Harmonic Oscillator ABCs

Consider the harmonic oscillator given by

Define

a) Show that [a, a^†] =1.

b) Show that H = a^†a + 1/2.

c) Show that [a^†a, H] = 0.

d) Show that if |n⟩ is an eigenstate of N = a^†a with eigenvalue n:

then a^† |n⟩ and a |n⟩ are also eigenstates of N with eigenvalues n + 1 and n – 1, respectively.

e) Define |0⟩ such that a |0⟩ = 0 What is the energy eigenvalue of |0⟩?

f) How can one construct other eigenstates of H starting from |0⟩?

g) What is the energy spectrum of H? Are negative eigenvalues possible?

SOLUTION

a) Here we took m = 1, h = 1, ω = 1.

(1)

since the commutator

b)

c)

(2)

d) In order to demonstrate that a^† |n⟩ and a |n⟩ are also eigenstates of N, compose the commutator

(3)

by (1). Similarly,

(4)

Now,

(5)

Substituting (4) into (5) and replacing N |n⟩ by n |n⟩ we have

(6)

Rearranging (6) yields

(7)

as required. A similar calculation gives

(8)

We see from the above results that the application of the operator a^† on a state |n⟩ has the effect of “raising” the state by 1, and the operator a  lowers the state by 1 (see (f) below).

e)

since, by assumption, a |0⟩ = 0.

f) Since by (c), the number operator N = a^†a and the Hamiltonian H = a^†a + 1/2 commute, they have simultaneous eigenstates. Starting with |0⟩, we may generate a number state whose energy eigenvalue is 1 + 1/2 by applying the raising operator a^†. Applying a^† again produces a state of eigenvalue 2+1/2. What remains to be done is to see that these eigenstates (number, energy) are properly normalized. If we assume that the state |n⟩ is normalized, then we may compose the inner product

(9)

Up to an arbitrary phase, we see that

Starting with the vacuum ket |0⟩, we can write an energy eigenket |n⟩:

g) The energy spectrum is n + 1/2, where takes all positive integer values and zero. From (9) and the fact that the norm of the eigenvectors is positive (actually, 1), we see that n cannot be negative, and so no negative eigenvalues are possible.