Past exam of the quantum theory course of the University of Oxford / 2021 / 2 / a / ii / Solution

First apply the commutator product rule in the form $[AB,C]=A[B,C]+[A,C]B$:Repeated use of the commutator product rule gives, for $n\ge 1$,This is the commutator with powers of an eigenoperator. For $n=0$, the commutator with the identity operator is zero, agreeing with the formula.

Each $a_k^{\ast }{a}_k$ is unchanged by taking the adjoint operator, so $N^{\ast }=N$ in this algebra. Taking the adjoint of a commutator yields $[N,(a_j^{\ast }{)}^n{]}^{\ast }=[a_j^n,N]=n{a}_j^n$. By antisymmetry,In particular, each creation operator raises the number operator eigenvalue by one, and each annihilation operator lowers it by one.