Timeline for Some questions about the spectral function
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2012 at 1:40 | vote | accept | Student | ||
| Nov 21, 2011 at 9:02 | comment | added | Jon | The point here is that you are working with spinors and the representation of the Lorentz group is not the same as for the scalar field. E.g. check Bjorken and Drell, vol. 1. So, if you write down your spinor in a standard form $(\xi\ 0)$, apart from some factor, the multiplication by $S$ moves this to the known coefficients of the solution of the free Dirac equation $u(p,s)$ and $v(p,s)$. Then, these have the properties I have said above and produce the right contributions to the propagator. | |
| Nov 21, 2011 at 0:59 | comment | added | Student | Thanks for your efforts. I have tried doing everything you are saying. When I insert that general $U$ in that matrix element then I will get those factors of $S$ but that doesn't help me understand as to why one should get the free-field $u_\alpha$ out of it? The main query of the question! And also why do you say that just translations should affect the spinor fields any non-trivially? I would think that only when one rotates does the non-triviality of the spin come into play - translations are always a phase! | |
| Nov 20, 2011 at 22:05 | comment | added | Jon | I added a clarification about in the answer. | |
| Nov 20, 2011 at 20:30 | comment | added | Student | I thought of this but am not clear as to how or where the spin contribution is going to come from. If you can show as to how the $\sum u\bar{u}$ - the free Dirac field term is going to emerge from the matrix elements I typed above. | |
| Nov 20, 2011 at 10:20 | comment | added | Jon | I do not know if you mean this. When you use translational invariance of the vacuum, that is you consider $\psi(x)=e^{ipx}\psi(0)e^{-ipx}$ and $e^{ipx}|0\rangle=|0\rangle$, you should also consider in this case the spin contribution with respect to the scalar field. This contribution, managed through standard formula like $\sum u\bar u=\gamma p+m$ will permit you to recover the standard contribution $iZ/\gamma p-m$ and in the end you will get the K-L representation for Fermions. | |
| Nov 18, 2011 at 0:38 | comment | added | Student | The "problem" is with the minus sign that time-ordering of the fermionic fields introduces. After one introduces a complete set one is left with two terms which look like, $<0|\psi_a(x)|n><n|\bar{\psi_b(y)}|0>$ and $<0|\bar{\psi_b(y)}|n><n|\psi_a(x)|0>$. In the corresponding terms of the calculation with scalar fields these terms are equal but for the Dirac field I don't know whether these are related or not..things would be simple if one could argue that these two terms differ by exactly a negative sign -- but i don't know either way. | |
| Nov 14, 2011 at 9:12 | history | answered | Jon | CC BY-SA 3.0 |