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 ⎤
2
(dimensionless)
⎢ − ∂ξ 2 + ξ − nω ⎥ ψ = 0
⎣
⎦
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
ψ v (x) = N v H v (ξ)e− ξ
2
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) ˆ
v′ integrals without
* semiclassical: good for pictures, insight, estimates of ψ *vOpψ
∫
solving Schrödinger equation
pE (x) = pclassical (x) = ⎡2
⎣ μ( E − V (x)) ⎦⎤
envelope of ψ(x) in classical region (classical mechanics)
1/4
2
⎞
⎛
⎤
1
1/2 ⎡ 2k / π
ψ(x)
ψ
*
ψdx
∝
,
for
H.
O.
=
2
⎜
⎟
⎢ E − V (x) ⎥
envelope
vv
⎣
⎦
⎟⎠
⎜⎝
velocity
spacing of nodes (quantum mechanics): # nodes between x1 and x2 is
2 x2
pE (x) dx (because λ(x) = h/p(x) and node spacing is λ/2)
h ∫x1
2 x+ (E )
# of levels below E: ∫
“Semi-classical quantization rule”
pE (x) dx
h x− (E )
“Action (h) integral.”
1/2
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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
ˆ aˆ † annihilation/creation or “ladder” or “step-up” operators
specific → a,
* 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 ⎤
x̂ = ⎢
⎥ xp̂,
⎣ μω ⎦
⎡ mf2t −1 ⎤
units = ⎢
⎣ mt −1 ⎦⎥
p̂
p̂ = [ nμω ] p,
units = [ mf2t −1mt −1 ] = mft −1
= p
1/2
1/2
=f
1/4
⎛
⎡ kμ ⎤ x ⎞
1/2
recall
ξ=α
x
=
⎟
⎝⎜
⎣⎢ n 2 ⎥⎦ ⎠
1/2
replace x̂ and p̂ by dimensionless operators
p̂ 2 1 2 nμω 2 k n 2
v
H=
+ kx̂ =
pp̂ +
xp̂
2μ 2
2μ
2 mω
�
�
� �
nω
nω
2
2
nω 2
⎡ pp̂ + xp̂ 2
⎤
⎣
⎦
2
nω
⎡( ipp̂ + xp̂ )( −ipp̂ + xp̂ ) ⎤⎦ ?
=
2 ⎣
↓
↓
=
21/2 aˆ
21/2 aˆ †
factor this?
does this work? No, this attempt at factorization
generates a term i ⎣⎡ p̂p , x̂p ⎤⎦ , which must be subtracted
v = nω ⎛ 2ââ − i ⎣⎡ p̂p , x̂p ⎦⎤⎞
out: H
� � ⎟⎠
2 ⎜⎝
=−i
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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 ⎤⎦
ˆ − x̂pf
ˆ = [ p̂, x̂ ] f ).
Recall [ p̂, x̂ ] = −in . (work this out by p̂xf
What is i ⎡⎣ p̂p , xp̂ ⎤⎦ ?
n ⎤
i ⎣⎡ ppˆ , xp̂ ⎤⎦ = i [ nmω ]−1/2 ⎢⎡
⎣ mω ⎥⎦
= i [n2 ]
−1/2
−1/2
[ p̂, x̂ ]
( −in ) = +1.
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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 ⎞
aa −
v
2⎟
H = nω ⎜
v
⎜
left ⎟
⎝
over ⎠
v is going to turn out to be very useful.
This form for H
* Another trick, what about [ â,â † ] = ?
i
−i
⎡⎣ â,av† ⎤⎦ = ⎡⎣ 2 −1/2 ( ip̂p + xp̂ ) ,2 −1/2 ( −ip̂p + xp̂ ) ⎤⎦ = ⎡⎣ ppˆ , xp̂ ⎤⎦ + ⎡⎣ xp̂, ppˆ ⎤⎦
2
2
1 1
= + = 1.
2 2
1
1
So we have some nice results. Hˆ = nω ⎡⎢ â †aˆ + ⎤⎥ = nω ⎡⎢ ââ † − ⎤⎥
⎣
2⎦
⎣
2⎦
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5.61 Fall, 2013
3.
Lecture #9
Page 5
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
v belonging to
If ψv is an eigenfunction of H
v
v
eigenvalue Ev + nω.
( )
v av†ψ = hω ⎡â †aˆ + 1 ⎤ aˆ
†ψ
H
v
v
⎢⎣
2 ⎥⎦
⎡ ˆ † 1 † ⎤
= hω ⎢â †aâ
+ â ⎥ ψ
v
2 ⎦
⎣
Factor â out front
1 ⎤
⎡
= â †hω ⎢ââ † + ⎥ ψ v
2⎦
⎣
ˆ â † ⎤⎦ + â †â = 1 + â †â
ââ † = ⎡⎣a,
†
v â †ψ = â † hω ⎡â †aˆ +1 + 1 ⎤
ψ
H
v
v
⎢⎣
2 ⎥
⎦


�
�
v
(
)
H +hω
and Ĥψ v = Ev ψ v , thus
v â †ψ = â † ( E + hω ) ψ = ( E + hω ) â †ψ
H
v
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).
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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
dx
Now we bring x̂ and p̂ back.
1/2
⎡
⎤
μω ⎞
−1/2
⎛
⎟⎠ x̂ ⎥ ψ min = 0
⎢i ( 2nμω ) pˆ + ⎜⎝
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 ⎞
= −⎜
⎟⎠ ⎜⎝
⎟ xψ min
⎝
⎠
dx
n
2n
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
2
de− cx
= −2cxe− cx
dx
μω
c=
2n
ψ min = N min e
−
μω 2
x
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.
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5.61 Fall, 2013
Lecture #9
Page 7
Now normalize:
∫
∞
−∞
dx ψ ψ min = 1 = N
�
�

*
min
give factor of
2 in exponent
2
min
∞
−
μω 2
x
n
dx e
∫�

�
– ∞
π1/2
( μω n )1/2
⎛ μω ⎞
ψ min ( x )=⎜
⎝ πn ⎠⎟
1/4 − μω x 2
e 2n
[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
∫
∞
∫
∞
∫
∞
∫
∞
0
0
0
0
2 2
dx e− r x =
2 2
π1/2
2r
dx xe− r x =
1
2r 2
2 2
dx x 2 e− r x =
2 2
π1/2
4r 3
dx x 2n+1e− r x =
n!
2r 2n+2
1⋅ 3⋅ 5·( 2n − 1)
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.
∫
∞
0
1.
2 2
dx x 2n e− r x = π1/2
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
min
2⎠
but âψ min = 0
v ψ = nω ⎛ 0 + 1 ⎞ ψ
so H
⎜⎝
⎟ min
min
2⎠
Emin =
1
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?
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5.61 Fall, 2013
Lecture #9
Page 9
v = nω ⎛ â †â + 1 ⎞ = nω ⎛ ââ † − 1 ⎞
H
⎜⎝
⎟
⎜⎝
⎟
2⎠
2⎠
v 1
v 1
H
H
ˆ
− = â †a,
+ = ââ †
nω 2
nω 2
v 1⎞
⎛ H
1 1⎞
⎛
†
− ⎟ ψ v = ⎜⎝ v + − ⎟⎠ ψ v = â âψ v
⎜⎝
2 2
nω 2 ⎠
â † â ψv = vψv
v.
â † â is “number operator”, N
for ââ † we use the trick
v +1
ââ † = â †aˆ + [ a,
ˆ â † ] = N
� �
+1
Now
∫
dx ψ *vââ †ψ v = ∫ dx â †ψ v
2
because ââ † is Hermitian
Prescription for operating to the left is ψ *v â = ( â*ψ v ) = ( a †ψ v )
*
v + 1 = cv
similarly for dv in âψv = dv ψv−1
∫
∫
dx âψ v = dv
dv = v1/2
2
cv = [ v + 1]1/2
dx ψ *v â †âψ v = v
2
*
Make phase choice and then verify by putting
in x̂ and p̂ .
2
Again, verify phase choice
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5.61 Fall, 2013
Lecture #9
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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 )
μω ⎤
x̂ = ⎡⎢
⎣ n ⎥⎦
−1/2
2 μω ⎤
xp̂ = ⎡⎢
⎣ n ⎥⎦
−1/2
( ↠+ â )
1/2
nμω ⎤
1/2
p̂ = [ nμω ] ppˆ = ⎡⎢
i ( â † − â )
⎣ 2 ⎥⎦
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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
nμω †2
−nμω †2
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
zero when the selection rule is satisfied.
m
revised 9/20/13 2:04 PM
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