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Theme 2 — Hilbert Spaces and Dirac Notation

Algebra of creation and annihilation operators

Keywords: harmonic oscillator · creation · annihilation · commutator · occupation number · Heisenberg

Exercise 1 : Algebra of creation and annihilation operators

Question 1
Let HH be a Hilbert space. Suppose that two operators aa and a†a^\dagger on HH satisfy the canonical commutation relation [a, a†]=1.[a,\, a^\dagger] = \mathbf{1}.

We define the number operator N=a†aN = a^\dagger a.

Question 2
Show that [N,a]=−a[N, a] = -a and [N,a†]=a†[N, a^\dagger] = a^\dagger.

Solution
Expanding and using [a,a†]=1[a, a^\dagger] = \mathbf{1} gives:

[N,a]=[a†a, a]=a†[a,a]+[a†,a] a=0−1⋅a=−a.[N, a] = [a^\dagger a,\, a] = a^\dagger[a, a] + [a^\dagger, a]\,a = 0 - \mathbf{1}\cdot a = -a.

Similarly:

[N,a†]=[a†a, a†]=a†[a,a†]+[a†,a†] a=a†⋅1+0=a†.[N, a^\dagger] = [a^\dagger a,\, a^\dagger] = a^\dagger [a, a^\dagger] + [a^\dagger, a^\dagger]\,a = a^\dagger \cdot \mathbf{1} + 0 = a^\dagger.

Question 3
Let ∣n⟩|n\rangle be an eigenvector of NN with eigenvalue λ∈C\lambda \in \mathbb{C}, where ∣n⟩≠0|n\rangle \neq 0. Show that if a∣n⟩≠0a|n\rangle \neq 0, then a∣n⟩a|n\rangle is an eigenvector of NN with eigenvalue λ−1\lambda - 1, and that if a†∣n⟩≠0a^\dagger|n\rangle \neq 0, then a†∣n⟩a^\dagger|n\rangle is an eigenvector of NN with eigenvalue λ+1\lambda + 1.

Solution
Suppose that N∣n⟩=λ∣n⟩N|n\rangle = \lambda|n\rangle. Then:

N(a∣n⟩)=[N,a]∣n⟩+aN∣n⟩=−a∣n⟩+λ a∣n⟩=(λ−1) a∣n⟩.N\bigl(a|n\rangle\bigr) = [N,a]|n\rangle + a N|n\rangle = -a|n\rangle + \lambda\, a|n\rangle = (\lambda - 1)\,a|n\rangle.

If a∣n⟩≠0a|n\rangle \neq 0, it is therefore an eigenvector of NN with eigenvalue λ−1\lambda - 1. Similarly:

N(a†∣n⟩)=[N,a†]∣n⟩+a†N∣n⟩=a†∣n⟩+λ a†∣n⟩=(λ+1) a†∣n⟩,N\bigl(a^\dagger|n\rangle\bigr) = [N, a^\dagger]|n\rangle + a^\dagger N|n\rangle = a^\dagger|n\rangle + \lambda\, a^\dagger|n\rangle = (\lambda+1)\,a^\dagger|n\rangle,

so a†∣n⟩a^\dagger|n\rangle, if non-zero, is an eigenvector with eigenvalue λ+1\lambda + 1.

Question 4
Show that λ≥0\lambda \geq 0. Deduce that the spectrum of NN is N\mathbb{N} and that the eigenvalues are integers.

Solution
For every ∣ψ⟩∈H|\psi\rangle \in H:

⟨ψ∣N∣ψ⟩=⟨ψ∣a†a∣ψ⟩=∥a∣ψ⟩∥2≥0,\langle \psi | N | \psi \rangle = \langle \psi | a^\dagger a | \psi \rangle = \| a|\psi\rangle \|^2 \geq 0,

so NN is a positive operator and every eigenvalue λ\lambda satisfies λ≥0\lambda \geq 0.

Suppose that λ∉N\lambda \notin \mathbb{N}. Repeatedly applying aa produces the sequence of eigenvalues λ,λ−1,λ−2,...\lambda, \lambda-1, \lambda-2, ..., which eventually becomes strictly negative, contradicting λ≥0\lambda \geq 0. Thus the spectrum is contained in N\mathbb{N}; by applying a†a^\dagger starting from ∣0⟩|0\rangle (an eigenvector with eigenvalue 00, whose existence is guaranteed by the positivity argument), all the integers are constructed.

Question 5
Show that

∥a∣n⟩∥2=n,∥a†∣n⟩∥2=n+1.\|a|n\rangle\|^2 = n, \qquad \|a^\dagger|n\rangle\|^2 = n+1.

Hence, choosing the phases so that the coefficients are real and positive, deduce that:

a∣n⟩=n ∣n−1⟩,a†∣n⟩=n+1 ∣n+1⟩.a|n\rangle = \sqrt{n}\,|n-1\rangle, \qquad a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle.

Solution
Using [a,a†]=1[a, a^\dagger] = \mathbf{1} and N=a†aN = a^\dagger a gives:

∥a∣n⟩∥2=⟨n∣a†a∣n⟩=⟨n∣N∣n⟩=n.\|a|n\rangle\|^2 = \langle n | a^\dagger a | n \rangle = \langle n | N | n \rangle = n.

For a†a^\dagger:

∥a†∣n⟩∥2=⟨n∣a a†∣n⟩=⟨n∣(a†a+1)∣n⟩=n+1.\|a^\dagger|n\rangle\|^2 = \langle n | a\, a^\dagger | n \rangle = \langle n | (a^\dagger a + \mathbf{1}) | n \rangle = n + 1.

Since a∣n⟩a|n\rangle is proportional to ∣n−1⟩|n-1\rangle (an eigenvector with eigenvalue n−1n-1, unique up to a phase in an irreducible representation), the phase can be chosen so that:

a∣n⟩=n ∣n−1⟩,a†∣n⟩=n+1 ∣n+1⟩.a|n\rangle = \sqrt{n}\,|n-1\rangle, \qquad a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle.