1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Also, naïvely, I would think that term $A$ would resemble term $D$ in some way (because both $(k+1, k)$ and $(k, k-1)$ are only one apart, just shifted. Is there anything in the physics which references absolute positional indexing in $k$? $\endgroup$ – MetaPhysic99 Feb 25 at 22:15
  • $\begingroup$ @MetaPhysic99 Thanks for replying. Please check the updated post where I showed what I'm concerned about regarding the Kernel at the center of eq.(7) and the two kernels adjacent to it. Also, I dropped $\epsilon^2$ term because when you have a limit $\epsilon\rightarrow0$, such term is negligible. In eq.(12)-(16), the $\epsilon^2$ term is retained because you need to multiply $A,B,C,D$ with $m^2/\epsilon^2$. $\endgroup$ – Lonitch Feb 25 at 23:52
  • $\begingroup$ Deleted some of my old comments since they weren't that helpful. Feynman's (7.97) has a $p^2 \sim \frac{\partial^2}{\partial x^2}$ operator, and $pp \sim (Hx-xH)(Hx-xH) = HxHx -xHHx -HxxH + xHxH$ (up to prefactors). We have something that looks like $A-B-C+D$, so I can see only one way this happens. Only the $\epsilon^2$ terms should contribute, with the rest canceling. The commentary before (7.99) in Feynman may also be helpful. $\endgroup$ – MetaPhysic99 Feb 26 at 1:50
  • $\begingroup$ @MetaPhysic99 Yeah, I'm stuck at how to get $xHHx$ and $HxxH$ terms $\endgroup$ – Lonitch Feb 26 at 2:14
  • $\begingroup$ I think these two have to come from $B_{k+1,k-1}$ since $C_{k,k} \sim x^2$ only. I also don't yet have an answer for why Feynman is able to treat $(x\psi)$ as a wavefunction object that can evolve with the propagator. As you first noted, the answer to this may give more insight for how to get all the terms of $p^2$. Are we missing a product/commutator rule of operators somewhere? $\endgroup$ – MetaPhysic99 Feb 26 at 2:25
0
+50
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Thanks for your reply. I guess what I don't understand is that how to keep double integral at RHS of Eqn. (1) if we use $(1\pm\frac{i\epsilon}{\hbar}H)$ in the derivation. In Feynman's book, we get (7.94) and (7.93) by integrating over $y$, one of. Also, could you please elaborate on what do you mean by "So having different variables gives a function of those new variables"? $\endgroup$ – Lonitch Mar 3 at 1:58
  • $\begingroup$ I had not noticed the $y$ integration in Feynman and Hibbs. That is a misprint. There needs to be a delta(x-y) for that to make sense. If you write the propagator for the $(1\pm \frac{i\epsilon}{\hbar} H)$ in space as $K(y,t\pm \epsilon.x,t)$, the limit $\epsilon\rightarrow 0$ gives a delta function. You can also verify that the units of the left and right sides of the Feynman and Hibbs expression are not correct. The left side has units of momentum squared. The right side has units of momentum squared times length. $\endgroup$ – user200143 Mar 3 at 17:45
  • $\begingroup$ I just meant the trivial comment that $K(x,t,x',t')$ is a function of the 4 arguments, so substituting different arguments gives a function of those new arguments. $\endgroup$ – user200143 Mar 3 at 17:47
  • $\begingroup$ Thanks for your reply again! However, I'm not very comfortable with your argument regarding the two states $\mid\psi(x,t_{k-1})\rangle$ and $\langle\chi(x_{k+1})\mid$. In fact, $\epsilon$ does matter, because we need to multiply $A,B,C,D$ with $m^2/\epsilon^2$. $\endgroup$ – Lonitch Mar 4 at 5:53
  • $\begingroup$ I added the explicit expansion in $\epsilon$ to show you that the terms I dropped go to zero in the limit. $\endgroup$ – user200143 Mar 4 at 17:43
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.