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I'm deriving the action of the squeezing operator on a non-vacuum Fock state and I have almost finished but can't work out how to apply an operator that is stuck in the power of a constant, not an exponential. I am trying to work out the result of this action:

$$\bigg(\frac{1}{\cosh{|\zeta }|}\bigg)^{\hat{a^{\dagger}} \hat{a} +\hat{b^{\dagger}}\hat{b} + 1} |{0,b}\rangle $$

I suspect the solution is:

$$\frac{1}{\cosh{|\zeta }|}\sqrt{b}|0,b\rangle,$$

but I'm not certain how to show this. I know you can Taylor expand and exponentiated operator but that doesn't apply here.

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Note that $c^x = \exp[x\ln(c)]$, so for an operator $\hat A$ you would have $$c^\hat A = \exp[\ln(c) \hat A] = \sum_{n=0}^\infty \frac{[\ln(c)]^n}{n!} \hat A^n$$

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Write the operator in its eigenbasis. In the eigenbasis, treat each eigenvalue as a number. Compute the function written. Then, the eigenbasis of the resulting operator is the same, and the eigenvalues are the corresponding functions of the eigenvalues.

In formulas: For a hermitian operator $A=\sum a_i\vert i\rangle\langle i\vert$, we have that $$ f(A)=\sum f(a_i)\vert i\rangle\langle i\vert\ . $$

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