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3 votes

Fermionic propagator

Hint: For eq. (1) to be non-trivial we must assume that $G^{ij}$ is antisymmetric in $i\leftrightarrow j$.
Qmechanic's user avatar
  • 203k
0 votes

Non-homogenous Helmholtz equation in 3+1D: Green's function and solution

The equation has solutions for both cases: Positive $\alpha^2$: The radial Bessel function solution combined with a $k_z$ where $k_z^2 > \frac{\omega^2}{c^2}$. Negative $\alpha^2$: The radial ...
Jos Bergervoet's user avatar
5 votes
Accepted

Why are 2-point functions Green's functions?

It follows from the Schwinger-Dyson equations. Let's prove it for the case you're interested in (though the equations are more general). I will do it for a scalar field and let you work out the ...
Prahar's user avatar
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1 vote
Accepted

Why does a singularity imply the need for a distribution?

OP's linked quote mentions that the free massless scalar propagators with Euclidean spacetime signature are singular functions for $d\geq 2$. These singular functions can be injectively imbedded into ...
Qmechanic's user avatar
  • 203k
7 votes

Wrong solution for Green function of one-dimensional Poisson equation

The general solution of $$G^{\prime \prime}(r)=-\delta(r) \tag{1} \label{1}$$ for the Green function $G(r)$ can indeed be obtained from your ansatz $$G(r)=(ar+b) \theta(r)+(cr+d) \theta(-r). \label{...
Hyperon's user avatar
  • 6,163
7 votes
Accepted

Wrong solution for Green function of one-dimensional Poisson equation

Integration over the singularity not only yields $$ \Big[\frac{d}{dr} G \Big]_{-\epsilon}^{\epsilon} = -1 $$ but also: $$ \big[G \big]_{-\epsilon}^{\epsilon} = 0. $$ That will fix $b$ and $d$ to be ...
Jos Bergervoet's user avatar

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