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What is a Feynman propagator

  1. Jul 24, 2014 #1

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

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

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

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


    [tex]x\text{ is a 4-vector: }x=(\mathbf{x},t)[/tex]

    [tex]\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}}[/tex]

    Step function:
    [tex]\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[/tex]

    Feynman propagator:
    [tex]\Delta_F(x)\ =\ i(\theta(x)\Delta_+(x)\ +\ \theta(-x)\Delta_+(-x))[/tex]

    [tex]\ =\ i\Delta_+(x)\text{ if }t > 0\ \text{ but }=\ i\Delta_+(-x)\text{ if }t < 0[/tex]

    [tex]=\ \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)[/tex]

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

    [tex]=\ \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)[/tex]

    [tex]=\ \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)[/tex]

    Extended explanation

    Re-calculation of ∆+(-x):

    [tex]\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}}[/tex]

    So, replacing [itex]\mathbf{q}[/itex] by [itex]-\mathbf{q}[/itex] and [itex]d^3\mathbf{q}[/itex] by [itex]-d^3\mathbf{q}[/itex]:

    [tex]\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}}[/tex]

    Calculation of the Feynman propagator:

    [tex]\Delta_F(x)\ =\ i(\theta(x)\Delta_+(x)\ +\ \theta(-x)\Delta_+(-x))[/tex]

    [tex]=\ \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)[/tex]

    [tex]=\ \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)[/tex]

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

    [tex]=\ \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)[/tex]

    which, writing [itex]q[/itex] as the 4-vector [itex](\mathbf{q},q_0)[/itex], is:

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

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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