How Does the Feynman Propagator Define Particle Movement in Quantum Mechanics?

[Greg Bernhardt]

Definition/Summary

The Feynman propagator $\Delta_F(x)$ is the propagator (the probability amplitude) for a scalar particle of non-zero mass, $m$, to travel over a space-time interval $x$.

It is obtained by integrating, over all possible 3-momentums $\mathbf{q}$ of a particle of mass $m$, the function $\Delta_+(x)$ if $x$ is "forward in time" or the function $\Delta_+(-x)$ if $x$ is "backward in time".

This is the same as integrating, over all possible 4-momentums $q$ (of any mass, and including those with negative energy), the function $e^{iq\cdot x}/(q^2\ +\ m^2\ -\ i\varepsilon)$

The propagator for a non-scalar particle is $P(-i\frac{\partial}{\partial x})\Delta_F(x)$ where P is a polynomial dependent on the spin of the particle.

Equations

DEFINITIONS:
$$x\text{ is a 4-vector: }x=(\mathbf{x},t)$$

$$\Delta_+(x)\ =\ \frac{1}{(2\pi)^3}\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} d^3\mathbf{q}\ \frac{e^{i(\mathbf{q}\cdot\mathbf{x}\ -\ \sqrt{\mathbf{q}^2\ +\ m^2}\,t)}}{2\sqrt{\mathbf{q}^2\ +\ m^2}}$$

Step function:
$$\theta(t)\ =\ \frac{-1}{2\pi i}\int_{-\infty}^{\infty} ds\,\frac{e^{-ist}}{s\ +\ i\,\varepsilon}\ =\ 1\text{ if }t > 0\ \text{ but }=\ 0\text{ if }t < 0$$

Feynman propagator:
$$\Delta_F(x)\ =\ i(\theta(x)\Delta_+(x)\ +\ \theta(-x)\Delta_+(-x))$$

$$\ =\ i\Delta_+(x)\text{ if }t > 0\ \text{ but }=\ i\Delta_+(-x)\text{ if }t < 0$$

$$=\ \frac{1}{(2\pi)^4}\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\ d^4q\ e^{iq\cdot x}\ \left(\frac{1}{q^2\ +\ m^2\ -\ i\varepsilon}\right)$$

Propagator for spin-1/2 particle:
$$[(-i\gamma_{\mu}\frac{\partial}{\partial x^{\mu}}\ +\ m)\beta]\Delta_F(x)$$

$$=\ \frac{1}{(2\pi)^4}\ \int\int\int\int\ d^4q\ e^{iq\cdot x}\ \left(\frac{[(-i\gamma_{\mu}q^{\mu}\ +\ m)\beta]}{q^2\ +\ m^2\ -\ i\varepsilon}\right)$$

$$=\ \frac{1}{(2\pi)^4}\ \int\int\int\int\ d^4q\ e^{iq\cdot x}\ \left(\frac{1}{\gamma_{\mu}q^{\mu}\ -\ m\ -\ i\varepsilon}\right)$$

Extended explanation

Re-calculation of ∆₊(-x):

$$\Delta_+(-x)\ =\ \frac{1}{(2\pi)^3}\ \int d^3\mathbf{q}\ \frac{e^{i(\mathbf{q}\cdot(\mathbf{-x})\ -\ \sqrt{\mathbf{q}^2\ +\ m^2}\,(-t))}}{2\sqrt{\mathbf{q}^2\ +\ m^2}}$$

So, replacing $\mathbf{q}$ by $-\mathbf{q}$ and $d^3\mathbf{q}$ by $-d^3\mathbf{q}$:

$$\Delta_+(-x)\ =\ \frac{-1}{(2\pi)^3}\ \int d^3\mathbf{q}\ \frac{e^{i(\mathbf{q}\cdot\mathbf{x}\ +\ \sqrt{\mathbf{q}^2\ +\ m^2}\,t)}}{2\sqrt{\mathbf{q}^2\ +\ m^2}}$$

Calculation of the Feynman propagator:

$$\Delta_F(x)\ =\ i(\theta(x)\Delta_+(x)\ +\ \theta(-x)\Delta_+(-x))$$

$$=\ \frac{-1}{(2\pi)^4}\ \int\ d^3\mathbf{q}\ \frac{e^{i\,\mathbf{q}\cdot\mathbf{x}}}{ 2\sqrt{\mathbf{q}^2\ +\ m^2}}\ \left(\int ds\, \frac{e^{-i(\sqrt{\mathbf{q}^2\ +\ m^2}\ +\ s)\,t}}{s\ +\ i\,\varepsilon}\ -\ \int ds\,\frac{e^{i(\sqrt{\mathbf{q}^2\ +\ m^2}\ +\ s)\,t}}{s\ +\ i\,\varepsilon}\right)$$

$$=\ \frac{-1}{(2\pi)^4}\ \int\ d^3\mathbf{q}\ \frac{e^{i\,\mathbf{q}\cdot\mathbf{x}}}{ 2\sqrt{\mathbf{q}^2\ +\ m^2}}\ \int dq_0\ \left(\frac{e^{-iq_0t}}{q_0\ - \sqrt{\mathbf{q}^2\ +\ m^2}\ +\ i\varepsilon}\ \ +\ \ \frac{e^{-iq_0t}}{-q_0\ -\ \sqrt{\mathbf{q}^2\ +\ m^2}\ +\ i\varepsilon}\right)$$

where a new "energy" variable $q_0$ has been substituted for $s+\sqrt{(\mathbf{q}^2\ +\ m^2)}$ in the left part, and for $-s-\sqrt{(\mathbf{q}^2\ +\ m^2)}$ in the right part

$$=\ \frac{1}{(2\pi)^4}\ \int\int\ d^3\mathbf{q}\ dq_0\ e^{i(\,\mathbf{q}\cdot\mathbf{x}\ -\ q_0t)}\ \left(\frac{1}{\mathbf{q}^2\ -\ q_0^2\ +\ m^2\ -\ i\varepsilon}\right)$$

which, writing $q$ as the 4-vector $(\mathbf{q},q_0)$, is:

$$=\ \frac{1}{(2\pi)^4}\ \int\ d^4q\ e^{iq\cdot x}\ \left(\frac{1}{q^2\ +\ m^2\ -\ i\varepsilon}\right)$$

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[AaronS]

Thanks for the overview of Feynman Propagator