I'm new on path integral methods and bra ket notation and, while as was reading these Berkeley 221A Lecture Notes on the Path Integral by Hitoshi Murayama (2006), I found

The propagator for a short interval of time is given by the Lagrangian $$ \langle x_1,t+\Delta t|x_0,t\rangle =K_1e^{i(L(t)\Delta t+O(\Delta t)^2)/\hbar}, \tag{5} $$ where $K_1$ is a normalitzation constant.

Considering an easy Hamiltonian such as $$H=\frac{p^2}{2m}+V,\tag{6}$$ $$ \langle x_1,t+\Delta t|x_0,t\rangle =\langle x_1|e^{iH\Delta t/\hbar}|x_0\rangle =\int dp \langle x_1|p\rangle \langle p|e^{iH\Delta t/\hbar}|x_0\rangle .\tag{7} $$

And, focusing on the second term (I know that $\langle x_1|p\rangle = \frac{ e^{-ipx_1/\hbar} }{\sqrt{2\pi\hbar}}$) one can obtain

\begin{align} \langle p|e^{iH\Delta t/\hbar}|x_0\rangle & =\langle p|1-iH\Delta t/\hbar+O(\Delta t)^2|x_0\rangle \\ & =\left[1-{\frac{p^2}{2m}\frac{i}{\hbar}\Delta t-\frac{i}{\hbar}V(x_0)\Delta t+O(\Delta t)^2}\right] \frac{e^{-ipx_0/\hbar}}{\sqrt{2\pi\hbar}} \\ & = \frac{1}{\sqrt{2\pi\hbar}}\exp\frac{-i}{\hbar}\left(px_0+\frac{p^2}{2m}\Delta t+V(x_0)\Delta t+O(\Delta t)^2\right). \tag{8} \end{align}

I can follow evertything except the last step, can someone explain it with more detail? (or only what's the key idea to understand)

  • 1
    $\begingroup$ The lecture notes are just Taylor expanding the exponential function in the very last step. $\endgroup$ – Qmechanic Dec 20 '17 at 15:44
  • $\begingroup$ But why tehre's still and exponential function if you expand it with Taylor (I'm really confused why is all inside an exponential) $\endgroup$ – MatMorPau22 Dec 20 '17 at 15:50

All that he has done is rewrite, $$(1-\frac{i}{\hbar}\frac{p^2\Delta t}{2}-\frac{i}{\hbar}V(x_0)\Delta t + O(\Delta t^2))=exp(-\frac{i}{\hbar}\frac{p^2\Delta t}{2}-\frac{i}{\hbar}V(x_0)\Delta t + O(\Delta t^2)).$$ Then simply moved out the $-i\hbar$, $$exp(-\frac{i}{\hbar}\frac{p^2\Delta t}{2}-\frac{i}{\hbar}V(x_0)\Delta t + O(\Delta t^2))=exp\bigg[\frac{-i}{\hbar}\bigg(+\frac{p^2\Delta t}{2}+V(x_0)\Delta t + O(\Delta t^2)\bigg)\bigg]$$ If you multiply this with, $$\frac{e^{-ipx_0/\hbar}}{\sqrt{2\pi\hbar}},$$ then you get the expression you are looking for.

  • $\begingroup$ I feel stupid right now, it's all about $e^x\cdot e^y=e^{x+y}$, I wasn't able to see, I was looknig for something more complicated $\endgroup$ – MatMorPau22 Dec 20 '17 at 22:23

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