Product of Dirac Spinors

  • Thread starter tommy01
  • Start date
  • #1
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Hi togehter.

I encountered the following problem:

The timeordering for fermionic fields (here Dirac field) is defined to be (Peskin; Maggiore, ...):

[tex]
T \Psi(x)\bar{\Psi}(y)= \Psi(x)\bar{\Psi}(y) \ldots x^0>y^0
[/tex]
[tex]
= -\bar{\Psi}(y)\Psi(x) \ldots y^0>x^0
[/tex]

where [tex]\Psi(x)[/tex] is a Dirac spinor and [tex]\bar{\Psi}(y) = \Psi(y)^\dagger \gamma^0[/tex] it's Dirac adjoint so that

[tex]
S(x-y) = \langle 0|T{ \Psi(x)\bar{\Psi}(y)}|0 \rangle
[/tex]

is the Feynman propagator wich is a 4x4 matrix.
But there is my problem: while it is clear that [tex]\Psi(x)\bar{\Psi}(y)}[/tex] is a 4x4 matrix, [tex]\bar{\Psi}(y)\Psi(x)[/tex] is a scalar.

I would be glad for an explanation.
Thanks.
Tommy
 

Answers and Replies

  • #2
Avodyne
Science Advisor
1,396
88
The indices are not contracted, and the propagator is a 4x4 matrix:

[tex]
T \Psi_\alpha(x)\bar{\Psi}_\beta(y)= \Psi_\alpha(x)\bar{\Psi}_\beta(y) \hbox{\ if\ } x^0>y^0
\hbox{\ and\ } -\bar{\Psi}_\beta(y)\Psi_\alpha(x) \hbox{\ if\ } y^0>x^0
[/tex]

[tex]
S_{\alpha\beta}(x-y) = \langle 0|T{ \Psi_\alpha(x)\bar{\Psi}_\beta(y)}|0 \rangle
[/tex]
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
12,993
546
In other words, the equations must be read component-wise, or you can think of a tensor product of the 2 spinors. That's the only way you can make sense of a product in which the barred spinor appears to the right of an un-barred one.
 
  • #4
40
0
Thanks a lot ...
 

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