Dirac spinors and commutation

[earth2]

Hey guys,

i'm stuck (yet again! :) )

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

$$u^r u^s\bar{u}^s$$ Is this the same as $$u^s\bar{u}^s u^r$$? Probably not because u^r is a vector while the other thing is a matrix, right?

Cheers,
earth2

 

[dextercioby]

So it's a sum after the spinor index s ? Well, then $u^r u^s\bar{u}^s = u^s\bar{u}^s u^r$, because the sum of products is a scalar wrt the Lorentz transformations and is a bosonic variable, as it has Grassmann parity 0.

 

[earth2]

Thanks! But i don't get it if i look at it in terms of vectors and matrices...

So, $$u^s\bar{u}^s$$=4x4 matrix where $$u^r$$ is a 1x4vector. How can i then have vector times matrix = matrix times vector?

 

[dextercioby]

Well, actually, No, actually the $u^s\bar{u}^s$ is not well defined, the barred spinor should always be put on the left, so that $$\bar{u}^s u^s$$ 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.

 

[earth2]

But look for instance at the completeness relation for spinors. It is nothing but

\slashed{P}= $$u^s\bar{u}^s$$ with a sum over s. I.e. it is a matrix :) See Peskin Schröder in the beginning... :)

 

[dextercioby]

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.

 

[earth2]

:) Thanks!