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How can one show that $\hat{U}^\dagger\hat{a}\hat{U}$ (with $\hat{U} =e^{-i\hat{H}t}$) involves only linear orders of the ladder operator, when $H$ is the general quadratic Hamiltonian $(\hat{H} = \alpha (\hat{a}^\dagger)^2+ \beta \hat{a}^\dagger\hat{a}+\alpha^*\hat{a}^2)?$

I have been trying to do it using the Baker-Campbell-Hausdorff formula.

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Hint: Instead of using the BCH formula in the form usually presented, for example at the top of this Wikipedia page, use this consequence of Hadamard's Lemma:

$$\tag{1} e^{iHt}\hat{a}e^{-iHt} = \hat{a} + [iHt,\hat{a}] + \frac{1}{2!}[iHt,[iHt,\hat{a}] + \cdots $$

Now substitute $H$ into the right-hand side and evaluate the commutators between $\hat{a}$ and each of the three terms in $H$.

Let's look at the following commutators that appear in the second term:

\begin{align}\tag{2} [(\hat{a}^\dagger)^2,\hat{a}] &= -2\hat{a}^\dagger\\\tag{3} [\hat{a}^\dagger \hat{a},\hat{a}] &= -\hat{a}\\\tag{4} [\hat{a}^2,\hat{a}] &= 0.\\ \end{align}

All commutators appearing in the third term will be commutators of the same form, or of the form $[(\hat{a}^\dagger)^2,\hat{a^\dagger}],[\hat{a}^\dagger \hat{a},\hat{a}^\dagger],[\hat{a}^2,\hat{a}^\dagger] $, which will again be linear.

In the end, you'll only be left with terms linear in $\hat{a}$ and $\hat{a^\dagger}$, but no "quadratic" terms like $\hat{a}^2$.

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Use the differential form of the time evolution, $$dO/dt=i[H, O]\ .$$

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Note that $$[(a^\dagger)^n,a] = -n(a^{\dagger})^{n-1}, \qquad [(a^\dagger)^n a^m,a] = -n (a^\dagger)^{n-1}a^m, \qquad [a^n,a]=0.$$ Consider an arbitrary function of the mode operators, that we assume be written in normal formal: $$f(a,a^\dagger) = \sum_{n,m=0}^\infty c_{n,m} (a^\dagger)^n a^m.$$ We know that $$e^{f(a,a^\dagger)}a e^{-f(a,a^\dagger)} = \sum_{k=0}^\infty \frac{1}{k!}\operatorname{ad}(f(a,a^\dagger))^k\cdot a,$$ where $\operatorname{ad}(A)^k\cdot B \equiv [\underbrace{A,...,[A}_k,B]\cdots]$ denotes the operation of taking the repeated commutator of $A$ with $B$. For example, for $k=2$, $\operatorname{ad}(A)^2\cdot B\equiv [A,[A,B]]$.

We know that $$[f(a,a^\dagger),a] = -\sum_{n,m\ge 0}c_{n,m} n (a^\dagger)^{n-1} a^m,$$ $$\operatorname{ad}(f(a,a^\dagger))^k\cdot a = (-1)^k \sum_{n,m\ge0} c_{n,m} \frac{n!}{(n-k)!} (a^{\dagger})^{n-k}a^m.$$ Thus $$e^{f(a,a^\dagger)}a e^{-f(a,a^\dagger)} = \sum_{n,m\ge0 } c_{n,m} (a^\dagger-1)^n a^m = f(a, a^\dagger-1). $$

This shows you explicitly that properties of $f$ such as its being quadratic permain in $e^f a e^{-f}$. Thus, in particular, you get your result by choosing $f=-iH t$.

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