# Proof for getting delta function on $t \to t_0$ from the equation of the propagator for the free particle in 1 dimension

From Sakurai's quantum mechanics equation 2.5.16 give propagator for a free particle in 1 dimension. Equation 2.5.16 is

$$K (x^",t;x',t_0)=\sqrt {m\over {2\pi i\hbar (t-t_0)}} \exp \Biggl [{im (x^"-x')^2 \over 2 \hbar (t-t_0) }\Biggr]\tag{2.5.16}$$

When $$\lim _{t \to t_0}$$ propagator $$K (x^",t;x',t_0)$$ must equal to a dirac delta function. Sakurai also states this condition in equation 2.5.9

$$\lim_{t \to t_ 0} K (x^",t;x',t_0) = \delta (x^"-x') .\tag{2.5.9}$$

Can any one spend some of your precious little time to mathematically show this?

Note that to make mathematical sense of the limit (2.5.9) it is implicitly assumed that $${\rm Re}(i\Delta t)~>~0\tag{1}$$ is slightly positive via the pertinent Feynman $$i\epsilon$$-prescription. After analytic continuation via a Wick rotation $$\Delta\tau~\equiv~ i\Delta t,\tag{2}$$ eq. (2.5.16) is just the heat kernel representation$$^1$$ $$\delta(x)~=~ \lim_{|\alpha|\to \infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2}, \qquad {\rm Re}(\alpha)~>~0, \tag{3}$$ of the Dirac delta distribution. Here we have identified $$\alpha~\equiv~\frac{m}{2\hbar\Delta\tau}.\tag{4}$$
$$^1$$ To prove the heat kernel representation (3), insert a test function on each side.
• You use $|\alpha | \to \infty$ this means if $\alpha$ is negative then it goes to $-\infty$ and if $\alpha$ is positive it goes to $\infty$.In the case of limit goes to $- \infty$ we get $e^{ positive number }$ this is not a delta function am I right? @Qmechanic Feb 29, 2020 at 6:17
• $\uparrow$ Yes. Feb 29, 2020 at 6:47
• If we take the case of only $\alpha$ is positive then we cannot get $\infty$ at $x=0$, Then how can you say we get a delta function?@Qmechanic Feb 29, 2020 at 7:13
• Note that the square root blows up, and if $x=0$ it is not tamed by the exponential. Feb 29, 2020 at 8:24