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Theme 2 — Hilbert Spaces and Dirac Notation

Truncation of the harmonic oscillator

Keywords: harmonic oscillator · truncation · Heisenberg algebra · commutator · finite dimension

Exercise 1 : Truncation of the harmonic oscillator

Question 1
We use the notation from the exercise on the creation and annihilation operators. Let HNH_N denote the NN-dimensional subspace spanned by {∣0⟩,∣1⟩,...,∣N−1⟩}\{|0\rangle, |1\rangle, ..., |N-1\rangle\}. We define the truncated operators aNa_N and aN†a_N^\dagger as the restrictions of aa and a†a^\dagger to HNH_N, with the convention aN∣0⟩=0a_N|0\rangle = 0 and aN†∣N−1⟩=0a_N^\dagger|N-1\rangle = 0.
Question 2
Write the matrices of aNa_N and aN†a_N^\dagger in the basis {∣0⟩,...,∣N−1⟩}\{|0\rangle,...,|N-1\rangle\} for N=4N = 4.

Solution
Using a∣n⟩=n∣n−1⟩a|n\rangle = \sqrt{n}|n-1\rangle and a†∣n⟩=n+1∣n+1⟩a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle (with vanishing action at the boundaries):

a4=(0100002000030000),a4†=(0000100002000030).\begin{aligned} a_4 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{pmatrix}, \qquad a_4^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & \sqrt{2} & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix}. \end{aligned}

Question 3
Verify that aN†=(aN)†a_N^\dagger = (a_N)^\dagger in the Hilbert-space sense on HNH_N.

Solution
By the result of the exercise on adjoints and projectors, (A†)ij=Aji∗(A^\dagger)_{ij} = A_{ji}^*. The two matrices obtained in the previous question are conjugate transposes of one another (their entries are real), so a4†=(a4)†a_4^\dagger = (a_4)^\dagger.

Question 4
Compute [aN,aN†][a_N, a_N^\dagger] and show that this commutator is no longer equal to 1HN\mathbf{1}_{H_N}. Interpret the result.

Solution
We compute:

[a4,a4†]=a4a4†−a4†a4.[a_4, a_4^\dagger] = a_4 a_4^\dagger - a_4^\dagger a_4.

A direct matrix calculation gives:

a4a4†=(1000020000300003),a4†a4=(0000010000200003),\begin{aligned} a_4 a_4^\dagger = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}, \qquad a_4^\dagger a_4 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}, \end{aligned}

and hence:

[a4,a4†]=(1000010000100000)=1H4−∣3⟩⟨3∣.\begin{aligned} [a_4, a_4^\dagger] = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \mathbf{1}_{H_4} - |3\rangle\langle 3|. \end{aligned}

The canonical commutation relation [a,a†]=1[a,a^\dagger]=\mathbf{1} is violated at the final basis state: the truncation breaks the Heisenberg algebra, because aN†∣N−1⟩=0a_N^\dagger|N-1\rangle = 0 while the space is “too small” to accommodate ∣N⟩|N\rangle.