⎤⎞
out: H
2 ⎜⎝ � � ⎟⎠
=−i

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 3

aˆ = 2 −1/2 ( xp̂ + ip̂p )

â † = 2 −1/2 ( xp̂ − ip̂p )
xp̂ = 2 −1/2 ( â + â † )
ppˆ = i2 −1/2 ( â † − â )

be careful about ⎡⎣ x,
p̂ p̂p ⎤⎦ ≠ 0

We will find that

âψ v = ( v )1/2 ψ v−1 annihilates one quantum

â †ψ v = ( v + 1)1/2 ψ v+1 creates one quantum
v = nω ( ââ † − 1 / 2 ) = nω ( â †aˆ + 1 / 2 ).
H

This is astonishingly convenient. It presages a form of operator algebra that proceeds without
ever looking at the form of ψ(x) and does not require direct evaluation of integrals of the form

Aij = ∫ dx ψ *i Âψ j .

2. v for the harmonic oscillator.

Now we must go back and repair our attempt to factor H

Instructive examples of operator algebra.

* What is ( ip̂p + xp̂ )( −ip̂p + xp̂ ) ?

p̂p 2 + xp̂ 2 + ip̂p xp̂ − ixp̂p̂p

�
 �
i⎡⎣ p̂p ,xp̂ ⎤⎦

Recall [ p̂, x̂ ] = −in . (work this out by p̂xf ˆ = [ p̂, x̂ ] f ).

ˆ − x̂pf

What is i ⎡⎣ p̂p , xp̂ ⎤⎦ ?

−1/2
n ⎤
i ⎣⎡ ppˆ , xp̂ ⎤⎦ = i [ nmω ]−1/2 ⎢⎡ [ p̂, x̂ ]
⎣ mω ⎥⎦
= i [n2 ]
−1/2
( −in ) = +1.

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 4

v . We have to subtract (1/2)nω:

So we were not quite successful in factoring H

⎛ ˆ v† 1 ⎞
v aa −
H = nω ⎜ 2⎟
v
⎜ left ⎟
⎝ over ⎠

v is going to turn out to be very useful.

This form for H

* Another trick, what about [ â,â † ] = ?

−i
⎡⎣ â,av† ⎤⎦ = ⎡⎣ 2 −1/2 ( ip̂p + xp̂ ) ,2 −1/2 ( −ip̂p + xp̂ ) ⎤⎦ = ⎡⎣ ppˆ , xp̂ ⎤⎦ + ⎡⎣ xp̂, ppˆ ⎤⎦
i
2 2
1 1
= + = 1.
2 2

So we have some nice results. Hˆ = nω ⎡⎢ â †aˆ + ⎤⎥ = nω ⎡⎢ ââ † − ⎤⎥

1 1
⎣ 2⎦ ⎣ 2⎦

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 5

3. Now we will derive some amazing results almost without ever looking at a wavefunction.

v with energy E , then â †ψ is an eigenfunction of H

If ψv is an eigenfunction of H v belonging to
v v

eigenvalue Ev + nω.

H ( )
v av†ψ = hω ⎡â †aˆ + 1 ⎤ aˆ †ψ
v ⎢⎣ 2 ⎥⎦
v

⎡ ˆ † 1 †⎤
= hω ⎢â †aâ + â ⎥ ψ v
†
Factor â out front
⎣ 2 ⎦
⎡ 1⎤
= â †hω ⎢ââ † + ⎥ ψ v
⎣ 2⎦
ââ † = ⎡⎣a,
ˆ â † ⎤⎦ + â †â = 1 + â †â

v â †ψ = â † hω ⎡â †aˆ +1 + 1 ⎤ ψ
H ( )
v ⎢⎣ 2 ⎥⎦
v
�

v
�

H +hω

and Ĥψ v = Ev ψ v , thus

H ( v )
v â †ψ = â † ( E + hω ) ψ = ( E + hω ) â †ψ
v v v v ( )
v with eigenvalue E + nω.
Therefore â †ψv is eigenfunction of H v

v and a new eigenvalue

So every time we apply â † to ψv, we get a new eigenfunction of H
increased by nω from the previous eigenfunction. â † creates one quantum of vibrational
excitation.

Similar result for â ψv.

v ( âψ ) = ( E − nω )( âψ ) .
H v v v

v that belongs to eigenvalue E – nω. â destroys one quantum of

â ψv is eigenfunction of H v
vibrational excitation.

We call â †, â “ladder operators” or creation and annihilation operators (or step-up, step-down).

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 6

Now, suppose I apply â to ψv many times. We know there must be a lowest energy eigenstate
for the harmonic oscillator because Ev ≥ V(0).

We have a ladder and we know there must be a lowest rung on the ladder. If we try to step
below the lowest rung we get
â ψmin = 0
−1/2
2 ⎣⎡ip̂p + xp̂ ⎦⎤ ψ min = 0
d
−in
Now we bring x̂ and p̂ back. dx

⎡ ⎛ μω ⎞
1/2
⎤
⎢i ( 2nμω ) pˆ + ⎜⎝
−1/2
⎟⎠ x̂ ⎥ ψ min = 0
⎣ 2n ⎦
⎡ ⎛ n ⎞ 1/2 d ⎛ μω ⎞ 1/2 ⎤
⎢+ ⎜ ⎟ +⎜ ⎟ x ⎥ ψ min = 0
⎣ ⎝ 2 μω ⎠ dx ⎝ 2n ⎠ ⎦
1/2 1/2
dψ min ⎛ 2 μω ⎞ ⎛ μω ⎞
= −⎜
⎝ n ⎟⎠ ⎜⎝ 2n ⎠⎟
xψ min
dx
μω
=− xψ min .
n
This is a first-order, linear, ordinary differential equation.

What kind of function has a first derivative that is equal to a negative constant times the variable
times the function itself?

2
de− cx 2
= −2cxe− cx
dx
μω
c=
2n
μω 2
− x
ψ min = N min e 2n
.
The lowest vibrational level has eigenfunction, ψmin(x), which is a simple Gaussian, centered at
x = 0, and with tails extending into the classically forbidden E < V(x) regions.

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 7

Now normalize:
μω 2
∞ ∞ − x
∫ dx ψ ψ min = 1 = N ∫� n
* 2
dx e
−∞ �
 � min min
–∞
�

give factor of
2 in exponent π1/2
( μω n )1/2
1/4 − μω x 2
⎛ μω ⎞
ψ min ( x )=⎜ e 2n
⎝ πn ⎠⎟

[recall asymptotic factor of ψ(x): e− ξ

2
/2
]

This is the lowest energy normalized wavefunction. It has zero nodes.

NON-LECTURE
Gaussian integrals

∞ π1/2
∫
2 2
dx e− r x =
0 2r

∞ 1
∫
2 2
dx xe− r x =
0 2r 2

∞ π1/2
∫
2 2
dx x 2 e− r x =
0 4r 3

∞ n!
∫
2 2
dx x 2n+1e− r x =
0 2r 2n+2

∞ 1⋅ 3⋅ 5·( 2n − 1)
∫
2 2
dx x 2n e− r x = π1/2
0 2 n+1 r 2n+1
By inspection, using dimensional analysis, all of these integrals seem OK.
We need to clean up a few loose ends.

1. Could there be several independent ladders built on linearly independent ψ min1 , ψ min2 ?

Assertion: for any 1-D potential it is possible to show that the energy eigenfunctions are
arranged so that the quantum numbers increase in step with the number of internal nodes.

particle in box n = 1, 2, …
# nodes = 0, 1, …, which translates into the general rule
# nodes = n – 1

harmonic oscillator v = 0, 1, 2, …

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 8

# nodes = v

We have found a ψmin that has zero nodes. It must be the lowest energy eigenstate. Call
it v = 0.

2. What is the lowest energy? We know that energy increases in steps of nω.

Ev+n – Ev = nnω.

We get the energy of ψmin by plugging ψmin into the Schrödinger equation.

BUT WE USE A TRICK:

v = nω ⎛ av†aˆ + 1 ⎞
H ⎜⎝ ⎟
2⎠
v ψ = nω ⎛ av†â + 1 ⎞ ψ
H ⎜⎝ ⎟ min
2⎠
min

but âψ min = 0

v ψ = nω ⎛ 0 + 1 ⎞ ψ
so H ⎜⎝ ⎟ min
2⎠
min

1
Emin = nω!
2
Now we also know
Emin +n − Emin = nnω
OR
E 0+v − E0 = vnω, thus Ev = nω(v + 1 / 2)
NON-LECTURE

3. We know

â †ψ v = cvψ v+1
âψ v = dvψ v−1
what are cv and dv?

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 9

v = nω ⎛ â †â + 1 ⎞ = nω ⎛ ââ † − 1 ⎞
H ⎜⎝ ⎟ ⎜⎝ ⎟
2⎠ 2⎠
v 1
H v 1
H
− = â †a,
ˆ + = ââ †
nω 2 nω 2
⎛ Hv 1⎞ ⎛ 1 1⎞
− ⎟ ψ v = ⎜⎝ v + − ⎟⎠ ψ v = â âψ v
†
⎜⎝
nω 2 ⎠ 2 2

â † â ψv = vψv

v.
â † â is “number operator”, N

for ââ † we use the trick

ââ † = â †aˆ + [ a,
ˆ â † ] = N
v +1
� �
+1

∫ dx ψ *vââ †ψ v = ∫ dx â †ψ v
2
Now because ââ † is Hermitian

Prescription for operating to the left is ψ *v â = ( â*ψ v ) = ( a †ψ v )

* *

2
v + 1 = cv
cv = [ v + 1]1/2
similarly for dv in âψv = dv ψv−1

∫ dx ψ *v â †âψ v = v Make phase choice and then verify by putting

in x̂ and p̂ .

∫
2 2
dx âψ v = dv
dv = v1/2 Again, verify phase choice

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 10

â †ψ v = ( v + 1)1/2 ψ v+1
âψ v = ( v )1/2 ψ v−1
v = â †â
N
v ψ = vψ
N v v

[ â, â ] = 1
†

Now we are ready to exploit the av† , aˆ operators.

Suppose we want to look at vibrational transition intensities.

use â,av†

μ(x) = μ0 + μ1 x̂ + μ2 x̂ 2 2 +…
More generally, suppose we want to compute an integral involving some integer power of x̂ (or
p̂ ).

â † = 2 −1/2 ( −ip̂p + xp̂ )

aˆ = 2 −1/2 ( ip̂p + xp̂ )
v = â †â
N (number operator)
xp̂ = 2 −1/2 ( â † + â )
ppˆ = 2 −1/2 i ( aˆ † − â )

−1/2 −1/2

x̂ = ⎡⎢
μω ⎤
xp̂ = ⎡⎢
2 μω ⎤
( ↠+ â )
⎣ n ⎥⎦ ⎣ n ⎥⎦
1/2
nμω ⎤
p̂ = [ nμω ] ppˆ = ⎡⎢ i ( â † − â )
1/2

⎣ 2 ⎥⎦

revised 9/20/13 2:04 PM

5.61 Fall, 2013 Lecture #9 Page 11

n n n
xv2 = ( ↠+ â )( ↠+ â ) = [â†2 + aˆ 2 + â†â + â↠] = [â†2 + aˆ 2 + 2â†â + 1]
2 μω 2 μω 2 μω

2 nμω †2 −nμω †2
p2 = − ( â + aˆ 2 − aˆ †aˆ − â↠) = [â + aˆ 2 − 2aˆ †aˆ − 1]
2 2

etc.
22
v = p + k xv2 = − nω ( â †2 + aˆ 2 − 2â †aˆ − 1) + nω ( â †2 + aˆ 2 + 2â †aˆ + 1) = nω ( â †aˆ + 1 / 2 )
H
2μ 2 4 4

v involving â †2 + aˆ 2 exactly cancel out.

as expected. The terms in H

Look at an ( â † ) ( â )n operator and, from m – n, read off the selection rule for Δv. Integral is not
m

zero when the selection rule is satisfied.

revised 9/20/13 2:04 PM

MIT OpenCourseWare
http://ocw.mit.edu

5.61 Physical Chemistry

Fall 2013

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.