Average of Two successive momenta $m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}$ using rules of path integral

Question

A Problem from Feynman's Path Integral Book

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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

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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)

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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.

Answer 1

I think your problem is simply not keeping track of the time differences for the states. That is using the notation that $t_k$ is the time when $x_k$ operates and Feynman and Hibbs notation that they used to write their Eq. 7.95 just above their statement of this problem (problem 7.15), \begin{equation} \langle \chi|(x_{k+1}-x_k)(x_{k}-x_{k-1}|\psi\rangle = \langle \chi(t_{k+1})| \left [x \left (1-\frac{i}{\hbar}H\epsilon\right) -\left (1-\frac{i}{\hbar}H\epsilon\right)x\right] \left [x\left (1-\frac{i}{\hbar}H\epsilon\right) -\left (1-\frac{i}{\hbar}H\epsilon\right)x\right]|\psi(t_{k-1})\rangle \end{equation} The terms in the brackets are both $\frac{i\epsilon }{h}[H,x] =\epsilon\frac{p}{m}$. The matrix element becomes \begin{equation} \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(t_{k+1})| \frac{\epsilon p}{m} \frac{\epsilon p}{m} \left (1+\frac{i}{\hbar}H\epsilon\right)|\psi(t_{k-1})\rangle =\langle\chi(t_k+\epsilon)|p^2|\psi(t_k-\epsilon)\rangle \,. \end{equation}

To get the two states at the same time, for example, $t_k$, we use \begin{equation} \langle \chi(t_{k+1})| = \langle \chi(t_k)|\left (1+\frac{i}{\hbar}H\epsilon\right) \end{equation} and \begin{equation} |\psi(t_{k-1})\rangle = \left (1+\frac{i}{\hbar}H\epsilon\right)|\psi(t_k)\rangle \,. \end{equation} Substituting, \begin{equation} \langle \chi| m\frac{x_{k+1}-x_k}{\epsilon} m\frac{x_{k}-x_{k-1}}{\epsilon}|\psi\rangle =\langle\chi(t_k)|p^2|\psi(t_k)\rangle + \frac{i\epsilon}{\hbar} \langle\chi(t_k)|Hp^2+p^2H|\psi(t_k)\rangle + O(\epsilon^2)\,. \end{equation}

But now taking $\epsilon$ to zero, these time differences don't matter, and the result is as in Feynman and Hibbs, which is equivalent to $\langle \chi(t_k)|p^2|\psi(t_k)\rangle$.

As for your question about the proagators, in Feynman and Hibbs, the $K$ is a function of those 4 variables once the Hamiltonian as a function of time is known. So having different variables gives a function of those new variables.

Answer 2

As the previous answer mentioned, there is a misprint in the problem. So the Eqn(1) in OP should really be written as

$$ \langle\chi|m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}|\psi\rangle=-\hbar^2\int\chi^*(x,t)\frac{\partial^2}{\partial x^2}\psi(x,t) dx\tag{1*}. $$

Also, It might seem to be that Eq.(5) is applied to Eq.(8) directly by replacing $f(y)$ in (5) with $x \psi$ in (8). However, Feynman was trying to motivate us here to build a relationship between the conventional operator formalism and the path integral formalism. So the $x$ can be understood as a coordinate operator or an eigenvalue of the operator. After a time period of $\epsilon$, the eigenvalue of the coordinate operator for the wavefunction becomes $x_{k+1}$, and thus the result (8). Using Eqn. (8) and its conjugate, and let $t_k=t$, $x_{k+1}=x$, $x_k=z$, and $x_{k-1}=y$, we have

$$ \begin{array}{ll} B&=\int\chi^*(x,t+\epsilon)xK(x,t+\epsilon;y,t-\epsilon)y\psi(y,t-\epsilon)dxdy\\ &=\int\chi^*(x,t+\epsilon)xK(x,t+\epsilon;z,t)K(z,t;y,t-\epsilon)y\psi(y,t-\epsilon)dxdydz\\ &=\int\chi^*(x,t+\epsilon)xK(x,t+\epsilon;z,t)(1-\frac{i\epsilon}{\hbar})z\psi(z,t-\epsilon)dxdz\\ &=\int\chi^*(z,t+\epsilon)z(1-\frac{i\epsilon}{\hbar})(1-\frac{i\epsilon}{\hbar})z\psi(z,t-\epsilon)dz\\ &=\int\chi^*(z,t)(1+\frac{i\epsilon}{\hbar}H)z(1-\frac{i\epsilon}{\hbar})(1-\frac{i\epsilon}{\hbar})z(1+\frac{i\epsilon}{\hbar})\psi(z,t)dz\\ &=\int\chi^*(z,t)(z^2+\frac{i\epsilon}{\hbar}zzH-2\frac{i\epsilon}{\hbar}zHz+2\frac{\epsilon^2}{\hbar^2}zHzH-\\ &\frac{\epsilon^2}{\hbar^2}zHHZ+\frac{i\epsilon}{\hbar}Hzz-\frac{\epsilon^2}{\hbar^2}HzzH+2\frac{\epsilon^2}{\hbar^2}HzHz)\psi(z,t)dz \tag{13} \end{array} $$

where we used the conjugate of Eqn. (8) at the fourth equivalence. The equations of (7.75) and (7.76) in Feynman's book were used at the fifth equivalence. Notice that we need to eliminate all the terms that have their orders higher than $\epsilon^2$ in the equation above. Because those terms are negligible at the limit of $\epsilon\rightarrow0$.

For $C$, it's simply

$$ C=\int\chi^*(x,t)x^2\psi(x,t)dx\tag{14}. $$

Finally, for $D$, we again let $x_{k}=x$ and $y=x_{k-1}$ to give

$$ \begin{array}{ll} D&=\int\chi^*(x,t)xK(x,t;y,t-\epsilon)y\psi(y,t-\epsilon)dxdy\\ &=\int\chi^*(x,t)x(1-\frac{i\epsilon}{\hbar}H)x\psi(x,t-\epsilon)dx\\ &=\int\chi^*(x,t)x(1-\frac{i\epsilon}{\hbar}H)x(1+\frac{i\epsilon}{\hbar}H)\psi(x,t)dx\\ &=\int\chi^*(x,t)(x^2+\frac{i\epsilon}{\hbar}xxH-\frac{i\epsilon}{\hbar}xHx+\frac{\epsilon^2}{\hbar^2}xHxH)\psi(x,t)dx\tag{15}. \end{array} $$

Combining (11)-(15), we have all the $\epsilon$ terms canceled out to give

$$ \begin{array}{ll} \langle\chi|m\frac{x_{k+1}-x_k}{\epsilon}m\frac{x_k-x_{k-1}}{\epsilon}|\psi\rangle&=\frac{m^2}{\hbar^2}\int\chi^*(x,t)(-xHxH\\ &-HxHx+xHHx+HxxH)\psi(x,t)dx\\ &=\int\chi^*(x,t)pp\psi(x,t)dx\tag{16}. \end{array} $$

which does match Feynman's results. Q.E.D.