Solution

In units with

ℏ = 1

H = ω (N + \frac{1}{2}), N = A^{†} A .

The nontrivial commutators among the creation and annihilation operators and

N

are

[A, A^{†}] = 1, [N, A] = - A, [N, A^{†}] = A^{†} .

Their reversed commutators have the opposite signs, while the commutator of any operator with itself is zero.

The operator

N

(A^{†} A)^{†} = A^{†} A .

Its eigenvalues are therefore real. Moreover, for every state

∣ ψ ⟩

⟨ ψ ∣ N ∣ ψ ⟩ = ⟨ A ψ ∣ A ψ ⟩ = ∥ A ∣ ψ ⟩ ∥^{2} \geq 0,

so every eigenvalue is nonnegative.

N ∣ ν ⟩ = ν ∣ ν ⟩

, the commutator with

A

gives

N (A ∣ ν ⟩) = (ν - 1) A ∣ ν ⟩,

and

∥ A ∣ ν ⟩ ∥^{2} = ν ∥ ∣ ν ⟩ ∥^{2} .

Repeated application of

A

lowers the eigenvalue by one. It must terminate before producing a negative eigenvalue. If

m

is the last occupied rung, then

A ∣ ν - m ⟩ = 0

, and the norm identity forces

ν - m = 0

. Hence

ν \in {0, 1, 2, \dots} .

Solved by gpt-5.6-sol high.

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