Suppose I have a time evolving quantum field:
$$\phi_2(t,x)=e^{iH_0 (t-t_0)} \phi_2 (t_0,x) e^{-iH_0 (t-t_0)} $$
I want to show that
$$e^{iH_0 (t-t_0)} (\nabla \phi_2(t_0 ,x))^2 e^{-iH_0 (t-t_0)} = (\nabla \phi_2(t ,x))^2.$$
To this end I start by taking the gradient:
$$ \nabla (\phi_2(t,x)) = \nabla(e^{iH_0 (t-t_0)}) \phi_2 (t_0,x) e^{-iH_0 (t-t_0)} + e^{iH_0 (t-t_0)} \nabla(\phi_2 (t_0,x)) e^{-iH_0 (t-t_0)} + e^{iH_0 (t-t_0)} \phi_2 (t_0,x) \nabla(e^{-iH_0 (t-t_0)})$$
$$ \nabla (\phi_2(t,x)) = i(t-t_0)\left[ \nabla(H_0)\phi_2(t ,x) – e^{iH_0 (t-t_0)} \phi_2 (t_0,x) \nabla (H_0)e^{-iH_0 (t-t_0)}\right] + e^{iH_0 (t-t_0)} \nabla(\phi_2 (t_0,x)) e^{-iH_0 (t-t_0)}$$
So it suffices to show:
$$\nabla(H_0)\phi_2(t ,x) = e^{iH_0 (t-t_0)} \phi_2 (t_0,x) \nabla (H_0)e^{-iH_0 (t-t_0)}$$
Or that:
$$\left[ \nabla H_0\ ,\ e^{iH_0 (t-t_0)} \phi_2 (t_0,x)\right] =0$$
$H_0$, meanwhile, is given by
$$H_0 = \int d^3 x\ \ \frac{1}{2}\pi_2^2 + \frac{1}{2}\phi_2^2 m^2 + \frac{1}{2} (\nabla \phi_2)^2$$
where $\phi_2$ and $\pi_2$ are subject to canonical quantization.
How can this commutator be evaluated? We know the commutator of $\phi, \pi$ from canonical quantization, but what about $[\nabla \pi, \phi ]?$ Is there a general formula for $[A, e^{B}]?$
Best Answer
$H_0$ does not depend on the space coordinate you take gradient with. The field operators inside the integral expression might be confusing, but note that they depend on the integrating variable $x$, different from every other “$x$” you write.
So your proof is actually complete, after noting $\nabla H_0=0$.
Regarding other commutators:
So $\nabla\pi$ doesn’t appear here, and I don’t clearly see where it would be relevant, but we do often encounter $[\nabla\phi,\pi]$. Or, we can more generally take $[\nabla_x O_1(x),O_2(y)]$. Reassuring that the gradient taken for $x$, not $y$, linearity of gradient implies $$[\nabla_x O_1(x),O_2(y)] = \nabla_x[O_1(x),O_2(y)]$$
We know $[A, e^B]$ appears frequently in quantum mechanics. Recall that the commutator can be decomposed term-by-term of the power series. $$[A,e^B] = \sum_{n=0}^\infty [A,\frac{B^n}{n!}]$$