path: root/source/know/concept/path-integral-formulation/index.md

---
title: "Path integral formulation"
sort_title: "Path integral formulation"
date: 2021-07-03
categories:
- Physics
- Quantum mechanics
layout: "concept"
---

The **path integral formulation** is an alternative description
of quantum mechanics, equivalent to the traditional Schrödinger equation.
Whereas the latter is based on [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
the former comes from [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).

It expresses the [propagator](/know/concept/propagator/) $$K$$
as the following "sum" over all possible paths $$x(t)$$
that take the particle from the starting point $$(x_0, t_0)$$
to the destination $$(x_N, t_N)$$:

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

Where $$A$$ is a normalization constant,
and $$S[x]$$ is the classical action of the path $$x(t)$$,
defined as shown below from the system's Lagrangian $$L$$,
and whose minimization would lead to the
[Euler-Lagrange equation](/know/concept/euler-lagrange-equation/)
of classical Lagrangian mechanics.
Let $$\dot{x}(t) = \idv{x}{t}$$:

$$\begin{aligned}
    S[x]
    \equiv \int_{t_0}^{t_N} L(x, \dot{x}, \tau) \dd{\tau}
\end{aligned}$$

Note that $$K$$'s sum gives each path an equal weight,
even unrealistic paths taking bigs detours.
This apparent problem solves itself as follows:
paths close to the classical optimum $$x_c(t)$$
have an action close to $$S_c = S[x_c]$$,
since $$S$$ is stationary there.
Meanwhile, for paths far away from $$x_c$$,
$$S$$ gives very different values,
which change by a lot if a small change is made to $$x$$.
Because $$S[x]$$ is inside a complex exponential,
paths close to $$x_c$$ therefore add more or less constructively,
while the others add destructively and cancel out.

Consequently, the "quantum path" is still close to $$x_c(t)$$.
An interesting way to think about this is by treating $$\hbar$$ as a parameter:
as its value decreases, small action changes result in bigger phase differences,
which makes the quantum wavefunction stay closer to $$x_c$$
for the aforementioned reasons.
In the limit $$\hbar \to 0$$, quantum mechanics simply turns into classical mechanics.

In reality, $$K$$'s sum is evaluated as an integral over all paths $$x(t)$$,
hence this is called the *path integral formulation*.
The proof that the propagator $$K$$'s Schrödinger-picture definition
can be rewritten as such an integral is given below.



## Time-slicing derivation

For a time-independent Hamiltonian $$\hat{H}$$,
we start from the definition of the propagator $$K$$,
and divide the time interval $$t_N \!-\! t_0$$ into $$N$$ "slices"
of equal width $$\Delta{t} \equiv (t_N \!-\! t_0) / N$$:

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

Between the exponentials we insert identity operators
$$\int_{-\infty}^\infty \Ket{x} \Bra{x} \dd{x}$$,
and define $$x_j \equiv x(t_j)$$ for an arbitrary path $$x(t)$$,
where $$t_j$$ is the endpoint of the $$j$$th slice.
This is equivalent to splitting $$K$$
into a product of all slices' individual propagators:

$$\begin{aligned}
    K
    &=  K(x_N, t_N; x_{N-1}, t_{N-1})
    \cdots K(x_2, t_2; x_1, t_1) \: K(x_1, t_1; x_0, t_0)
    \\
    &= \int \!\cdots \! \int
    \matrixel{x_N}{e^{- i \hat{H} \Delta{t} / \hbar}}{x_{N-1}}
    \cdots \matrixel{x_1}{e^{- i \hat{H} \Delta{t} / \hbar}}{x_0}
    \dd{x_1} \cdots \dd{x_{N - 1}}
\end{aligned}$$

For sufficiently small time steps $$\Delta{t}$$ (i.e. large $$N$$),
we can split the Hamiltonian
into its kinetic and potential terms $$\hat{H} = \hat{T} + \hat{V}$$.
Note that this is an approximation,
since $$\hat{T}$$ and $$\hat{V}$$ are operators that do not commute,
but it becomes exact in the limit $$\Delta{t} \to 0$$:

$$\begin{aligned}
    e^{- i \hat{H} \Delta{t} / \hbar}
    \approx e^{- i \hat{T} \Delta{t} / \hbar} \: e^{- i \hat{V} \Delta{t} / \hbar}
\end{aligned}$$

We substitute $$\hat{V} = V(x_j)$$, and apply it directly to $$\ket{x_j}$$,
such that we can take it out of the inner product as a constant factor:

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

In order to evaluate the remaining inner product,
we insert the identity operator again,
this time expanded in the momentum basis $$\int_{-\infty}^\infty \Ket{p} \Bra{p} \dd{p}$$,
and use $$\hat{T} = \hat{p}^2 / (2m)$$ to get:

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

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

$$\begin{aligned}
    \inprod{p}{x}
    = \frac{e^{- i x p / \hbar}}{\sqrt{2 \pi \hbar}}
\end{aligned}$$

Inserting this and looking up the resulting integral,
we arrive at:

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

Including the factor due to $$\hat{V}$$,
we find that the propagator of a single time slice is:

$$\begin{aligned}
    \matrixel{x_{j+1}}{e^{- i \hat{H} \Delta t / \hbar}}{x_j}
    = \sqrt{\frac{- i m}{2 \pi \hbar \Delta{t}}}
    \exp\!\bigg( \frac{i}{\hbar} \frac{m}{2} \frac{(x_{j+1} \!-\! x_j)^2}{\Delta{t}} - \frac{i}{\hbar} V(x_j) \: \Delta{t} \bigg)
\end{aligned}$$

This is a "local" result;
inserting it into the "global" propagator $$K(x_N, t_N; x_0, t_0)$$ yields:

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

It is worth noting that there are $$N\!-\!1$$ integrals,
but $$N$$ factors $$(-i m / 2 \pi \hbar \Delta{t})^{1/2}$$
i.e. one for each slice.
According to convention, $$N\!-\!1$$ of those factors
are said to belong to the integrals,
and then the remaining one belongs to the process as a whole.

In the limit $$\Delta{t} \to 0$$ (or $$N \to \infty$$),
the sum in the exponent becomes an integral:

$$\begin{aligned}
    \lim_{\Delta{t} \to 0}
    \sum_{j = 0}^{N - 1} \bigg( \frac{m}{2} \frac{(x_{j+1} \!-\! x_j)^2}{\Delta{t}^2} - V(x_j) \bigg) \Delta{t}
    \:\:&=\:\:
    \int_{t_0}^{t_N} \!\bigg( \frac{1}{2} m \dot{x}^2 - V(x) \bigg) \dd{\tau}
    \\
    \:\:&=\:\:
    \int_{t_0}^{t_N} L(x, \dot{x}, \tau) \dd{\tau}
    \\
    \:\:&=\:\:
    S[x]
\end{aligned}$$

Where we have recognized the Lagrangian $$L = T - V$$
and hence the action $$S[x]$$ of the path $$x(t)$$.
We thus arrive at the following formula for the global propagator $$K$$,
known as **Feynman's path integral**
or sometimes the **configuration space path integral**:

$$\begin{aligned}
    \boxed{
        K
        = \int e^{i S[x] / \hbar} \:\mathcal{D}{x}
    }
\end{aligned}$$

Where we have introduced the following notation
to indicate an integral over all paths,
because writing the factor and all those integrals can become tedious:

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

It is worth stressing that this is simply an abbreviation;
in practice, calculating $$K$$ in this way
still requires the individual slices to be taken into account.



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