 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2021/2/a/ii/solution

[A B, C] = A [B, C] + [A, C] B

[N, a_{j}^{*}] = \sum_{k = 1}^{2} (a_{k}^{*} [a_{k}, a_{j}^{*}] + [a_{k}^{*}, a_{j}^{*}] a_{k}) = a_{j}^{*} .

Repeated use of the commutator product rule gives, for

n \geq 1

[N, (a_{j}^{*})^{n}] = \sum_{r = 0}^{n - 1} (a_{j}^{*})^{r} [N, a_{j}^{*}] (a_{j}^{*})^{n - 1 - r} = n (a_{j}^{*})^{n} .

n = 0

, the commutator with the identity operator is zero, agreeing with the formula.

Each

a_{k}^{*} a_{k}

is unchanged by taking the adjoint operator, so

N^{*} = N

[N, (a_{j}^{*})^{n}]^{*} = [a_{j}^{n}, N] = n a_{j}^{n}

[N, (a_{j}^{*})^{n}] = n (a_{j}^{*})^{n}, [N, a_{j}^{n}] = - n a_{j}^{n}, n = 0, 1, 2, \dots .

In particular, each creation operator raises the number operator eigenvalue by one, and each annihilation operator lowers it by one.

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