Conjugation of creation operators
The identity
\[\begin{equation}
x^{\hat n_H} a_H^\dagger x^{-\hat n_H} = x\, a_H^\dagger
\end{equation}\]
is proven by evaluating the action of both sides on number states.
To begin, recall how exponentials of the number operator act on Fock states:
\[\begin{equation}
\label{eq:x pow n m}
x^{-\hat n_H} |m\rangle = x^{-m} |m\rangle.
\end{equation}\]
This follows directly from the fact that \(|m\rangle\) is an eigenstate of \(\hat n_H\).
Applying the creation operator to this result gives
\[\begin{equation}
a_H^\dagger x^{-\hat n_H} |m\rangle
= x^{-m} a_H^\dagger |m\rangle
= x^{-m} \sqrt{m+1}\, |m+1\rangle.
\end{equation}\]
Next, the operator \(x^{\hat n_H}\) is applied. Since \(|m+1\rangle\) is also an eigenstate of \(\hat n_H\), its action is straightforward:
\[\begin{equation}
\begin{aligned}
x^{\hat n_H} a_H^\dagger x^{-\hat n_H} |m\rangle
&= x^{-m} \sqrt{m+1}\, x^{m+1} |m+1\rangle \\
&= x \sqrt{m+1} |m+1\rangle.
\end{aligned}
\end{equation}\]
For comparison, the direct action of \(x\,a_H^\dagger\) on the same state is
\[\begin{equation}
x a_H^\dagger |m\rangle = x \sqrt{m+1} |m+1\rangle.
\end{equation}\]
Since both expressions produce the same result for arbitrary \(|m\rangle\), the operator identity follows:
\[\begin{equation}\label{eq:x pow n a dagger}
x^{\hat n_H} a_H^\dagger x^{-\hat n_H} = x\, a_H^\dagger.
\end{equation}\]
For the \(V\) mode, the commutation relation \([\hat n_H, a_V^\dagger]=0\) implies that the operators commute. As a result, the conjugation has no effect:
\[\begin{equation}
\label{eq:x pow n a dagger2}
x^{\hat n_H} a_V^\dagger x^{-\hat n_H} = a_V^\dagger.
\end{equation}\]