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 cuz 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!