Where an expression for the (1+1)-dimensional retarded Dirac propagator in position space can be found, especially including the generalized funcion supported on the light-cone?

In particular, is it true that the square of the absolute value of this distribution is ill-defined (hence cannot be interpreted as charge/probability density)?

An expression for the required propagator, i.e., the retarded fundamental solution of the Dirac equation in $(1+1)$-dimensions $$ \begin{pmatrix} -m & i\,\partial/\partial t-i\,\partial/\partial x \\ i\,\partial/\partial t+i\,\partial/\partial x & -m \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}=0 $$ inside the future light cone $t>|x|$ is (see e.g. Eq.~(13) in this paper) $$ \frac{m}{2} \begin{pmatrix} -\frac{t+x}{\sqrt{t^2-x^2}}\,J_1(m\sqrt{t^2-x^2}) & i\,J_0(m\sqrt{t^2-x^2}) \\ i\,J_0(m\sqrt{t^2-x^2}) & \frac{-t+x}{\sqrt{t^2-x^2}}\,J_1(m\sqrt{t^2-x^2}) \end{pmatrix}. $$ The propagator contains also some generalized function supported on the future light cone $t=|x|$, which particularly interested in. A receipt for computation has been given in this answer, but it seems tricky to identify all the resulting delta-functions and their derivatives (and also the retarded propagator has been confused with the Feynman propagator there). Thus a reference to the final expression would be preferrable.

Notice that searching for retarded Dirac propagator (vanishing outside the light cone), not just the 'Dirac propagator' or Feynman propagator (not vanishing outside the light cone).

  • $\begingroup$ If you have a retarded function, it is easy to obtain its advanced version by a simple variable change. Then you may construct the Feynman function, etc., and vice versa, no? $\endgroup$ – Vladimir Kalitvianski Oct 31 '19 at 8:20
  • $\begingroup$ The retarded propagator indeed can be obtained from the Feynman propator. But expression for neither the retarded nor Feynman propagator (including the generalized function!) is known to me for 1+1-dimensions and spin 1/2. $\endgroup$ – Mikhail Skopenkov Nov 5 '19 at 14:15

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