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

• I
• Golak Bage
Golak Bage
TL;DR Summary
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.

dextercioby
Golak Bage said:
##\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##.

dextercioby, Demystifier and Golak Bage
TSny said:
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.

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