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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)
* 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!
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)
* 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!