130_notes.dvi

Since we are dealing with harmonic oscillators, we want to find the analog of theraising and
lowering operators. We developed the raising and lowering operators by trying to write the
Hamiltonian asH=A†A ̄hω. Following the same idea, we get

ak,α =

1

√

2 ̄hω

(ωQk,α+iPk,α)

a†k,α =

1

√

2 ̄hω

(ωQk,α−iPk,α)

a†k,αak,α =

1

2 ̄hω

(ωQk,α−iPk,α)(ωQk,α+iPk,α)

=

1

2 ̄hω

(ω^2 Q^2 k,α+Pk,α^2 +iωQk,αPk,α−iωPk,αQk,α)

=

1

2 ̄hω

(ω^2 Q^2 k,α+Pk,α^2 +iωQk,αPk,α−iω(Qk,αPk,α+

̄h

i

))

=

1

2 ̄hω

(ω^2 Q^2 k,α+Pk,α^2 − ̄hω)

a†k,αak,α+

1

2

=

1

2 ̄hω

(ω^2 Q^2 k,α+Pk,α^2 )

(

a†k,αak,α+

1

2

)

̄hω =

1

2

(ω^2 Q^2 k,α+Pk,α^2 ) =H

H=

(

a†k,αak,α+

1

2

)

̄hω

This is just thesame as the Hamiltonian that we had for the one dimensional harmonic
oscillator. We therefore have the raising and lowering operators, as long as [ak,α,a†k,α] = 1, as we
had for the 1D harmonic oscillator.

[

ak,α,a†k,α

]

= [

1

√

2 ̄hω

(ωQk,α+iPk,α),

1

√

2 ̄hω

(ωQk,α−iPk,α)]

=

1

2 ̄hω

[ωQk,α+iPk,α,ωQk,α−iPk,α]

=

1

2 ̄hω

(−iω[Qk,α,Pk,α] +iω[Pk,α,Qk,α])

=

1

2 ̄hω

( ̄hω+ ̄hω)

= 1

So these are definitely the raising and lowering operators. Of course the commutator would be zero
if the operators were not for the same oscillator.

[ak,α,a†k′,α′] =δkk′δαα′

(Note that all of our commutators are assumed to be taken at equal time.) The Hamiltonian is
written in termsaanda†in the same way as for the 1D harmonic oscillator. Therefore, everything
we know about the raising and lowering operators (See section 10.3) applies here, including the
commutator with the Hamiltonian, the raising and lowering of energy eigenstates, and even the