Dirac spinors and commutation

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Hey guys,

i'm stuck (yet again! :) )

I am somewhat confused by Dirac spinors [tex]u,\bar{u}[/tex]. Take the product (where Einstein summation convention is assumed):

[tex]u^r u^s\bar{u}^s[/tex] Is this the same as [tex]u^s\bar{u}^s u^r[/tex]? Probably not cuz u^r is a vector while the other thing is a matrix, right?

Cheers,
earth2
 

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  • #2
dextercioby
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So it's a sum after the spinor index s ? Well, then [itex]u^r u^s\bar{u}^s = u^s\bar{u}^s u^r [/itex], because the sum of products is a scalar wrt the Lorentz transformations and is a bosonic variable, as it has Grassmann parity 0.
 
  • #3
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Thanks! But i don't get it if i look at it in terms of vectors and matrices...

So, [tex]u^s\bar{u}^s[/tex]=4x4 matrix where [tex]u^r[/tex] is a 1x4vector. How can i then have vector times matrix = matrix times vector?
 
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  • #4
dextercioby
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Well, actually, No, actually the [itex]u^s\bar{u}^s[/itex] is not well defined, the barred spinor should always be put on the left, so that [tex]\bar{u}^s u^s[/tex] becomes just an ordinary complex number which commutes with everything, that's why you can switch it around.

EDIT: It is well defined, as a tensor product. See the below comments.
 
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  • #5
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But look for instance at the completeness relation for spinors. It is nothing but

\slashed{P}= [tex]u^s\bar{u}^s[/tex] with a sum over s. I.e. it is a matrix :) See Peskin Schröder in the beginning... :)
 
  • #6
dextercioby
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Hmm, you're right, I guess. It's a tensor product. Why didn't I realize that ? :)) So in that case, the answer to your initial question is NO, you can't switch them around. A line is multiplied by a sq. matrix and not viceversa.
 
  • #7
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:) Thanks!
 

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