A Problem from Feynman's Path Integral Book
Let $x_i$ be coordinates at different time instances, prove that
$$ \langle\chi|m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}|\psi\rangle=\int\int \chi^* \hat{p}\hat{p}\psi dxdy=-\hbar^2\int\int\chi^*\frac{\partial^2}{\partial x^2}\psi dxdy\tag{1} $$
with $\hat{p}=-i\hbar\frac{\partial}{\partial x}$ being momentum operator.
Equations Developed by Feynman for Later Use
To show eqn.(1), I first noticed that, from the definition of path integral,
$$\langle\chi|1|\psi\rangle=\int\int\chi^*(x_a,t_b)K(x_a,t_b;x_b,t_a)\psi(x_b,t_a)dx_adx_b\tag{2},$$
with kernel $K(x,t_b;y,t_a)$ being a path integral with two ends of paths at $t_a$ and $t_b$. If the 1 in Eq.(2) is replaced with a function $F(x_c,t_c)$, we have a transition element as
$$ \langle\chi|F|\psi\rangle_S=\int\int\int_{x_a}^{x_b}\chi^*(x_b,t_b)F(x_c,t_c)e^{iS/\hbar}\mathcal{D}x(t)\psi(x_a,t_a)dx_adx_b\tag{3}, $$
The kernel is written out as a path integral explicitly, and the action $S$ describes the system's behavior.
Because $t_c$ is a different time instance, we can further split the path integral in Eq.(3) to get:
$$ \langle\chi|F|\psi\rangle_S=\int\int\int \chi^*(b)K(b;c)F(c)K(c;a)\psi(a) dx_cdx_adx_b\tag{4} $$
where $a=(x_a,t_a)$. Also, for an arbitrary wavefunction $f(y,t)$, we have
$$ \begin{align} \int_{\infty}K(x,t+\epsilon;y,t)f(y,t)dy&=f(x,t+\epsilon)=f(x,t)+\epsilon\frac{\partial f}{\partial t}\\ &=f(x,t)-\frac{i\epsilon}{\hbar}\hat{H}f(x,t)\tag{5} \end{align} $$
Shrodinger's equation of $f(x,t)$ is used at the last line of Eq. (5). The Hamiltonian operator $\hat{H}$ is related to the Lagrangian $\mathcal{L}$ in $S$. Using the first equivalence of eq. (5) in eq. (4) gives
$$ \langle\chi|F|\psi\rangle_S=\int\chi^*(c)F(c)\psi(c) dx_c\tag{6}. $$
Now, if we replace $F$ with a product of coordinates at two different time instances, $x_{k+1}(t+\epsilon)x_k(t)$ in eq.(4), we have
$$ \begin{align} \langle\chi|x_{k+1}x_k|\psi\rangle_S&=\int\int\int\int \chi^*(b)K(b;x_{k+1},t+\epsilon)x_{k+1}K(x_{k+1},t+\epsilon;x_k,t)x_k\times\\ &\quad\quad\quad K(x_k,t;a)\psi(a) dx_{k+1}dx_kdx_adx_b\\ &=\int\int \chi^*(x_{k+1},t+\epsilon)x_{k+1}K(x_{k+1},t+\epsilon;x_k,t)x_k\psi(x_k,t)dx_{k+1}dx_k.\tag{7} \end{align} $$
According to Feynman's argument, if we integrate over $x_k$, we can use eq. (5) to get
$$ \begin{align} \int K(x_{k+1},t+\epsilon;x_k,t)x_k\psi(x_k,t)dx_k&=x_{k+1}\psi(x_{k+1},t+\epsilon)\\ &=(1-\frac{i\epsilon}{\hbar}\hat{H})x_{k+1}\psi(x_{k+1},t)\tag{8}, \end{align} $$
Eq.(5) is applied to Eq.(8) directly by replacing $f(y)$ in (5) with $x\psi$ in (8). However, the kernels $K(x_k,t;a)$ and $K(b;x_{k+1},t+\epsilon)$ in (7) indicates that $\psi$ $\chi$ are states that satisfy
$$ i\hbar\frac{\partial \psi}{\partial t}=\hat{H}\psi $$
But $K(x_{k+1},t+\epsilon;x_k,t)$ in Eq.(7) and (8) implies instead $$ i\hbar\frac{\partial}{\partial t}(x\psi)=\hat{H}x\psi. $$
So my question is Is it alright to think that $K(x_{k+1},t+\epsilon;x_k,t)$ is different from $K(x_k,t;a)$ and $K(b;x_{k+1},t+\epsilon)$?
If we ignore this and move forward the eq. (7) now becomes
$$ \begin{align} \langle\chi|x_{k+1}x_k|\psi\rangle_S&=\int \chi^*(x_{k+1},t+\epsilon)x_{k+1}(1-\frac{i\epsilon}{\hbar}\hat{H})x_{k+1}\psi(x_{k+1},t)dx_{k+1}\\ &=\int \chi^*(x,t+\epsilon)x(1-\frac{i\epsilon}{\hbar}\hat{H})x\psi(x,t)dx\\ &=\int\chi^*(x,t)(1+\frac{i\epsilon}{\hbar}\hat{H})x(1-\frac{i\epsilon}{\hbar}\hat{H})x\psi(x,t)dx\\ &=\int\chi^*(x)x^2\psi(x)dx+\frac{i\epsilon}{\hbar}\int\chi^*(\hat{H}x-x\hat{H})x\psi(x)dx,\tag{9} \end{align} $$
where we dropped $\epsilon^2$ term at the limit of $\epsilon\rightarrow0$.Since $\frac{im}{\hbar}[H,x]=p$, the equation above gives
$$ \langle\chi|x_{k+1}x_k|\psi\rangle_S=\int\chi^*(x)x^2\psi(x)dx+\frac{\epsilon}{m}\int\chi^*px\psi(x)dx\tag{10}. $$
Coming back to eq. (1)
Notice that
$$ \begin{align} \langle\chi|m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}|\psi\rangle&=\frac{m^2}{\epsilon^2}\{\langle\chi|x_{k+1}x_k|\psi\rangle-\langle\chi|x_{k+1}x_{k-1}|\psi\rangle-\\ &\langle\chi|x_kx_k|\psi\rangle+\langle\chi|x_{k}x_{k-1}|\psi\rangle\}\\ &=\frac{m^2}{\epsilon^2}\{A-B-C+D\}\tag{11}. \end{align} $$
where
$$ \begin{align} A=\int\chi^*(x)x^2\psi(x)dx+&\frac{i\epsilon}{\hbar}\int\chi^*(\hat{H}x-x\hat{H})x\psi(x)dx\\ &+\frac{\epsilon^2}{\hbar^2}\int\chi^*(x,t)\hat{H}x\hat{H}x\psi(x,t)dt,\tag{12} \end{align} $$
$$ \begin{align} B&=\int\chi^*(x_{k+1},t+\epsilon)x_{k+1}K(x_{k+1},t+\epsilon;x_{k-1},t_{k-1})x_{k-1}\psi(x_{k-1},t-\epsilon)dx_{k+1}dx_{k-1}\\ &=\int\chi^*(x_{k+1},t^{\prime\prime}+2\epsilon)x_{k+1}K(x_{k+1},t^{\prime\prime}+2\epsilon;x_{k-1},t^{\prime\prime})x_{k-1}\psi(x_{k-1},t^{\prime\prime})dx_{k+1}dx_{k-1},\tag{13} \end{align} $$
$$ C=\int\chi^*(x)x^2\psi(x)dx\tag{14} $$
and
$$ \begin{align} D&=\int\int \chi(x_k,t)x_kK(x_k,t;x_{k-1},t-\epsilon)x_{k-1}\psi(x_{k-1},t-\epsilon)dx_{k}dx_{k-1}\\ &=\int\int \chi(x_k,t'+\epsilon)x_kK(x_k,t'+\epsilon;x_{k-1},t')x_{k-1}\psi(x_{k-1},t')dx_{k}dx_{k-1}\tag{15} \end{align} $$ according to equations(8)-(10). Combining (11)-(15), we have all the $\epsilon$ terms canceled out to give
$$ \langle\chi|m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}|\psi\rangle=-\frac{2m^2}{\hbar^2}\int\chi^*(x,t)\hat{H}x\hat{H}x\psi(x,t)dx\tag{16}, $$
Which does not match Feynman's results. Can anyone tell me which step in my derivation needs to be modified? I really appreciate any help you can provide.