Feynman's Quantum Mechanics and Path Integral has a vivid physical interpretation of the path integral formalism. But I was stumbled on some mathematical details while following his derivation of the Schrodinger equation of "kernel" from the "kernel integral equation". A kernel $K(b,a)=K(x_b,t_b;x_a,t_a)$ is defined as a path integral between two endpoints $(x_b,t_b)$ and $(x_a,t_a)$ with the time instances $t_b<t_a$. It follows the relation of


with the Schrodinger equation for the kernel being

$$\frac{\partial}{\partial t_b}K(b,a)=\frac{-i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K(b,a)+V(x_b,t_b)K(b,a)\right]=-\frac{i}{\hbar}\hat{H}_bK(b,a).\tag{2}$$

$V(x_b,t_b)$ in eqn.(2) is a potential function. Because $K(b,a)=0$ for $t_b<t_a$, and

$$\lim_{t_b\rightarrow t_a+0^+}K(b,a)=\delta(x_b-x_a),\tag{3}$$

Feynman then makes eqn. (2) complete by writing

$$\frac{\partial}{\partial t_b}K(b,a)=\frac{-i}{\hbar}\hat{H}_bK(b,a)+\delta(x_b-x_a)\delta(t_b-t_a).\tag{4}$$

Now, if we treat the potential $V(b)=V(x_b,t_b)$ as a perturbation to a system represented by $K(b,a)$, and write




we can expand $exp\{-\frac{i}{\hbar}\int_{t_a}^{t_b}V(x,t)dt\}$ in eqn.(5) to give

$$K(b,a)=K_0(b,a)-\frac{i}{\hbar}\int K_0(b,c)V(c)[K_0(c,a)-\frac{i}{\hbar}\int K_0(c,d)V(d)K_0(d,a)d\tau_a+\dots]d\tau_c\tag{7}$$

which gives the integral equation:

$$K(b,a)=K_0(b,a)-\frac{i}{\hbar}\int K_0(b,c)V(c)K(c,a)d\tau_c\tag{8},$$

where $d\tau=dxdt$. In the book, Feynman argued that the kernel $K(b,a)$ satisfies the following equation:

$$\frac{\partial}{\partial t_b}K(b,a)+\frac{i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K(b,a)+V(x_b,t_b)K(b,a)\right]=\delta(t_b-t_a)\delta(x_b-x_a)\tag{9},$$

which is just eqn.(4). Now, in the problem 6-3 of the book it says that the eqn.(9) can be derived by using "free particle Schrodinger equation",

$$\frac{\partial}{\partial t_b}K_0(b,a)+\frac{i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K_0(b,a)\right]=\delta(t_b-t_a)\delta(x_b-x_a),\tag{10}$$

and the integral equation (8). To do this, I simply take a derivative w.r.t. $t_b$ on both size of eqn.(8), and notice that

$$\int K_0(b,c)V(c)K(c,a)d\tau_c=\int_{t_a}^{t_b}\int_{-\infty}^{\infty}K_0(b,c)V(c)K(c,a)dx_cdt_c\tag{11}$$

should also be treated as a function of $t_b$. So I applied the Leibniz integral rule to the second term on the RHS of eqn.(8) to find

$$\frac{\partial}{\partial t_b}K(b,a)=\frac{\partial}{\partial t_b}K_0(b,a)-\frac{i}{\hbar}\int_{-\infty}^{\infty}K_0(x_b,t_b;x_c,t_b)V(c)K(c,a)dx_c-\frac{i}{\hbar}\int \frac{\partial}{\partial t_b}K_0(b,c)V(c)K(c,a)d\tau_c.\tag{12}$$

I use eqn.(3) in the second term above and replace $\partial K_0/\partial t_b$ with the terms in eqn. (10) I further find that

$$\frac{\partial}{\partial t_b}K(b,a)=\frac{\partial}{\partial t_b}K_0(b,a)-\frac{i}{\hbar}V(b)K(b,a)-\frac{i}{\hbar}\frac{i}{\hbar}\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}\int K_0(b,c)V(c)K(c,a)d\tau_c-\frac{i}{\hbar}V(b)K(b,a).\tag{13}$$

Because $K_0(b,a)-K(b,a)=\frac{i}{\hbar}\int K_0(b,c)V(c)K(b,a)d\tau_c$, I finally arrive at

$$\frac{\partial}{\partial t_b}K(b,a)+\frac{i}{\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_b^2}K(b,a)+2V(x_b,t_b)K(b,a)\right]=\delta(t_b-t_a)\delta(x_b-x_a).\tag{14}$$

Now, eqn.(14) is very much like eqn.(9), but the factor 2 in front of $V(x_b,t_b)$ really confused me. Can anybody tell me what's wrong with my derivation leading to eqn.(14)? Any help will deeply appreciated.


1 Answer 1


OP's factor of $2$ in eq. (14) can be traced back to the fact that Ref. 1 fails to clearly distinguish between the retarded Greens function/propagator

$$\begin{align} G(x_2,t_2;x_1,t_1)~=~&\theta(\Delta t)~K(x_2,t_2;x_1,t_1), \cr \Delta t~:=~&t_2-t_1,\end{align}\tag{G}$$

and the kernel/path integral

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\langle x_2,t_2 | x_1,t_1 \rangle\cr ~=~&\langle x_2|U(t_2,t_1)|x_1 \rangle\cr ~=~&\int_{x(t_1)=x_1}^{x(t_2)=x_2} \! {\cal D}x~ \exp\left[\frac{i}{\hbar}\int_{t_1}^{t_2} \!dt ~L\right] .\end{align}\tag{K} $$

Here $\theta$ denotes the Heaviside step function. The Schrödinger equation with a Dirac delta [OP's eqs. (4), (9) & (10)] refers to $G$, while the rest of OP's eqs. (1)-(11) refers to $K$.

  1. The simplest is to think of OP's integral equation (8) as consisting purely of $K$. Then there is no Dirac delta in the Schrödinger equation so that only $1$ of the $2$ $V$-terms in OP's eq. (14) survives.

  2. Alternatively, one may interpret OP's integral equation (8) as $G=G_0 + \int G_0VK$. Then there is a hidden $\theta(0)=\frac{1}{2}$ so that the factor $2$ again goes away.

See also this related Phys.SE post.


  1. R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.

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