2.4. The path integral: from quantum mechanics to classical mechanics

In the previous section we have considered how we can arrive from classical mechanics at the Schrödinger equation. This formalism can be generalised in the sense that for each system for which we can write down a Lagrangian, we have a way to find a quantum formulation in terms of the path integral. Whether a Schrödinger-like equation can be found is not sure: sometimes we run into problems which are beyond the scope of these notes. In this section we assume that we have a system described by some Hamiltonian and show that the time-evolution operator has the form of a path integral as found in the previous section.

The starting point is the time evolution operator, or propagator, which, for a time-independent Hamiltonian, takes the form $$U({\bf r}_{\text{f}}, t_{\text{f}}; {\bf r}_{\text{i}}, t_{\text{i}}) = \left\langle {\bf r}_{\text{f}} \right| e^{-\frac{{\rm i}}{\hslash}(t_{\text{f}}-t_{\text{i}}) \hat{H}} \left| {\bf r}_{\text{i}} \right\rangle.$$ The matrix element is difficult to evaluate -- the reason is that the Hamiltonian which, for a particle in one dimension, takes the form $$\hat{H} = - \frac{\hslash^2}{2m} \frac{d^2}{dx^2} + V(x)$$ is the sum of two noncommuting operators. Although it is possible to evaluate the exponents of the separate terms occurring in the Hamiltonian, the exponent of the sum involves an infinite series of increasingly complicated commutators. For any two noncommuting operators $\hat{A}$ and $\hat{B}$ we have $$e^{\hat{A}+\hat{B}} = e^{\hat{A}} e^{\hat{B}} e^{- 1/2 [\hat{A},\hat{B}] - 1/12 ([\hat{A},[\hat{A},\hat{B}]] + [\hat{B},[\hat{B},\hat{A}]]) + 1/24 [\hat{A},[\hat{B},[\hat{A},\hat{B}]]] +\ldots}$$ This is the so-called Campbell--Baker--Hausdorff (CBH) formula. The cumbersome commutators occurring on the right can only be neglected if the operators $A$ and $B$ are small in some sense. We can try to arrive at an expression involving small commutators by applying the time slicing procedure of the previous section: $$e^{-\frac{{\rm i}}{\hslash}(t_{\text{f}}-t_{\text{i}}) \hat{H}} = e^{-\frac{{\rm i}}{\hslash} \Delta t \hat{H}} e^{-\frac{{\rm i}}{\hslash} \Delta t \hat{H}} e^{-\frac{{\rm i}}{\hslash} \Delta t \hat{H}} \ldots$$ Note that no CBH commutators occur because $\Delta t \hat{H}$ commutes with itself.

Having this, we can rewrite the propagator as (we omit the hat from the operators) $$U(x_{\text{f}}, t_{\text{f}}; x_{\text{i}}, t_{\text{i}}) = \int dx_{1} \ldots dx_{N-1} \left\langle x_{\text{f}} \right| e^{-{\rm i}\Delta t H/\hslash} \left| x_{N-1} \right\rangle \left\langle x_{N-1} \right| e^{-i \Delta t H/\hslash} \left| x_{N-2} \right\rangle \cdots \left\langle x_1 \right| e^{-{\rm i}\Delta t H/\hslash} \left| x_{\text{i}} \right\rangle$$ Now that the operators occurring in the exponents can be made arbitrarily small by taking $\Delta t$ very small, we can evaluate the matrix elements explicitly: $$\left\langle x_j \right| e^{-{\rm i}\Delta t H} \left| x_{j+1} \right\rangle = \left\langle x_j \right| e^{-\frac{{\rm i}\Delta t}{\hslash} [p^2/(2m) + V(x)]} \left| x_{j+1} \right\rangle = e^{-{\rm i}\Delta t V(x_j) / \hslash} \left\langle x_j \right| e^{-\frac{{\rm i}\Delta t}{\hslash} p^2/(2m)} \left| x_{j+1} \right\rangle.$$ The last matrix element can be evaluated by inserting two unit operators formulated in terms of integrals over the complete sets $\left| p \right\rangle$: $$\left\langle x \right|e^{-\frac{{\rm i}\Delta t}{\hslash} p^2/(2m)} \left| x' \right\rangle = \int \int \left\langle x | p \right\rangle \left\langle p \right| e^{-\frac{{\rm i}\Delta t}{\hslash} \hat{p}^2/(2m)} \left| p' \right\rangle \left\langle p' | x' \right\rangle \; dp \, dp'.$$ We have seen that $\left\langle x | p \right\rangle= \exp({\rm i}px/\hslash)/\sqrt{2\pi\hslash}$. Realising that the operator $\exp\left[-\frac{{\rm i}\Delta t}{\hslash} \hat{p}^2/(2m)\right]$ is diagonal in $p$ space, we find, after integrating over $p$: $$\left\langle x | e^{-\frac{{\rm i}\Delta t}{\hslash} p^2/(2m)}| x' \right\rangle = \sqrt{\frac{m}{2\pi \hslash \Delta t}} \exp\left[{\rm i}m(x-x')^2/(2\Delta t\hslash)\right].$$ All in all we have $$\left\langle x_j | e^{-{\rm i}\Delta t H/\hslash} | x_{j+1} \right\rangle = \sqrt{\frac{m}{2\pi \hslash \Delta t}} e^{-{\rm i}\Delta t V(x_j) / \hslash} \exp\left[{\rm i}m(x-x')^2/(2\Delta t\hslash)\right].$$ Note that we have evaluated matrix elements of operators. The result is expressed completely in terms of numbers, and we no longer have to bother about commutation relations. Collecting all terms together we obtain $$U(x_{\text{f}}, t_{\text{f}}; x_{\text{i}}, t_{\text{i}}) = \left( {\frac{m}{2\pi \hslash \Delta t}} \right)^{(N-1)/2} \int dx_1 \ldots dx_{N-1} \exp\left\{ \frac{{\rm i}}{\hslash} \sum_{j=0}^{N} \left[ \frac{m(x_{j+1}-x_j)^2}{2} - V(x_j) \right] \Delta t\right\}.$$ The expression in the exponent is the discrete form of the Lagrangian; the integral over all intermediate values $x_j$ is the sum over all paths. We therefore have shown that the time evolution operator from $x_{\text{i}}$ to $x_{\text{f}}$ is equivalent to the sum of the phase factors of all possible paths from $x_{\text{i}}$ to $x_{\text{f}}$.

In conclusion, we have seen that the idea that the probability to a find a particle starting off from $x_0$ at $t_0$ at $x_1$ at time $t_1$ is given by a sum of all the phase factors corresponding to all paths from the starting to the end point, gives us the quantum theory. On the other hand, once we have the Hamiltonian of a quantum theory, we can reformulate this as a path integral involving the corresponding Lagrangian.