# If a Hamiltonian is quadratic in the ladder operator, why is its time evolution linear in the ladder operator?

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.

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 $$\hat{a}^2$$.

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

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 $$f=-iH t$$.