Categories: Physics, Quantum mechanics.

Path integral formulation

In quantum mechanics, the path integral formulation is an alternative description of quantum mechanics, which is equivalent to the “traditional” Schrödinger equation. Whereas the latter is based on Hamiltonian mechanics, the former comes from Lagrangian mechanics.

It expresses the propagator $K$ using the following sum over all possible paths $x(t)$ , which all go from the initial position $x_0$ at time $t_0$ to the destination $x_N$ at time $t_N$ :

\begin{aligned} \boxed{ K(x_N, t_N; x_0, t_0) = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar) } \end{aligned}

Where $A$ normalizes. $S[x]$ is the classical action of the path $x$ , whose minimization yields the Euler-Lagrange equation from Lagrangian mechanics. Note that each path is given an equal weight, even unrealistic paths that make big detours.

This apparent problem solves itself, thanks to the fact that paths close to the classical optimum $x_c(t)$ have an action close to $S_c = S[x_c]$ , while the paths far away have very different actions. Since $S[x]$ is inside a complex exponential, this means that paths close to $x_c$ add contructively, and the others add destructively and cancel out.

An interesting way too look at it is by varying $\hbar$ : as its value decreases, minor action differences yield big phase differences, which make the quantum wave function stay closer to $x_c$ . In the limit $\hbar \to 0$ , quantum mechanics thus turns into classical mechanics.

Time-slicing derivation

The most popular way to derive the path integral formulation proceeds as follows: starting from the definition of the propagator $K$ , we divide the time interval $t_N - t_0$ into $N$ “slices” of equal width $\Delta t = (t_N - t_0) / N$ , where $N$ is large:

\begin{aligned} K(x_N, t_N; x_0, t_0) &= \matrixel{x_N}{e^{- i \hat{H} (t_N - t_0) / \hbar}}{x_0} = \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar} \cdots e^{- i \hat{H} \Delta t / \hbar}}{x_0} \end{aligned}

Between the exponentials we insert $N\!-\!1$ identity operators $\hat{I} = \int \Ket{x} \Bra{x} \dd{x}$ , and define $x_j = x(t_j)$ for an arbitrary path $x(t)$ :

\begin{aligned} K &= \int\cdots\int \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar}}{x_{N-1}} \cdots \matrixel{x_1}{e^{- i \hat{H} \Delta t / \hbar}}{x_0} \dd{x_1} \cdots \dd{x_{N - 1}} \end{aligned}

For sufficiently small time steps $\Delta t$ (i.e. large $N$ we make the following approximation (which would be exact, were it not for the fact that $\hat{T}$ and $\hat{V}$ are operators):

\begin{aligned} e^{- i \hat{H} \Delta t / \hbar} = e^{- i (\hat{T} + \hat{V}) \Delta t / \hbar} \approx e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar} \end{aligned}

Since $\hat{V} = V(x_j)$ , we can take it out of the inner product as a constant factor:

\begin{aligned} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}}{x_j} = e^{- i V(x_j) \Delta t / \hbar} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j} \end{aligned}

Here we insert the identity operator expanded in the momentum basis $\hat{I} = \int \Ket{p} \Bra{p} \dd{p}$ , and commute it with the kinetic energy $\hat{T} = \hat{p}^2 / (2m)$ to get:

\begin{aligned} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j} = \int_{-\infty}^\infty \Inprod{x_{j+1}}{p} \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \Inprod{p}{x_j} \dd{p} \end{aligned}

In the momentum basis $\Ket{p}$ , the position basis vectors are represented by plane waves:

\begin{aligned} \Inprod{p}{x_j} = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( \!-\! i \frac{x_j p}{\hbar} \Big) \qquad \Inprod{x_{j+1}}{p} = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( i \frac{x_{j+1} p}{\hbar} \Big) \end{aligned}

With this, we return to the inner product and further evaluate the integral:

\begin{aligned} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j} &= \frac{1}{2 \pi \hbar} \int_{-\infty}^\infty \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \exp\!\Big(i \frac{(x_{j+1} - x_j) p}{\hbar}\Big) \:dp \\ &= \frac{1}{2 \pi \hbar} \sqrt{\frac{2 \pi m \hbar}{i \Delta t}} \exp\!\Big( i \frac{m (x_{j+1} - x_j)^2}{2 \hbar \Delta t} \Big) \end{aligned}

Inserting this back into the definition of the propagator $K(x_N, t_N; x_0, t_0)$ yields:

\begin{aligned} K = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} \int\cdots\int \exp\!\bigg(\! \sum_{j = 0}^{N - 1} i \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \hbar \Delta t} - \frac{V(x_j) \Delta t}{\hbar} \Big) \!\bigg) \dd{x_1} \cdots \dd{x_{N-1}} \end{aligned}

For large $N$ and small $\Delta t$ , the sum in the exponent becomes an integral:

\begin{aligned} \frac{i}{\hbar} \sum_{j = 0}^{N - 1} \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \Delta t^2} - V(x_j) \Big) \Delta t \quad \to \quad \frac{i}{\hbar} \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau} \end{aligned}

Upon closer inspection, this integral turns out to be the classical action $S[x]$ , with the integrand being the Lagrangian $L$ :

\begin{aligned} S[x(t)] = \int_{t_0}^{t_N} L(x, \dot{x}, \tau) \dd{\tau} = \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau} \end{aligned}

The definition of the propagator $K$ is then further reduced to the following:

\begin{aligned} K = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} \int\cdots\int \exp(i S[x] / \hbar) \dd{x_1} \cdots \dd{x_{N-1}} \end{aligned}

Finally, for the purpose of normalization, we define the integral over all paths $x(t)$ as follows, where we write $D[x]$ instead of $\dd{x}$ :

\begin{aligned} \int D[x] \equiv \lim_{N \to \infty} \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} \int\cdots\int \dd{x_1} \cdots \dd{x_{N-1}} \end{aligned}

We thus arrive at Feynman’s path integral, which sums over all possible paths $x(t)$ :

\begin{aligned} K = \int \exp(i S[x] / \hbar) \:D[x] = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar) \end{aligned}

References

R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.
L.E. Ballentine, Quantum mechanics: a modern development, 2nd edition, World Scientific.