**Free Particle Kernel from the Path Integral Formulation**

What were doing with our functional integration is integrating over every possible position that the particle can occupy at $t=j\epsilon$. Hence, at each time integration, we are integrating over an infinite number of positions, each of which are weighted by $\exp(iS/\hbar)$, which is minimum for the path (and hence the set of possible positions) that the particle will actually take. We should therefore be convinced that integration over $\exp{(iS/\hbar)}$ will cancel out for all positions that are not near the set of positions that extremize (minimize) $S$. Passing to a discrete functional domain, our integral for the kernel goes to $$\langle x_a|x_b\rangle = \lim_{\epsilon\rightarrow 0}\sqrt{\frac{m}{i2\pi\hbar \epsilon}}\int_{-\infty}^{\infty} \times \cdots $$ $$\cdots\times \int_{-\infty}^{\infty}\textrm{d}x_1\cdots \textrm{d}x_{n-1}\exp{\bigg[i\sum_{j=0}^{n-1}\frac{m}{2}\frac{(x_{j+1}-x_j)^2}{\hbar\epsilon}}\bigg],$$ where I have suppressed the limit. Now we perform a variable change that will make our full calculation of this integral easier to see. The following variable change was done in Shankar and hence much of the following derivation goes along the same lines as Shankar does in his "Principles of Quantum Mechanics" book. In fact, our future exploration of the harmonic oscillator wave function's evolution in free space is motivated by Shankar's chapter on 1-d problems in quantum mechanics. Hence, let us switch variables to $$y_j = \bigg(\frac{m}{2\hbar \epsilon}\bigg)^{1/2}x_j,$$ where $N^n$ now goes as $(2\hbar\epsilon/m)^{(n-1)/2}N^n=A$. With this variable substitution, our path integral is now $$\langle x_b|x_a\rangle =A\int_{-\infty}^{\infty}\times \cdots \times \int_{-\infty}^{\infty}\textrm{d}x_1\cdots \textrm{d}x_{n-1} \exp{\bigg[-\sum_{j=0}^{n-1}\frac{(y_{j+1}-y_j)^2}{i}\bigg]}.$$ Let us consider an integration over $\textrm{d}x_j$ after $j-1$ integrals have already been performed. This will evaluate to $$\xi\int_{-\infty}^{\infty}\times \cdots\times\int_{-\infty}^{\infty}\textrm{d}x_j\cdots \textrm{d}x_{n-1}\exp{\bigg[-\frac{(y_{j+1}-y_j)^2+(y_j-{y_{0}})^2}{i}\bigg]}\cdots$$ $$\cdots\exp\bigg[-\frac{(y_{n}-y_{n-1})^2}{i}\bigg]$$ $$=A\int_{-\infty}^{\infty}\times\cdots\times\int_{-\infty}^{\infty}\textrm{d}x_{j+1}\cdots\textrm{d}x_{n-1}\sqrt{\frac{(i\pi)^j}{j+1}}\exp\bigg[-\frac{(y_{j+1}-y_{0})^2}{(j+1)i}\bigg]\cdots$$ $$\cdots\exp\bigg[-\frac{(y_{j+2}-y_{j+1})^2}{i}\bigg]\cdots\exp\bigg[-\frac{(y_{n}-y_{n-1})^2}{i}\bigg]$$ using another Gaussian integral. We have included $\xi$ to remind the reader that we still have to worry about $A$, but also that our various $j-1$ integrations gives us an overall factor of $\sqrt{(i\pi)^{j-1}/j}$ before we perform our integral. Shankar gives an explicit calculation of this in his text up to $j=2$ (the above calculation is not completely obvious and I highly recommend taking a look at Shankar's approach or the approach taken in Feynman and Hibbs' "Quantum Mechanics and Path Integrals"). We should therefore expect that, if we go over our integrations for each value $x_j$ between $j=1$ and $j=n$ that our final integral will evaluate to $$\langle x_b|x_a\rangle =A \frac{(i\pi)^{(n-1)/2}}{\sqrt{n}}\exp\bigg[-\frac{(y_{n}-y_0)^2}{ni}\bigg].$$ Now if we make our variable change back from $y_j$ to $x_j$, we obtain $$\langle x_b|x_a\rangle = \bigg(\frac{m}{i2\pi \hbar n\epsilon}\bigg)^{1/2}\exp\bigg[\frac{im(x_{n}-x_0)}{2\hbar n\epsilon}\bigg].$$ This is, of course, is not our final answer, as we have suppressed the limit. As $\epsilon\rightarrow 0$, $n\rightarrow \infty$, so $n\epsilon\rightarrow t_b-t_a$. Moreover, $x_{n}-x_0\rightarrow x_b-x_a$, appropriately. Therefore, taking the limit: $$\langle x_b|x_a\rangle =\bigg(\frac{m}{i2\pi\hbar (t_b-t_a)}\bigg)^{1/2}\exp\bigg[\frac{im(x_b-x_a)^2}{2\hbar (t_b-t_a)}\bigg].$$ Great, so there are other, much more simple, ways that we could have obtained this bad boy without the path integral, but we will not discuss them in this blog post. We can now find the solution to the free particle Shrodinger equation, $$\hat{H}|\psi\rangle = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}|\psi\rangle=\frac{p^2}{2m}|\psi\rangle,$$ as an initial value problem using the kernel that we have obtained, such that $$\psi(x_b,t_b)=\int \textrm{d}x\textrm{ } \langle x_b,t_b|x,t\rangle \psi(x,t),$$ where $\psi(x,t)$ is the initial wave function that we must specify.

Resources:

Feynman, Richard Phillips, et al.

*Quantum mechanics and path integrals*. Dover Publications, 2014.

Schwartz, Matthew Dean.

*Quantum field theory and the standard model*. Cambridge University Press, 2017.

Shankar, Ramamurti.

*Principles of quantum mechanics*. Springer, 2014.