In the last post, I presented a plausibility argument for the Feynman path integral. Central to this argument is the identification of a relation between the transition amplitude $$\left< \vec r, t \right. \left| \vec r_0, t_0 \right>$$ and the classical action given by
\[ \left< \vec r, t \right. \left| \vec r_0, t_0\right> \sim N e^{\frac{i S}{\hbar}} \;. \]
However, because of the composition property of the quantum propagator, we were forced into evaluating the action not just for the classical path but also for all possible paths in that obey causality (i.e. were time ordered).
In this post I will evaluate the closed form for the free particle propagator and will show how to get the same result from the path integral. Along the way, it will also be noted that the same result obtains when only the classical path and action are used. This strange property holds for a variety of systems more complex than the free particle as is proven in Shankar’s book. My presentation here follows his discussion in Chapter 8.
To truly appreciate the pros and cons of working with the path integral, let’s start by first deriving the quantum propagator for the free particle using a momentum space representation. To keep the computations clearer, I will work only in one dimension. The Hamiltonian for a free particle is given in the momentum representation by
\[ H = \frac{p^2}{2m} \]
and in the position representation by
\[ H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \; .\]
Since the Hamiltonian is purely a function of momentum and is an algebraic function in the momentum representation it is a easier to work with than if it were expressed in the position representation.
Because the momentum operator $$\hat P$$ commutes with the Hamiltonian, the momentum eigenkets for a natural diagonal basis for the Hamiltonian with the energy being given by
\[ E(p) = p^2/2m \; \]
This basis is clearly doubly degenerate for each given value of $$E_p$$ with a right-going momentum $$p_R = +\sqrt{2mE_p}$$ and a left-going momentum $$p_L = -\sqrt{2mE_p}$$ both having the same energy.
The Schrodinger equation in the momentum representation is
\[ \frac{p^2}{2m} \psi(p,t) = i \hbar \partial_t \psi(p,t) \; , \]
which has easily-obtained solutions
\[ \psi(p,t) = e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t-t_0)} \; .\]
The quantum evolution operator can now be expressed in the momentum basis as
\[ U(t_f,t_0) = \int_{-\infty}^{\infty} dp \, \left| p \right> \left< p \right| e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t-t_0)} \; \] By sandwiching the evolution operator between two position eigenkets \[ K^+(x_f,t_f;x_0,t_0) = \left< x_f \right| U(t_f,t_0) \left| x_0 \right> \theta(t_f-t_0)\]
we arrive at the expression for the forward-time propagator
\[ K^+(x_f,t_f;x_0,t_0) = \int_{-\infty}^{\infty} dp \, \left< x_f \right. \left| p \right> \left< p \right. \left| x_0 \right> e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t_f-t_0)} \theta(t_f-t_0)\; .\]
In the remaining computations, it will be understood that $$t_f > t_0$$ and so I will drop the explicit reference to the Heaviside function. This equation can be can be evaluated by using
\[ \left< p | x \right> = \frac{1}{\sqrt{2 \pi \hbar}} e^{-\frac{ipx}{\hbar} } \]to give
\[ K^+(x_f,t_f;x_0,t_0) = \frac{1}{2 \pi \hbar} \int_{-\infty}^{\infty} dp \, e^{\frac{i}{\hbar}p(x_f-x_0)} e^{-\frac{i}{\hbar} \frac{p^2}{2m} (t_f-t_0)}\; .\]
The integral for the propagator can be conveniently written as
\[ K^+(x_f,t_f;x_0,t_0) = \frac{1}{2 \pi \hbar} \int_{-\infty}^{\infty} dp \, e^{-a p^2 + b p} \; , \]
where
\[ a = \frac{i (t_f – t_0)}{2 m \hbar} \]
and
\[ b = \frac{i(x_f – x_0)}{\hbar} \; .\]
Using the standard Gaussian integral
\[ \int_{-\infty}^{\infty} dx \, e^{-ax^2+bx} = e^{b^2/2a} \sqrt{\frac{\pi}{a}} \; ,\]
we arrive at the exact answer for the free-particle, forward-time quantum propagator
\[ K^+(x_f,t_f;x_0,t_0) = \sqrt{ \frac{m}{2\pi i\hbar(t_f-t_0)} } e^{\frac{i m}{2\hbar}\frac{ (x_f-x_0)^2}{(t_f-t_0)}} \; .\]
Now we turn to performing the same computation using the path integral approach. The first step is to express the classical action
as a function of the path. The Lagrangian only consists of the kinetic energy and so
\[ S = \int_{t_0}^{t_f} L[x(t)] dt = \int_{t_0}^{t_f} \frac{1}{2} m {\dot x} ^2 \; .\]
The basic idea of the path integral is to look at the quantum evolution across many small time steps so that each step can be handled more easily. In keeping with this idea, the action integral can be approximated as a sum by the expression
\[ S = \sum_{i=0}^{N-1} \frac{m}{2} \left( \frac{x_{i+1} – x_{i}}{\epsilon} \right)^2 \epsilon \; ,\]
where $$\epsilon$$ is the time step. The forward-time propagator is now written as
\[ K^+(x_f,t_f;x_0,t_0) = \lim_{N\rightarrow \infty, \epsilon \rightarrow 0} Q \int_{-\infty}^{\infty} dx_1 \int_{-\infty}^{\infty} dx_2 … \\ \exp \left[ \frac{i m}{2 \hbar} \sum_{i=0}^{N-1} \frac{(x_{i+1} – x_{i})^2}{\epsilon} \right] \; ,\]
where $$Q$$ is a normalization that will have to be determined at the end. The form of the action gives us hope that these integrals can be evaluated, since the term $$x_{i+1}-x_i$$ connects the positions on only two time slices. For notational convenience we’ll define an intermediate set of integrals
\[ I = \lim_{N\rightarrow\infty} Q \int_{-\infty}^{\infty} dx_1…dx_n \\ \exp \left[ i q \left( (x_N-x_{N-1})^2 + … + (x_2 – x_1)^2 + (x_1 – x_0)^2 \right) \right] \; \]
with $$q = \frac{m}{2 \hbar \epsilon}$$.
To start, let’s work on the $$x_1$$ integral. Since it only involves $$x_0$$ and $$x_2$$ this amounts to solving
\[ I_1 = \int_{-\infty}^{\infty} dx_1 \exp \left\{ i q \left[ 2 x_1^2 – 2(x_2 + x_0) x_1 + (x_2^2 + x_0^2) \right] \right\} \; ,\]
which can be done using essentially the same Gaussian integral as above
\[ \int_{-\infty}^{\infty} e^{-ax^2 + bx + c} = exp(b^2/4a +c) \sqrt{\frac{\pi}{a}} \; .\]
This results in
\[ I_1 = \frac{1}{\sqrt{2}} \left( \frac{i\pi}{q} \right)^{1/2} e^{i q \frac{(x_2-x_0)}{2}} \; .\]
Now the next integral to solve is
\[ I_2 = \int_{-\infty}^{\infty} dx_2 \exp \left\{ i q \left[ (x_3-x_2)^2 + (x_2-x_0)^2/2 \right] \right\} \; .\]
Rather than go through this in detail, I wrote some Maxima code to carry these integrals out
calc_int(integrand,ivar) := block([f,a,b,c],
f : integrand,
a : coeff(expand(f),ivar^2),
b : coeff(expand(f),ivar),
c : ratsimp(f-a*ivar^2-b*ivar),
a : -1*a,
sqrt(%pi/a)*exp(factor(ratsimp(b^2/(4*a) + c))
and the results for up through $$I_4$$ are
\[ I_2 = \frac{1}{\sqrt{3}} \left( \frac{i\pi}{q} \right)^{2/2} e^{i q \frac{(x_3-x_0)}{3}} \; , \]
\[ I_3 = \frac{1}{\sqrt{4}} \left( \frac{i\pi}{q} \right)^{3/2} e^{i q \frac{(x_4-x_0)}{4}} \; ,\]
and
\[ I_4 = \frac{1}{\sqrt{5}} \left( \frac{i\pi}{q} \right)^{4/2} e^{i q \frac{(x_5-x_0)}{5}} \; \]
yielding the result for $$N$$
\[ I = Q \frac{1}{\sqrt{N}} \left( \frac{i\pi}{q} \right)^{\frac{N-1}{2}} e^{i q \frac{(x_N-x_0)}{N}} \; .\]
With all the factors in $$q$$ now fully restored, we get
\[ I = \frac{Q}{\sqrt{N}} \left( \frac{2 \pi i \hbar \epsilon}{m} \right)^{\frac{N-1}{2}} e^{\frac{i m (x_N-x_0)^2}{2 \hbar N \epsilon}} \; .\]
Setting
\[ Q = \left( \frac{m}{2 \pi i \hbar \epsilon} \right)^{\frac{N}{2}} \]
gives
\[ I = \left( \frac{m}{2 \pi i \hbar N \epsilon} \right) e^{\frac{i m (x_N-x_0)^2}{2 \hbar N \epsilon}} \; .\]
Taking the limit as $$N \rightarrow \infty$$, $$\epsilon \rightarrow 0$$, and $$N \epsilon = (t_f – t_0)$$ yields
\[ K^+(x_f,t_f;x_0,t_0) = \sqrt{ \frac{m}{2\pi\hbar i (t_f-t_0)} } e^{i m (x_f-x_0)^2/2\hbar (t_f-t_0)} \; ,\]
which is the exact answer that was obtained above.
While this was a lot more work than the momentum-representation path, it is interesting to note that Shankar proves that
any potential with the form
\[ V = a + bx + c x^2 + d \dot x + e x \dot x \]
yields immediately the forward-time propagator
\[K^+(x_f,t_f;x_0,t_0) = e^{i S_{cl}/h} Q(t) \]
where $$S_{cl}$$ is the classical action and $$Q(t)$$ is a function solely of time that generally cannot be determined. Shankar shows, in the case of a free particle, that
\[ S_{cl} = \int_{t_0}^{t_f} L[x(t)] dt = \frac{m v_{av}^2}{2} (t_f-t_0) = \frac{m}{2} \frac{(x_f-x_0)^2}{t_f-t_0} \]
yielding
\[ K^+(x_f,t_f;x_0,t_0) = Q(t) \exp \left[ \frac{i m (x_f – x_0)^2}{2 \hbar (t_f – t_0)} \right] \; , \]
where $$Q(t)$$ can be determined from the requirement that $$K^+(x_f,t_f;x_0,t_0) = \delta(x_f – x_0)$$ when $$t_f = t_0$$. The applicable identity is
\[ \delta(x_f – x_0) = \lim_{\sigma \rightarrow 0} \frac{1}{\sqrt{\pi \sigma^2}} \exp \left[ -\frac{(x_f – x_0)^2}{\sigma^2} \right] \]
from which we immediately get
\[ Q(t) = \sqrt{ \frac{m}{2\pi \hbar i (t_f-t_0)}} \; .\]
These results mean that a host of practical problems that have intractable propagators in any given representation (due to the need to find the eigenfunctions and then plug them into a power series representing the exponential) can now be calculated with relative ease.