Harmonic Oscillator Energies and Wavefunctions
via Raising and Lowering Operators
We can rearrange the Schrödinger equation for the HO into an interesting form ...
1 ⎡⎛ ! d ⎞ 2⎤
2
1
⎢⎜ ⎟ + ( mω x ) ⎥ ψ = ⎡ p 2 + ( mω x )2 ⎤ ψ = Eψ
2m ⎢⎣⎝ i dx ⎠ ⎥⎦ 2m ⎣ ⎦
with
1
H= ⎡ p 2 + ( mω x )2 ⎤
2m ⎣ ⎦
which has the same form as
u 2 + v 2 = ( iu + v )( −iu + v ) .
We now define two operators
1
a± ≡ ( ∓ip + mω x )
2!mω
that operate on the test function f(x) to yield
( a a ) f ( x ) = ⎛⎝ ( ip + mω x )( −ip + mω x )⎞ f ( x )
1
− +
2!mω ⎠
1
⎡ p 2 + ( mω x ) − imω (xp − px) ⎤ f (x)
2
=
2!mω ⎣ ⎦
= { 1
2!mω
( i
)
p 2 + ( mω x ) − [ x, p ] f ( x )
2
2! }
a− a+ =
1
2!mω
( 1
p 2 + ( mω x ) + =
2 1
2 !ω
H+ )
1
2
Which leads to a new form of the Schrödinger equation in terms of a+ and a- …
⎛ 1⎞
H ψ = !ω ⎜ a− a+ − ⎟ ψ
⎝ 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
⎛ 1⎞
H ψ = !ω ⎜ a+ a− + ⎟ ψ
⎝ 2⎠
or
⎛ 1⎞
!ω ⎜ a± a∓ ± ⎟ ψ = Eψ
⎝ 2⎠
and the interesting relation
a− a+ − a+ a = [ a a ] = 1
− − +
A CLAIM: If ψ satisfies the Schrödinger equation with energy E, then a+ψ
satisfies it with energy (E+!ω ) !
⎛ 1⎞ ⎛ 1 ⎞
H ( a + ψ ) = ! ω ⎜ a + a − + ⎟ ( a + ψ ) = !ω ⎜ a + a − a + + a + ⎟ ψ
⎝ 2⎠ ⎝ 2 ⎠
⎛ 1⎞ ⎧ ⎛ 1⎞ ⎫ ⎧ ⎛ 1⎞ ⎫
= !ω a+ ⎜ a− a+ + ⎟ ψ = a+ ⎨!ω ⎜ a+ a− + 1 + ⎟⎠ ψ ⎬ = a+ ⎨!ω ⎜⎝ a+ a− + ⎟⎠ + !ω ⎬ψ
⎝ 2⎠ ⎩ ⎝ 2 ⎭ ⎩ 2 ⎭
= a+ ( H + !ω )ψ = ( E + !ω )( a+ψ )
H ( a+ψ ) = (E + !ω ) ( a+ψ )
Likewise, a-ψ satisfies the Schrödinger equation with energy (E-!ω ) …
⎛ 1⎞ ⎛ 1 ⎞ ⎛ 1⎞
H ( a−ψ ) = !ω ⎜ a− a+ − ⎟ ( a−ψ ) = !ω ⎜ a− a+ a− − a− ⎟ ψ = a !ω ⎜ a+ a− − ⎟ ψ
⎝ 2⎠ ⎝ 2 ⎠ ⎝ −
2⎠
⎧ ⎛ 1⎞ ⎫
= a− ⎨!ω ⎜ a− a+ − 1 − ⎟ ψ ⎬ = a− ( H − !ω )ψ = a− ( E − !ω )ψ
⎩ ⎝ 2⎠ ⎭
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 ψ0 so that
a−ψ 0 = 0