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I'm reading Feynman and Hibbs, Quantum Mechanics and Path Integrals. How do I show that the kernel

$$\tag{2-25} K(x_2 t_2;x_1 t_1)=\int e^{\frac{i}{\hbar}S[2,1]}\mathcal{D}x$$

satisfies the differential equation

$$\tag{4-29} \frac{\partial K(2,1)}{\partial t_2}+\frac{i}{\hbar}H_2K(2,1)=\delta(x_2-x_1)\delta(t_2-t_1)?$$

I know that the kernel satisfies the Schrodinger equation for $t_2 > t_1$, but how do I show the delta's on the RHS as $t_2 \to t_1$. Firstly, it is clear $<x_2 t_1|x_1 t_1>=\delta(x_2-x_1)$, so that $\lim_{t_2 \to t_1}=\delta (x_2-x_1)$. But is there any method to prove this w/o using the transition amplitude interpretation i.e. directly from the definition as the integral over exponent of actions? Can I evaluate this integral in the $t_2 \to t_1$ limit, and show it as the delta function.

Secondly, why does the LHS acting on the delta function of $x$, give the RHS. It seems to me that the derivative wrt. time, introduces the delta function of time, and the Hamiltonian leaves it unchanged? This is clearly wrong IMO, for e.g. the free particle where the Laplacian should give a term $\delta^{\prime \prime}(x_2-x_1)$. What happens to this term?

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1 Answer 1

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Consider non-relativistic quantum mechanics of a point particle in 1 dimension with the classical Lagrangian

$$\tag{A} L~:=~\frac{m}{2}\dot{x}^2-V(x).$$

Let $\Delta t:=t_f-t_i$ and $\Delta x:=x_f-x_i$. OP's question concerns the following properties (B) and (C) of the kernel $K(x_f,t_f;x_i,t_i)$:

$$\tag{B} K(x_f,t_f;x_i,t_i) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^{+}, $$

and

$$\tag{C} D_f K(x_f,t_f;x_i,t_i) ~=~\delta(\Delta t)~\delta(\Delta x), $$

where

$$D_f~:= ~\frac{\partial}{\partial t_f} + \frac{i}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_f^2}+V(x_f)\right)$$ $$\tag{D} ~=~\frac{\partial}{\partial t_f} + \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_f^2}+\frac{i}{\hbar}V(x_f).$$

Perhaps the following heuristic derivation is the most convincing. For sufficiently short times $|\Delta t| \ll \tau$, where $\tau$ is some characteristic time scale, the particle only has time to feel an averaged effect of the potential $V$. So in that limit $|\Delta t| \ll \tau$, the kernel reads$^1$

$$\tag{E} K(x_f,t_f;x_i,t_i) ~=~\theta(\Delta t) \sqrt{\frac{m}{2\pi i\hbar \Delta t}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, $$

where the averaged potential is of the form

$$\tag{F} \langle V\rangle ~=~ V\left(\frac{x_i+x_f}{2}\right)+{\cal O}(\Delta x) ~=~ V(x_f)+{\cal O}(\Delta x)~=~ V(x_i)+{\cal O}(\Delta x). $$

Equation (B) follows directly from eq. (E) via the heat kernel representation

$$\tag{G} \delta(x)~=~ \lim_{|\alpha|\to \infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2}, \qquad {\rm Re}(\alpha)~>~0, $$

of the Dirac delta function.

Next let us consider eq. (C). For sufficiently small times $|\Delta t| \ll \tau$, it is a straightforward to check that eq.(E) satisfies the eq. (C) modulo contributions that vanish as $\Delta t\to 0$, cf. the following Lemma.

Lemma. For sufficiently small times $|\Delta t| \ll \tau$, one has $$\tag{H} D_f K(x_f,t_f;x_i,t_i) ~=~\delta(\Delta t)~\delta(\Delta x)+{\cal O}(\Delta t). $$

Sketched proof of eq. (H): Straightforward differentiation yields

$$ \frac{\partial}{\partial t_f} K(x_f,t_f;x_i,t_i)~\stackrel{(E)}{=}~\delta(\Delta t) \sqrt{\frac{m}{2\pi i\hbar \Delta t}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\} $$ $$\tag{I} -\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar}\left[ \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2+ \langle V\rangle +{\cal O}(\Delta t) \right]\right\} K(x_f,t_f;x_i,t_i). $$

$$ \tag{J} \frac{\hbar}{i}\frac{\partial}{\partial x_f} K(x_f,t_f;x_i,t_i) ~\stackrel{(E)}{=}~\left\{m \frac{\Delta x}{\Delta t} +{\cal O}(\Delta t) \right\} K(x_f,t_f;x_i,t_i). $$

$$ \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_f^2} K(x_f,t_f;x_i,t_i) \qquad $$ $$\tag{K}\qquad ~\stackrel{(E)}{=}~\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar} \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2 +{\cal O}(\Delta t) \right\} K(x_f,t_f;x_i,t_i). $$

Also note that

$$ \tag{L} \left\{V(x_f)-\langle V\rangle \right\}K(x_f,t_f;x_i,t_i) ~\stackrel{(F)}{=}~{\cal O}(\Delta x) K(x_f,t_f;x_i,t_i) ~\stackrel{(B)}{=}~{\cal O}(\Delta t), $$

due to eqs. (B) and (F). End of sketched proof.

The Lemma explains the delta functions on the rhs. of eq. (C). For the remainder of this answer we will assume that $\Delta t\neq 0$ is strictly different from zero, and we will not count delta function contributions. For negative $\Delta t<0$, the eq. (C) is trivially satisfied, cf. footnote $1$. For positive $\Delta t>0$, we use the kernel property

$$\tag{2-31} K(x_f,t_f;x_i,t_i)~=~\int_{\mathbb{R}} \! dx_{m}~ K(x_f,t_f;x_m,t_m)~K(x_m,t_m;x_i,t_i), $$

which is independent of the intermediate time $t_m\in ]t_i,t_f[$. We now pinch the intermediate time $t_m$ sufficiently close to the final time $t_f$, so that we can approximate the kernel $K(x_f,t_f;x_m,t_m)$ by the analog of eq.(E). If we apply the operator $D_f$ on the kernel, we get

$$ D_f K(x_f,t_f;x_i,t_i)~\stackrel{(2-31)}{=}~\int_{\mathbb{R}} \! dx_{m}~ D_f K(x_f,t_f;x_m,t_m)~K(x_m,t_m;x_i,t_i)$$ $$\tag{M} ~\stackrel{(H)}{=}~\int_{\mathbb{R}} \! dx_{m}~ {\cal O}(t_f-t_m)~K(x_m,t_m;x_i,t_i)~=~{\cal O}(t_f-t_m) .$$

Since the lhs. of eq. (M) does not depend on $t_m$, we conclude it is zero. Hence eq.(C) also holds for positive $\Delta t>0$.

References:

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

--

$^1$ Feynman defines the kernel $K(x_f,t_f;x_i,t_i)$ to be zero for negative $\Delta t<0$, see Ref. 1 between eq. (4-27) and eq. (4-28). Note that it is implicitly assumed in eq. (E) that ${\rm Re}(i\Delta t)>0$ is slightly positive for ${\rm Re}(\Delta t)>0$ via pertinent $i\epsilon$-prescription from Wick rotation.

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Thanks for the fantastic answer.However, I am not being able to prove statement H. Could you guide me as to from which term does the $\delta(\Delta t)$ factor comes from. After applying the chain rule my $\frac{1}{(\Delta t)^2}$ terms are cancelling, and I am getting terms involving $\frac{\partial^2 V}{\partial x_f^2}$. How do I handle these? $\delta(\Delta t)$ is nowhere in site. I could show you my full working out, if this is not clear.Please tell me if this is so. –  ramanujan_dirac May 23 '13 at 5:09
    
@ramanujan_dirac: Right, such terms are absorbed into the last term ${\cal O}(\Delta t)$ in eq. (H). –  Qmechanic May 23 '13 at 21:05
1  
I do not agree. The kernel K satisfies Schrodinger equation, so it cannot satisfies OP equation or equations (C) or (H) in the answer. The RHS is simply zero: $D_f K(f, i)$ = 0 –  Trimok May 24 '13 at 9:50
    
@Trimok: Did you remember to differentiate the Heaviside step function $\theta(\Delta t)$ in eq. (E) wrt. $t_f$? –  Qmechanic May 24 '13 at 10:48
    
I updated the answer. –  Qmechanic May 24 '13 at 19:54

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