Polarization 4-vectors to get matrix element in QFT

[jameson2]

I'm working through the start of the Quantum Field Theory book by Peskin and Schroeder. The first section deals with an electron and positron colliding to give a positive and negative muon traveling along a line at an angle theta to the line of the e,p collision.(This is using center of mass coordinates). Calculating the matrix elements involves working out the polarization vectors of each particle. For the case in the book it gives (0,1,i,0) for the elctron and (0,cos theta, i, sin theta). I don't know how they get these though. They also give the other matrix elements without proof. Could anyone explain to me how to work out these vectors?
Thanks alot

 

[strangerep]

[...] Calculating the matrix elements involves working out the polarization vectors of each particle. For the case in the book it gives (0,1,i,0) for the electron and (0,cos theta, i, sin theta). I don't know how they get these though.

Actually, the polarization vector \epsilon^\mu = (0,1,i,0) refers to the intermediate photon, not the electron. So it's just a standard result from classical electrodynamics for circularly polarized light (corresponding to a angular momentum eigenstate in the quantum case). See, e.g., Jackson's "Classical Electrodynamics" for more detail on the classical aspects.

To get the other vector (0,\cos\theta, i, -\sin\theta), just involves a standard rotation matrix in the xz plane. E.g., (suppressing the t,y components),
<br /> \pmatrix{\cos\theta &amp; \sin\theta \cr -\sin\theta &amp; \cos\theta} \pmatrix{1 \cr 0}<br /> ~=~ \pmatrix{\cos\theta \cr -\sin\theta} <br />

They also give the other matrix elements without proof.

Remember that ch1 is only meant to be an "invitation", or a "taste". Such cases are worked out in far more detail later in the book.