130_notes.dvi

10.2 Commutators ofA,A†andH.

We will use the commutator betweenAandA†to solve the HO problem. The operators are defined
to be

A =

(√

mω

2 ̄h

x+i

p

√

2 m ̄hω

)

A† =

(√

mω

2 ̄h

x−i

p

√

2 m ̄hω

)

.

Thecommutatoris

[A,A†] =

mω

2 ̄h

[x,x] +

1

2 m ̄hω

[p,p]−

i

2 ̄h

[x,p] +

i

2 ̄h

[p,x]

=

i

2 ̄h

(−[x,p] + [p,x]) =

i

̄h

[p,x] = 1.

Lets use this simple commutator
[A,A†] = 1

to computecommutators with the Hamiltonian. This is easy ifHis written in terms ofAand
A†.

[H,A] = ̄hω[A†A,A] = ̄hω[A†,A]A=− ̄hωA

[H,A†] = ̄hω[A†A,A†] = ̄hωA†[A,A†] = ̄hωA†

10.3 Use Commutators to Derive HO Energies

We have computed the commutators

[H,A] = − ̄hωA

[H,A†] = ̄hωA†

Apply [H,A] to the energy eigenfunctionun.

[H,A]un=− ̄hωAun

HAun−AHun=− ̄hωAun

H(Aun)−En(Aun) =− ̄hωAun

H(Aun) = (En− ̄hω)(Aun)

This equation shows thatAunis an eigenfunction ofHwith eigenvalueEn− ̄hω. Therefore,Alowers
the energyby ̄hω.