# Feynman Path Integral

Exploring Quantum Physics – Week 2, Lecture 3

Developed in Feynman’s 1948 paper: Space-Time Approach to Non-Relativistic Quantum Mechanics

## Summary

For a particle traveling from a coordinate $\vec{R}_i \longrightarrow \vec{R}_f$
The probability of going to $\vec{R}_f$ is the sum of the probability (or weight) of taking a particular path, over all possible paths:
$\displaystyle \vec{R}_f = \omega_{i\shortrightarrow f} = \left| \sum_l e^{i \frac{S_l}{\hbar}} \right|^2$
$\displaystyle \mathrm{where}\; S_l = \mathrm{action\; of\; trajectory\;} l = \int_0^t [K - V] \;dt$

## Trajectories

• Consider a particle localized at $\vec{R}_i$ at time t =0, that is $| \Psi(0) \rangle = | \vec{R}_i \rangle$
• Evolve according to Schrödinger:
$\displaystyle i \hbar \partial_t | \Psi \rangle = \hat{H} | \Psi \rangle = \left[ \frac{\hat{\vec{p}}^2}{2m} + V(\hat{\vec{R}}) \right] |\Psi\rangle$
• What “part” will propagate to $\vec{R}_f$?
i.e. what is the overlap with $\Psi$, $\langle \vec{R}_f | \Psi{t} \rangle = ??$
• How will it get there? (In Q.M. it follows all possible trajectories)

## Evolution Operator

$\displaystyle i \hbar | \Psi(t) \rangle = \hat{H} | \Psi(t) \rangle \\ | \Psi(0) \rangle = | \Psi_0 \rangle = | \vec{R}_i \rangle \\ | \Psi(t) \rangle = \hat{U}(t) | \Psi_0 \rangle$
Where $\hat{U}(t)$ is the evolution operator

Evolution operator must preserve the vector norm (according to the Born interpretation), so it looks like a rotation.

Plug into Schrödinger:
$\displaystyle i\hbar \partial_t \hat{U}(t) = \hat{H}\hat{U}(t), \qquad \hat{U}(0) = \hat{1}$
i.e. $\hat{U}$ does not depend on the initial condition of $| \Psi_0 \rangle$
$\displaystyle \hat{U}(t) = e^{-\frac{i}{\hbar}\hat{H}t}$
Verify:
$\displaystyle i \hbar \partial_t \left(e^{-\frac{i}{\hbar}\hat{H}t}\right) = \hat{H} \left(e^{-\frac{i}{\hbar}\hat{H}t} \right) = \hat{H}\hat{U}(t)$

$\Rightarrow | \Psi(t) \rangle = e^{-\frac{i}{\hbar}\hat{H}t} | \vec{R}_i \rangle$

Asside: define the exponential of an operator, $\hat{X}$ with it’s Taylor series:
$e^{\hat{X}} = 1 + \hat{X} + \frac{\hat{X}^2}{2!} + \frac{\hat{X}^3}{3!} + \ldots$

## Propagator

a.k.a. the transition amplitude

$\displaystyle \langle \vec{R}_f | \Psi(t) \rangle = \langle \vec{R}_f | e^{-\frac{i}{\hbar}\hat{H}t} | \vec{R}_i \rangle = G(\vec{R}_i, \vec{R}_f; t) = \mathrm{"Green function"}$

Exponential of an operator is messy (infinite sum, see above) so simplify using:

• $\int | x \rangle \langle x | \; dx = \hat{1}$

• $\begin{array}{ll} | \Psi(t) \rangle &= \hat{U}(t) | \Psi(0) \rangle \\ &= \hat{U}(t - t_1) | \Psi(t_1) \rangle \\ &= \hat{U}(t - t_1) \hat{U}(t_1) | \Psi(0) \rangle \end{array}$

Combining the above:
$\displaystyle \langle x_f | \hat{U}(t) | x_i \rangle = \langle x_f | \hat{U}(t/2) \hat{U}(t/2) | x_i \rangle = \int \langle x_f | \hat{U}(t/2) | x \rangle \langle x | \hat{U}(t/2) | x_i \rangle \; dx$

Consider all possible intermediate points on the path from x_i to x_f

Reapplying this N times gives:
$\displaystyle \hat{U}(t) = \hat{U}(t/N)^N = \hat{U}(t/N) \times \hat{U}(t/N) \times \ldots \times \hat{U}(t/N)$

Let $N \rightarrow \infty, \quad \Delta t = t/N$ becomes small, and using a first order approximation:
$\displaystyle e^{-\frac{i}{\hbar N} \hat{H}t} \approx 1 - \frac{i}{\hbar} \hat{H} \Delta t$
$\displaystyle \mathrm{i.e.}\; e^{-\frac{i}{\hbar}\hat{H}t} = \left( 1 - \frac{i}{\hbar} \hat{H} \frac{t}{N} \right)^N, \quad N \rightarrow \infty$

Propagator becomes a product:
$\displaystyle \langle x_f | \hat{U}(t) | x_i \rangle = \int dx_1 \ldots x_{N-1} \prod_{k=0}^N \langle x_{k+1} | 1 - \frac{i}{\hbar N} \hat{H} t | x_k \rangle$

Asside: difference between $| \Psi \rangle, \; \Psi(x)$

$| \Psi \rangle$ – Represents an “abstract” state of the system, coordinate-independent
$\Psi(x)$ – Represents $| \Psi \rangle$ in a coordinate system. $\Psi(x) = \langle x | \Psi \rangle$

Asside: the Dirac delta-function

• $\delta_{ij} = \vec{e}_i \cdot \vec{e}_j = \left\{ \begin{array}{ll}0 & i \neq j \\ 1 & i = j \end{array} \right.$
• Extension to a “continuum” basis $\{ | x \rangle \}$
$\displaystyle \langle x | x' \rangle = \delta(x - x') = \left\{ \begin{array}{ll} 0 & x \neq x' \\ C & x=x' \end{array} \right.$
where C normalizes: $\int_{-\infty}^{\infty} dx \; \delta(x - x') = 1$
$\Rightarrow C = \infty$
• Fourier transform:
$\displaystyle \delta(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} e^{ikx}$

## Breaking down the propagator equation

• $\displaystyle \langle x | P \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{i}{\hbar}px}$
• $\displaystyle \langle x_{k+1} | x_k \rangle = \delta(x_{k+1} - x_k)$
• $\displaystyle \langle x_{k+1} | V(\hat{x}) | x_k \rangle = V(x_k) \delta (x_{k+1} - x_k)$
because $| x_k \rangle$ is an eigen state of the coordinate system
• $\displaystyle \langle x_{k+1} | \frac{\hat{p}^2}{2m} | x_k \rangle = \int dp_k \langle x_{k+1} | \frac{\hat{p}^2}{2m} | p_k \rangle \langle p_k | x_k \rangle = \int \frac{1}{2\pi\hbar} \frac{p_k^2}{2m} e^{\frac{ip_k}{\hbar}(x_{k+1} - x_k)}$

## Putting it all together

Illustration of 2 possible trajectories.

$\displaystyle \begin{array}{lll} \langle x_{k+1} | 1 - \frac{i}{\hbar} \hat{H} \Delta t | x_k \rangle &\propto \int dp_k\; e^{\frac{i}{\hbar}p_k \Delta x_k} \left[ 1 - \frac{i}{\hbar} \left( \frac{p_k^2}{2m} + V(x_k) \right) \Delta t \right] & \\ &= \int dp_k\; e^{\frac{i}{\hbar}p_k \Delta x_k} \exp \left[ -\frac{i}{\hbar} \left( \frac{p_k^2}{2m} + V(x_k) \right) \Delta t \right] & (\mathrm{using}\; e^\epsilon = 1 + \epsilon) \\ &= \exp \left[ \frac{i}{\hbar} \left( \frac{m}{2} \frac{\Delta x_k^2}{\Delta t^2} - V(x_k) \right) \Delta t \right] & (\mathrm{recognizing\; Gaussian\; integral}) \\ &= e^{\frac{i}{\hbar} \mathscr{L}_k \Delta t} \end{array}$

where $\mathscr{L}$ is the Lagrangian

$\displaystyle \begin{array}{ll} \Rightarrow \langle x_f | \hat{U}(t) | x_i \rangle &= \int dx_1\;dx_2 \ldots\; e^{\frac{i}{\hbar} \mathscr{L}_1 \Delta t} e^{\frac{i}{\hbar} \mathscr{L}_2 \Delta t} \ldots \\ &= \int e^{\frac{i}{\hbar} \sum_k \mathscr{L}_k \Delta t} = \int e^{\frac{i}{\hbar} \int_0^t \mathscr{L} dt} = e^{\frac{i}{\hbar}S} \\ &\equiv \int \mathscr{D}x(t) e^{\frac{i}{\hbar} \int_0^t \mathscr{L} dt} \end{array}$