Boson operator algebra - unitary transformation

Question

What is the simple way to evaluate

$$ e^{i \alpha n(n-1)}a^{\dagger} e^{-i \alpha n(n-1)}$$

and

$$ e^{i \alpha n(n-1)}a e^{-i \alpha n(n-1)}$$

where $n=a^\dagger a$ and $a$ and $a^\dagger$ are the boson annihilation/creation operators for bosons and $\alpha$ is a constant?

Answer 1

You can evaluate them on states with definite particle number. Let $|k\rangle$ be a state such that $n|k\rangle=k|k\rangle$. Then

\begin{eqnarray*} \mathrm e^{\mathrm i \alpha n(n-1)}a^\dagger\mathrm e^{-\mathrm i \alpha n(n-1)}|k\rangle &=& \mathrm e^{\mathrm i \alpha n(n-1)}a^\dagger\mathrm e^{-\mathrm i \alpha k(k-1)}|k\rangle \\ &=& \mathrm e^{\mathrm i \alpha (k+1)k}a^\dagger\mathrm e^{-\mathrm i \alpha k(k-1)}|k\rangle \\ &=& \mathrm e^{2\mathrm i \alpha k}a^\dagger|k\rangle \\ &=& a^\dagger\mathrm e^{2\mathrm i \alpha n}|k\rangle\;. \end{eqnarray*}

Since $|k\rangle$ was arbitrary and the states with definite particle number span the state space, this implies

$$ \mathrm e^{\mathrm i \alpha n(n-1)}a^\dagger\mathrm e^{-\mathrm i \alpha n(n-1)}=a^\dagger\mathrm e^{2\mathrm i \alpha n}\;. $$

Analogously, you can obtain

$$ \mathrm e^{\mathrm i \alpha n(n-1)}a\mathrm e^{-\mathrm i \alpha n(n-1)}=a\mathrm e^{-2\mathrm i \alpha(n-1)}\;. $$