Obtaining quantum Hamiltonian for charged particle from path integral formulation

I was working on Shankar 8.6.4, which is about obtaining in one dimension the Hamiltonian operator of a charged particle from the path integral formulation.

First, I get the propagator over a time slice $\epsilon$, which is

$$U(x,\epsilon;x') = \sqrt{m \over 2\pi i\hbar\epsilon} \exp\left({i\over\hbar}\left( {m\eta^2\over 2\epsilon} - {q\eta\over c}A(x+\alpha\eta,0) - \epsilon q\phi(x+\alpha\eta,0)\right)\right)$$

where $\eta = x'-x$, $A$ is the vector potential, and $\phi$ is the scalar potential.

Then, I obtain an expression for the time evolution of the initial wavestate in a time slice:

\begin{align} \psi(x,\epsilon) &= \int_{-\infty}^\infty U(x,\epsilon;x') \psi(x',0) \, dx'\\ &= \sqrt{m \over 2\pi i\hbar\epsilon} \int_{-\infty}^\infty \exp\left({i\over\hbar}\left( {m\eta^2\over 2\epsilon} - {q\eta\over c}A(x+\alpha\eta,0) - \epsilon q\phi(x+\alpha\eta,0)\right)\right) \psi(x+\eta,0) \, d\eta \end{align}

I expand the following expressions into series:

\begin{align} \exp\left(-{iq\eta\over\hbar c}A(x+\alpha\eta,0)\right) &= 1 - {iq\eta\over\hbar c}A(x+\alpha\eta,0) - {1\over 2}\left({q\eta\over\hbar c}A(x+\alpha\eta,0)\right)^2+\cdots\\ &= 1 - {iq\eta\over\hbar c}A(x,0) + \left(- {1\over 2}\left({q\over\hbar c}A(x,0)\right)^2-{i\alpha q\over\hbar c}{\partial A(x,0)\over\partial x}\right)\eta^2+\cdots \end{align}

\begin{align} \exp\left(-{i\epsilon q\over\hbar}\phi(x+\alpha\eta,0)\right) &= 1 - {i\epsilon q\over\hbar}\phi(x+\alpha\eta,0) + \cdots\\ &= 1 - {i\epsilon q\over\hbar}\phi(x,0) + \cdots \end{align}

\begin{align} \psi(x+\eta,0) &= \psi(x,0) + \eta{\partial\psi(x,0)\over\partial x} + {\eta^2\over 2}{\partial^2\psi(x,0)\over\partial x^2} + \cdots \\ \end{align}

The wavestate after the time slice is

\begin{align} \psi(x,\epsilon) =& \sqrt{m \over 2\pi i\hbar\epsilon} \int_{-\infty}^\infty \exp\left(-{m\eta^2\over 2i\hbar\epsilon}\right) \\ & \cdot \exp\left(-{iq\eta\over\hbar c}A(x+\alpha\eta,0)\right) \exp\left(-{i\epsilon q\over\hbar}\phi(x+\alpha\eta,0)\right) \psi(x+\eta,0) \, d\eta \\ =& \sqrt{m \over 2\pi i\hbar\epsilon} \int_{-\infty}^\infty \exp\left(-{m\eta^2\over 2i\hbar\epsilon}\right) \\ & \cdot \left( \left(1-{i\epsilon q\phi_x\over\hbar}\right)\psi_x + \eta^2\left(- {1\over 2}\left({q A_x\over\hbar c}\right)^2\psi_x-{i\alpha q\over\hbar c}{\partial A_x\over\partial x}\psi_x-{i q\over\hbar c}A_x{\partial\psi_x\over\partial x}+{1\over 2}{\partial^2\psi_x\over\partial x^2}\right) \right) \, d\eta\\ =& \left(1-{i\epsilon q\phi_x\over\hbar}\right)\psi_x + \left({i\epsilon\hbar\over m}\right)\left(- {1\over 2}\left({q A_x\over\hbar c}\right)^2\psi_x-{i\alpha q\over\hbar c}{\partial A_x\over\partial x}\psi_x-{i q\over\hbar c}A_x{\partial\psi_x\over\partial x}+{1\over 2}{\partial^2\psi_x\over\partial x^2}\right)\\ =& \psi_x - {i\epsilon\over\hbar}\left( q\phi_x + {1\over 2m}\left({q A_x\over c}\right)^2+{i\hbar\alpha q\over m c}{\partial A_x\over\partial x}+{i\hbar q\over m c}A_x{\partial\over\partial x}-{\hbar^2\over 2m}{\partial^2\over\partial x^2} \right)\psi_x \end{align}

where for fields and states, $\omega_x=\omega(x,0)$.

According to the above,

$$\newcommand{\ket}[1]{\left| #1\right\rangle} i\hbar\dot{\ket{\psi}} = \left(q\phi + {1\over 2m}\left({q A\over c}\right)^2 - {\alpha q \over m c} P A - {q \over m c} A P + {P^2\over 2m} \right)\ket{\psi}$$

but after the midpoint prescription $\alpha={1\over 2}$, the Hamiltonian should be

$$q\phi + {1\over 2m}\left({q A\over c}\right)^2 - {q \over 2 m c} P A - {q \over 2 m c} A P + {P^2\over 2m}$$

What happened with the coefficient of $AP$?

• You must be careful with your notation. By the product rule, $A'\psi/2 + A \psi' = (A\psi)'/2 + A\psi'/2$. This is proportional to $p(A\psi) + Ap \psi$, where the prime denotes derivative w.r.t. $x$. On the l.h.s. $p$ acts on either $A$ or $\psi$, while in the first term of the r.h.s., $p$ acts on BOTH $A$ and $\psi$ via the product rule. Jun 9, 2015 at 14:24