[QFT-Schwartz Page. 256] Violation of operator exponentiation rule ?

[Golak Bage]

TL;DR

Schwartz derives path integral formulation from 'non-relativistic QM'.

When computing the projection of time-evoluted state $|x_j>$ on $|x_{j+1}>$ it uses the 'completeness' of momentum basis $\int \frac{dp}{2\pi} |p><p|$. Next it explicitly states the form of Hamiltonian $\hat{H} = \frac{\hat{p}^2}{2m}+\hat{V}(\hat{x_j},t_j)$. Thereafter i believe it uses the relation $$ e^{\frac{\hat{p}^2}{2m}+\hat{V}(\hat{x_j},t_j)} = e^{\frac{\hat{p}^2}{2m}}\times e^{\hat{V}(\hat{x_j},t_j)}.$$ This pre-supposes that $[\hat{p},\hat{V}(\hat{x_j},t_j)]=0$. In QM for any two operators (say $\hat{A}\ \&\ \hat{B}$) $e^{\hat{A}+\hat{B}}=e^{\hat{A}}\times e^{\hat{B}}\times e^{-\frac{1}{2}[\hat{A}, \hat{B}]}$, therefore above relation doesn't appear general (it's more specific). I'd like some feedback on my thought.

 

[TSny]

$\large e^{\hat{A}+\hat{B}}=e^{\hat{A}}\times e^{\hat{B}}\times e^{-\frac{1}{2}[\hat{A}, \hat{B}]}$

Apply this to $\large e^{-i[\frac{\hat{p}^2}{2m} + V(\hat{x}_j, t_j) ] \delta t }. \,\,\,\,$ So, $\hat{A} = -i\frac{\hat p^2}{2m} \delta t$ and $\hat B =-iV(\hat x_j, t_j)\delta t$.

Note that $[\hat A, \hat B]$ is proportional to $\delta t ^2. \,\,$ If $\delta t$ is assumed to be very small, perhaps we can neglect terms of second order in $\delta t$ and approximate $e^{-\frac{1}{2}[\hat{A}, \hat{B}]} \approx 1$.

 

[Golak Bage]

Apply this to $\large e^{-i[\frac{\hat{p}^2}{2m} + V(\hat{x}_j, t_j) ] \delta t }. \,\,\,\,$ So, $\hat{A} = -i\frac{\hat p^2}{2m} \delta t$ and $\hat B =-iV(\hat x_j, t_j)\delta t$.

Note that $[\hat A, \hat B]$ is proportional to $\delta t ^2. \,\,$ If $\delta t$ is assumed to be very small, perhaps we can neglect terms of second order in $\delta t$ and approximate $e^{-\frac{1}{2}[\hat{A}, \hat{B}]} \approx 1$.

Thank you. It actually makes sense.