5.61 Fall 2007 12-15 Lecture Supplement Page
1
We can rearrange the Schrödinger equation for the HO into an interesting form ... with
1
⎢⎣
⎢
⎡ ⎛ !
d ⎞
2 m
⎝⎜ i dx
⎠⎟
2
+ ( m
ω x
)
2 ⎥
⎤
⎥⎦
ψ =
1
2 m
⎡⎣ p 2 + ( m
ω x
)
2 ⎤⎦ ψ = E
ψ
H =
1
2 m
⎣ p 2 + ( m
ω x
)
2
⎦ which has the same form as u 2 + v 2 = ( iu
+ v
)( − iu
+ v
)
.
We now define two operators a
±
≡
1
2 !
m
ω
(
∓ ip
+ m
ω x
) that operate on the test function f(x) to yield
( a
− a
+
) f
( x
)
=
⎝
⎛ 1
2 !
m
ω
( ip
+ m
ω x
)(
− ip
+ m
ω x
)
⎞
⎠ f
( )
=
1
2 !
m
ω
p
2 +
( m
x
) 2
i m
ω ( xp
px )
f ( x )
=
{
1
2 !
m
ω
( p
2 + ( m
ω x
) 2
)
−
2 i
!
[ x , p
}
] f
( ) a
− a
+
=
1
2 !
m
ω
( p
2 + ( m
ω x
) 2
)
+
1
=
2 !
1
ω
H
+
1
2
Which leads to a new form of the Schrödinger equation in terms of a
+
and a
-
…
H
= !
a
− a
+
1
2
If we reverse the order of the operators-a
− a
+
a
+ a
−
-- we obtain …

5.61 Fall 2007 12-15 Lecture Supplement Page
2
H
ψ = !
ω
⎝⎜
⎛ a
+ a
−
+
1 ⎞
2
⎠⎟
ψ or
!
ω
⎝⎜
⎛ a
± a
∓
1 ⎞
2
⎠⎟
ψ
E
ψ and the interesting relation a
− a
+
a
+ a
−
=
[ a a
+
]
= 1
A CLAIM: If ψ satisfies the Schrödinger equation with energy E, then a satisfies it with energy (E+ !
ω ) !
+
ψ
H
( a
+
ψ
)
= !
ω
⎝⎜
⎛ a
+ a
−
+
1
2
⎠⎟
⎞ ( a
+
ψ
)
= !
ω
⎝⎜
⎛ a
+ a
− a
+
+
1 a
2
+ ⎠⎟
⎞
ψ
= !
ω a
+
⎛
⎝⎜ a
− a
+
+
1
2
⎞
⎠⎟
ψ = a
+
⎨ !
ω
= a
+
⎛
⎝⎜ a
+ a
−
+
1
+
1
2
⎞
⎠⎟
ψ
⎫
= a
⎭
+
(
H
+ !
)
= (
E
+ !
)
⎧
!
ω
( a
⎛
⎝⎜
+
) a
+ a
−
+
1
2
⎞
⎠⎟
+ !
ω
⎫
ψ
⎭
H
( a
+
ψ )
=
( E
+ !
ω
)
( a
+
ψ )
Likewise, a
-
ψ satisfies the Schrödinger equation with energy (E!
ω ) …
H
( a
−
ψ
)
= !
ω
⎝⎜
⎛ a
− a
+
−
1
2
⎠⎟
⎞ ( a
−
ψ
)
= !
ω
⎝⎜
⎛ a
− a
+ a
−
−
1
2 a
− ⎠⎟
⎞
ψ = a
−
!
ω
⎝⎜
⎛ a
+ a
−
= a
−
⎧
⎩
!
ω
⎛
⎝⎜ a
− a
+
−
1
−
1
2
⎠⎟
⎭
= a
−
(
H
− !
ω ) ψ = a
−
(
E
− !
ω ) ψ
−
1
2
⎠⎟
⎞
ψ
H
( a
−
ψ
)
= (
E
− !
ω ) ( a
−
ψ
)
So, these are operators connecting states and if we can find one state then we can use them to generate other wavefunctions and energies. In the parlance of the trade the a
± are known as LADDER operators or a
+
= RAISING and a
-
= LOWERING operators .
We know there is a bottom rung on the ladder
0 so that a
−
ψ
0
= 0

5.61 Fall 2007 12-15 Lecture Supplement Page
3
2 !
1 m
ω ⎝⎜
⎛
!
d dx
+ m
ω x
⎠⎟
⎞
ψ
0
=
0
Integrating this equation yields d
ψ
0 dx
= − m
ω
!
x
ψ
0
d
ψ
0
ψ
0
= − m
ω
!
xdx
⇒ ln
ψ
0
= − m
ω
2 h x 2 +
A
0
0
( x
) =
A
0 e
− m ω
2 !
x
2 and E
0
=
1
!
2
E
0
comes from plugging
ψ
0 below. into H
ψ
= E
ψ
. We will perform the normalization
Now that we are firmly planted on the bottom rung of the ladder, we can utilize a
+ repeatedly to obtain other wavefunctions, ψ n
, and energies, E n
. That is,
n
( x
) =
A n
( a
+
) n e
− m
2
ω
!
x 2
,
Thus, for ψ
1 we obtain with E
= n
⎝⎜
⎛ n
+
1
2
⎠⎟
⎞
!
ω
ψ
1
( x
) =
A
1
π !
⎠⎟ m
ω
2 !
x
2
!
⎠⎟ where you still have to determine the normalization constant A
1
.

5.61 Fall 2007 12-15 Lecture Supplement Page
4
Algebraic Normalization of the Wavefunctions: We can perform the normalization algebraically. We know that a
+
ψ n
= c n
ψ n
+
1 a
−
ψ n
= d n
ψ n
−
1
What are the proportionality factors c n
∞
−∞
and d n
? For any functions f ( x )
and g ( x ) f
∗
( a
±
) gdx
=
∞
−∞
( a
∓ f )
∗ g dx here a
∓
is the Hermitian conjugate of a
±
Proof:
∞
−∞ f
∗
( a
± g ) dx
=
1
(2 !
m
ω
)
1 2
∞
−∞ f
∗
⎝⎜
⎛
∓!
d dx
+ m
ω x
⎠⎟
⎞ g dx recall that a
±
=
1
(2 !
m
ω
) 1 2
[
∓ ip + mwx
]
=
1
(2 !
m
ω
) 1 2
⎡
⎢⎣ ∓ i
⎛ !
⎝⎜ d i dx
⎞
⎠⎟
+ mwx
⎤
⎦
⎥
=
1
(2 !
m
ω
) 1 2
⎡
⎣⎢
∓!
d dx
+ mwx
⎤
⎦⎥
Integrate by parts
∫ f
∗
( a
± g ) dx
=
1
(2 !
mw ) dx
= ∫
∞
− ∞
⎡
⎣
⎛
⎝⎜
± !
d dx
+ mwx
⎞
⎠⎟ f
⎤
⎦⎥
∗ gdx
= ∫
∞
− ∞
( a
∓ f )
∗ g dx
So we can write
∫ ∞
−∞
( a
±
ψ n
) * ( a
±
ψ n
) dx
= ∫ ∞
−∞
( a
∓ a
±
ψ n
) *
ψ n dx
We now use
!
ω
⎝⎜
⎛ a
± a
∓
±
1
2
⎠⎟
⎞
ψ n
⎝⎜
⎛ a
+ a
−
= E n
ψ n
± and
E n
=
⎝⎜
⎛ n +
1
2
⎠⎟
⎞
!
w
1
2
⎠⎟
⎞
ψ n
=
⎛
⎝⎜ n
+
1
2
⎠⎟
⎞
ψ n and therefore a
+ a
−
ψ n
= n
ψ n
And
!
ω
⎝⎜
⎛ a
− a
+
−
1
2
⎠⎟
⎞
ψ n
= !
ω
⎝⎜
⎛ n +
1
2
⎠⎟
⎞
ψ n

5.61 Fall 2007 12-15 Lecture Supplement Page
5 a
− a
+
ψ n
= ( n
+
1
) ψ n
We can now calculate c n
:
∫ ∞
−∞
( a
+
ψ n
) * ( a
+
ψ n
) dx
= c n
2
∫ ∞
−∞
ψ * n
+
1
ψ n
+
1 dx
= ∫ ∞
−∞
( a
− a
+
n
)
*
ψ n dx
=
( n
+
1)
∫ ∞
−∞
ψ * n
ψ n dx c n
= n
+
1
The calculation for
∫ ∞
−∞
( d n proceeds in a similar manner : a
−
ψ n
) * ( a
−
ψ n
) dx = d n
2
∫ ∞
−∞
ψ * n
−
1
ψ n
−
1 dx = ∫ ∞
−∞
( a
+ a
−
ψ n
) * ψ n dx = n
∫ ∞
−∞
ψ * n
ψ n dx d n
= n
Thus we obtain the two normalization constants for the a
± a
+
ψ n
= ( n
+
1
) 1 2 ψ n + 1 a
−
ψ n
= n 1 2 ψ n − 1
HO Wavefunctions: Rearranging the equations to a slightly more useful form yields
ψ n + 1
=
1
( n
+
1
)
1 2 a
+
ψ n
ψ n − 1
=
1 n 1 2 a
−
ψ n
We can now use these equations to generate other wavefunctions. Thus, if we start with ψ
0
we obtain: n = 0 n
=
1 n
=
2 n
=
3
ψ
1
=
(
1
0
+
1
)
1 2 a
+
ψ
0
= a
+
ψ
0
ψ
2
ψ
3
=
ψ
4
=
=
1 a
+
ψ
1
=
2
2
1
+ 1 a
+
ψ
2
=
1
2
( a
+
)
2 ψ
0
3
1
⋅ 2 a 3
+
ψ
0
1
3
+
1 a
+
ψ
3
=
1
4
⋅
3
⋅
2 a
+
4 ψ
0
So that
n is
ψ n
=
1 n !
( a
+
) n ψ
0

5.61 Fall 2007 12-15 Lecture Supplement Page
6
Orthogonality of the HO Wavefunctions: Recall the orthogonality condition for two wave functions is
∞
−∞
* n
m dx
=
nm
Using the a
± operators we can show this condition also holds for the HO wavefunctions. The proof is as follows.
ψ m
*
( a
+ a
−
)
ψ n dx
= n
ψ m
* ψ n dx
∫ ( a
−
ψ m
) ∗ ( a
−
ψ m
) dx = ∫ ( a
+ a
−
ψ m
) ∗ ψ n dx = m
∫ ψ m
* ψ n dx
( n
m
)
* m
n dx
=
0
The trivial case occurs when n
= m
; but when n
≠ m
then
ψ m
* ψ n dx
=
0
Potential Energy of the Harmonic Oscillator: We can now use the a
± operators to perform some illustrative calculations. Consider the potential energy associated with the HO. and therefore
V
=
1
2 kx
2 =
1
2 m
ω 2 x
2
V
=
1
2 m
ω 2 x 2 =
1
2 m
ω 2
ψ n
* x ˆ 2 ψ n dx
First, we express x
and p in terms of a
± operators … x ˆ =
⎝⎜
⎛ !
2 m
ω ⎠⎟
⎞
1 2
( a
+
+ a
−
) p ˆ = i
⎝⎜
⎛ !
m
2
ω
⎠⎟
⎞
1 2
( a
+
− a
−
) and x 2 =
⎛ !
⎞
⎝⎜
2 m
ω ⎠⎟
( a
+
+ a
−
)
2 =
⎛
⎝⎜
2
!
m
ω ⎠⎟
⎞ ( a
+ a
+
+ a
+ a
−
+ a
− a
+
+ a
− a
−
)

5.61 Fall 2007 12-15 Lecture Supplement Page
7
Dirac Notation: Before we evaluate this expression let’s introduce some new notation that will make life simpler for us on future occasions. Instead of writing the integral between
we use brackets to denote this integral. The first half is called a “bra” and the second a “ket”. That is, “bra”-c-“ket” notation is bra = and ket = and for the probability density we would have an expression such as where the present of
∫ ∞
−∞
ψ * m
ψ n dx
= m n
* m
is understood. Using this notation, integrals for x , p , and x 2 assume the form
∫ ∞
−∞
ψ * m x ˆ ψ n dx
= m x ˆ n
and ∫ ∞
−∞
ψ * m p
ψ n dx
= m ˆ n
V
=
1
2 m
ω 2 x
2 =
1 m
ω
2
2
ψ n
* x ˆ 2 ψ n dx
=
1 m
ω
2
2 n x ˆ 2 n yielding
V
=
⎛ !
⎞ 1
⎝⎜
2 m
ω ⎠⎟
2 m
ω 2 ⎡⎣ n a
+ a
+ n
+ n a
+ a
− n
+ n a
− a
+ n
+ n a
− a
− n
⎤⎦
V
= h
4
⎡⎣ n
+ n
+
1 ⎤⎦ =
!
2
⎛
⎝⎜ n
+
1
2
⎞
⎠⎟
It is important to not get totally embroiled in the equations and neglect the chemistry and physics. Accordingly, we should ask the question as to the physical significance of this formula ?