130_notes.dvi

Now, apply [H,A†] to the energy eigenfunctionun.

[H,A†]un= ̄hωA†un

HA†un−A†Hun= ̄hωA†un

H(A†un)−En(A†un) = ̄hω(A†un)

H(A†un) = (En+ ̄hω)(A†un)

A†unis an eigenfunction ofHwith eigenvalueEn+ ̄hω.A†raises the energyby ̄hω.

We cannot keep lowering the energy becausethe HO energy cannot go below zero.

〈ψ|H|ψ〉=

1

2 m

〈p ψ|p ψ〉+

1

2

mω^2 〈x ψ|x ψ〉≥ 0

The only way to stop the lowering operator from taking the energy negative, is for the lowering to
give zero for the wave function. Because this will be at the lowest energy, this must happen for the
ground state.When we lower the ground state, we must get zero.

Au 0 = 0

Since the Hamiltonian containsAin a convenient place, we candeduce the ground state energy.

Hu 0 = ̄hω(A†A+

1

2

)u 0 =

1

2

̄hωu 0

The ground state energy isE 0 =^12 ̄hωand the states in general have energies

E=

(

n+

1

2

)

̄hω

since we have shown raising and lowering in steps of ̄hω. Only a state with energyE 0 =^12 ̄hωcan
stop the lowering so theonly energies allowedare

E=

(

n+

1

2

)

̄hω.

It is interesting to note that we have anumber operatorforn

H =

(

A†A+

1

2

)

̄hω

Nop = A†A

H = (Nop+

1

2

) ̄hω

10.3.1 Raising and Lowering Constants

We know thatA†raises the energy of an eigenstatebut we do not know what coefficient it
produces in front of the new state.
A†un=Cun+1