Green’s function of Dirac operator

Pouramat
Messages
27
Reaction score
1
Homework Statement
My question comes from the textbook by Peskin & Schroeder,

If $$S_F(x-y)$$ is Green’s function of Dirac operator, how we should verify
$$ (i {\partial}_{\mu} \gamma^{\mu} -m)S_F (x-y)= i \delta^{(4)} (x-y) . $$
!!Didn’t know how to write slashed partial!!
all of $$\partial _x$$ in my solution are slashed but I did not know how to write it.
Relevant Equations
Using $$S_F(x-y)$$ definition:
\begin{align}
S_F(x-y) &= < 0|T \psi (x) \bar\psi (y) |0> \\
& = \theta(x^0-y^0) <0|\psi (x) \bar\psi (y) |0>- \theta(y^0-x^0) <0|\bar\psi (y) \psi (x) |0>
\end{align}
I started from eq(3.113) and (3.114) of Peskin and merge them with upper relation for $S_F$, as following:
\begin{align}
S_F(x-y) &=
\theta(x^0-y^0)(i \partial_x +m) D(x-y) -\theta(y^0-x^0)(i \partial_x -m) D(y-x) \\
&= \theta(x^0-y^0)(i \partial_x +m) < 0| \phi(x) \phi(y)|0 > -\theta(y^0-x^0)(i \partial_x -m) < 0| \phi(y) \phi(x)|0 >
\end{align}
Now we can calculate Green's Function of Dirac operator using this form of $S_F$
\begin{align}
(i \partial_x -m) S_F =& [(i \partial_x -m) \theta(x^0-y^0)][(i \partial_x +m) < 0| \phi(x) \phi(y)|0 >]\\
&+\theta(x^0-y^0)[(\partial^2-m^2) <0| \phi(x) \phi(y)|0>] \\
&-[(i \partial_x -m) \theta(y^0-x^0)][(i \partial_x-m) <0| \phi(y) \phi(x)|0 >] \\
&- \theta(y^0-x^0)[(i \partial_x -m)(i \partial_x -m)< 0| \phi(y) \phi(x)|0 >]
\end{align}

All of the terms are fine except the last line.The 1st and 3rd terms simplify as following The 2nd term is zero using klein-Gordon equation
The 1st term :
\begin{equation}
[(i \partial_x -m) \theta(x^0-y^0)][(i \partial_x +m) < 0| \phi(x) \phi(y)|0 >] = [-\partial_0 \theta(x^0-y^0)][<0| \pi(x) \phi(y)|0>]
\end{equation}
The 3nd term:
\begin{equation}
[(i \partial_x -m) \theta(y^0-x^0)][(i \partial_x +m) < 0| \phi(y) \phi(x)|0 >] = [-\partial_0 \theta(y^0-x^0)][< 0| \phi(x) \pi(y)|0 >]
\end{equation}
if the 4th term like the 2nd term was Klein-Gordon equation the problem gets solved, but it isn't.
 
Last edited:
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top