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I figured out this one, just thought it was quite nice...
We start with the only requirement that the Green's function of the propagator is causal in the sense that it propagates stricktly forward in time, so that the Green's function is zero at t<0. Using the Heaviside step function we can express this as follows.
<br /> {\cal D}(r,t)\ =\ {\cal D}(r,t)\ \theta(t) <br />A multiplication with the Heaviside function amounts to a convolution with it's Fourier transform in momentum space
<br /> {\cal D}(\omega,k)\ =\ {\cal D}(\omega,k)\ * \ \frac{1}{2}\left(\delta(\omega)-\frac{i}{\pi \omega}\right)<br />
The convolution with the delta gives us the function itself, while the convolution with 1/\omega is a standard transformation known as the Hilbert transform. This transformation and its inverse are defined by:
<br /> {\cal H}\left\{\ \textit{f}\ (\omega)\ \right\}\ = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{1}{\eta-\omega}\ \textit{f}\ (\omega)\ d\eta<br /> \qquad \mbox{and} \quad<br /> {\cal H}^{-1} = -{\cal H}<br />
This allows us to write for the propagator in momentum space.
<br /> {\cal D}(\omega,k)\ =\ \frac{1}{2}\left(\ {\cal D}(\omega,k)\ +\ i {\cal H}\left\{\ {\cal D}(\omega,k)\ \right\}\ \right)<br />
From the right hand side we see that the propagator can be expressed as a complex sum of a function and its own Hilbert transform. The two halves together form a so-called Hilbert pair. The closest we find in terms of the usual propagators involving the d'Alembertian is: (From Stephan L. Hahn's book on the Hilbert transform)
<br /> \frac{a}{\omega^2-a^2} \quad <br /> \begin{array}{c} {\cal H}_\omega \\ \mbox{\Large $\Leftrightarrow$} \end{array}<br /> \quad -\frac{i\omega}{\omega^2-a^2}<br />
We already know what the denominators should be, but what about the numerators? Especially the factor a which should be the square root of a^2. Well, of course somebody already solved this almost 80 years ago. So,<br /> \mbox{using:}\ \ \ \left\{ <br /> \begin{array}{lcl}<br /> \omega^2 & \Rightarrow & E^2 \\ <br /> a^2 & \Rightarrow & p_x^2 + p_y^2 + p_z^2 + m^2 \\<br /> \omega & \Rightarrow & E,\ \ \mbox{or better:}\ \ \sqrt{I}\ E \\<br /> a & \Rightarrow & \alpha_1 p_x + \alpha_2 p_x + \alpha_3 p_x + \beta m<br /> \end{array}<br /> \right.<br />We can now write down for our propagator:
<br /> {\cal D}(E,p)\ =\ \frac{E+\vec{\alpha}\cdot\vec{p}+\beta m}{E^2-p^2-m^2} <br />Regards, Hans
We start with the only requirement that the Green's function of the propagator is causal in the sense that it propagates stricktly forward in time, so that the Green's function is zero at t<0. Using the Heaviside step function we can express this as follows.
<br /> {\cal D}(r,t)\ =\ {\cal D}(r,t)\ \theta(t) <br />A multiplication with the Heaviside function amounts to a convolution with it's Fourier transform in momentum space
<br /> {\cal D}(\omega,k)\ =\ {\cal D}(\omega,k)\ * \ \frac{1}{2}\left(\delta(\omega)-\frac{i}{\pi \omega}\right)<br />
The convolution with the delta gives us the function itself, while the convolution with 1/\omega is a standard transformation known as the Hilbert transform. This transformation and its inverse are defined by:
<br /> {\cal H}\left\{\ \textit{f}\ (\omega)\ \right\}\ = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{1}{\eta-\omega}\ \textit{f}\ (\omega)\ d\eta<br /> \qquad \mbox{and} \quad<br /> {\cal H}^{-1} = -{\cal H}<br />
This allows us to write for the propagator in momentum space.
<br /> {\cal D}(\omega,k)\ =\ \frac{1}{2}\left(\ {\cal D}(\omega,k)\ +\ i {\cal H}\left\{\ {\cal D}(\omega,k)\ \right\}\ \right)<br />
From the right hand side we see that the propagator can be expressed as a complex sum of a function and its own Hilbert transform. The two halves together form a so-called Hilbert pair. The closest we find in terms of the usual propagators involving the d'Alembertian is: (From Stephan L. Hahn's book on the Hilbert transform)
<br /> \frac{a}{\omega^2-a^2} \quad <br /> \begin{array}{c} {\cal H}_\omega \\ \mbox{\Large $\Leftrightarrow$} \end{array}<br /> \quad -\frac{i\omega}{\omega^2-a^2}<br />
We already know what the denominators should be, but what about the numerators? Especially the factor a which should be the square root of a^2. Well, of course somebody already solved this almost 80 years ago. So,<br /> \mbox{using:}\ \ \ \left\{ <br /> \begin{array}{lcl}<br /> \omega^2 & \Rightarrow & E^2 \\ <br /> a^2 & \Rightarrow & p_x^2 + p_y^2 + p_z^2 + m^2 \\<br /> \omega & \Rightarrow & E,\ \ \mbox{or better:}\ \ \sqrt{I}\ E \\<br /> a & \Rightarrow & \alpha_1 p_x + \alpha_2 p_x + \alpha_3 p_x + \beta m<br /> \end{array}<br /> \right.<br />We can now write down for our propagator:
<br /> {\cal D}(E,p)\ =\ \frac{E+\vec{\alpha}\cdot\vec{p}+\beta m}{E^2-p^2-m^2} <br />Regards, Hans
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