130_notes.dvi

1.14 Harmonic Oscillator Solution with Operators

We can solve the harmonic oscillator problem using operator methods(See section 10). We write
the Hamiltonian in terms of the operator

A≡

(√

mω

2 ̄h

x+i

p

√

2 m ̄hω

)

.

H=

p^2

2 m

+

1

2

mω^2 x^2 = ̄hω(A†A+

1

2

)

We compute thecommutators

[A,A†] =

i

2 ̄h

(−[x,p] + [p,x]) = 1

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

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

If we apply the the commutator [H,A] to the eigenfunctionun, we get [H,A]un=− ̄hωAunwhich
rearranges to the eigenvalue equation

H(Aun) = (En− ̄hω)(Aun).

This says that (Aun) is an eigenfunction ofHwith eigenvalue (En− ̄hω) so itlowers the energy
by ̄hω. Since the energy must be positive for this Hamiltonian, the lowering must stop somewhere,
at the ground state, where we will have
Au 0 = 0.

This allows us to compute theground state energylike this

Hu 0 = ̄hω(A†A+

1

2

)u 0 =

1

2

̄hωu 0

showing that the ground state energy is^12 ̄hω. Similarly,A†raises the energyby ̄hω. We can
travel up and down the energy ladder usingA†andA, always in steps of ̄hω. The energy eigenvalues
are therefore

En=

(

n+

1

2

)

̄hω.

A little more computation shows that

Aun=

√

nun− 1

and that
A†un=

√

n+ 1un+1.

These formulas are useful for all kinds ofcomputationswithin the important harmonic oscillator
system. Bothpandxcan be written in terms ofAandA†.

x=

√

̄h

2 mω

(A+A†)

p=−i

√

m ̄hω

2

(A−A†)