How to efficiently exponentiate a differential operator with non-commuting terms, acting on an infinite series?

Question

My question is related to this one, but I am more explicit with the form of the operator in the exponential and the function $f$.

Let us have the expression:

$$ e^{\mathrm{i} \left( f(x)(\hat{q} \hat{p} +\hat{p} \hat{q}) + g(x) \hat{p}^2\right)} \sum_{n=0}^\infty \frac{a^n}{\sqrt{n!}} h_n(x) $$

where $f,g,h_n$ are some real functions, $a$ is a real constant and $\hat{q},\hat{p}$ are quantum mechanical operators of position and momentum, i.e. $\hat{q}=q$ and $\hat{p}=-i\mathrm{d}/\mathrm{d}x$ in the coordinate representation.

How to handle this efficiently? I've tried to use the above post to tackle this problem, with not much success though. Later, I can try to update this post to include what I've done so far.

Answer 1

The answer seems to be simply a math issue. What you're looking for can be obtained by solving a differential equation instead of doing the operator exponential. I will simply call the entire series $h(x)$ because it doesn't really matter whether it's a series or not. Think of it as treating each term separately.

Let's call the operator in the exponent $\hat{H}$. The result you're looking for is

$u(x) \equiv e^{i\hat{H}} h(x)$.

Then introduce a control parameter ("time") and a function $u(x, t)$ defined as

$u(x, t) \equiv e^{i t \hat{H}} h(x)$

so that

$u(x, 0) = h(x)$

at $t = 0$, and the desired result is obtained for $t = 1$:

$u(x, 1) = u(x)$

Now differentiate the definition of $u(x, t)$ with respect to time, yielding

$$\frac{\partial}{\partial t}u(x, t) = i\hat{H} u(x, t)$$

Finally, integrate this differential equation from $t=0$ with the initial condition $u(x, 0) = h(x)$ to the final time $t = 1$, and you have the result.

You can find clever ways of solving this differential equation. In your case, it could involve finding the spectrum of the operator $\hat{H}$ and expanding the function $h(x)$ in the eigenfunctions.