Dirac Spinor Algebra: Simplifying Expressions

[maverick280857]

Hi,

In a calculation I am doing, I encounter terms of the form

$$\bar{u}^{s_1}(\boldsymbol{\vec{p}})\gamma^{\mu}{v}^{s_2}(\boldsymbol{\vec{q}})$$

where $u$ and $v$ are the electron and positron spinors. Is there any recipe for simplifying this expression, using the spin sums or other identities? I am unable to figure anything out except if I break u and v into components and consider various cases depending on what $s_1$, $s_2$ and $\mu$ are...which is too tedious.

(I have to compute the product of this with $A_{\mu}(x)$)

Suggestions would be greatly appreciated.

Thanks in advance.

PS -- I am not looking at the amplitude squared, so I probably cannot use the trace methods directly..

 

[maverick280857]

Okay, so to be more specific, I am interested in simplifying the quantity

$$G^{\mu\nu} = Tr\left[u^{s_1}(\boldsymbol{p}_1)\bar{u}^{s_1}(\boldsymbol{p}_1)\gamma^{\mu}v^{s_2}(\boldsymbol{p}_2)\bar{v}^{s_2}(\boldsymbol{p}_2)\gamma^{\nu}\right]$$

There is no sum over the spins (so the spin sum identities won't help). Basically I am taking the trace of a 4x4 matrix (for given $\mu$ and $\nu$).

Any ideas?

 

[tom.stoer]

An idea (only an idea!) is to use Fierz identities

 

[Avodyne]

Expressions such as $u^{s}(\boldsymbol{p})\bar{u}^{s}(\boldsymbol{p})$ can be simplified by first doing the spin sum, then applying a spin projection matrix that removes the spin you don't want. For example, in Srednicki's conventions, the result is

$$u^{s}(\boldsymbol{p})\bar{u}^{s}(\boldsymbol{p})={1\over2}(1-s\gamma_5\rlap{z}\slash)(-\rlap{p}\slash+m)$$

where $z^\mu$ is the spin-quantization axis. For v-type spinors, the result is the same, except +m becomes -m.

 

[tom.stoer]

but doesn't want to sum over the spins ...

 

[Avodyne]

This is NOT a sum over spins, this is for a particular value of s (plus or minus one).

 

[maverick280857]

Thanks for the replies, Tom and Avodyne. Yes, it's not a spin sum..I think I have to leave it as it is (and wait till some part of the calculations actually requires me to sum over all spins ). Or if I want, I can split u and v into their components in some basis, and compute the 4x4 matrix 'by hand'. I don't think there is much point to it, so I'm just leaving it as it is.

Avodyne, your idea assumes that the 3-momenta arguments of both spinors are identical, which is not the case in my problem. In fact, an overall delta function ensures that $\boldsymbol{p}_1 = -\boldsymbol{p}_2$ under a very special case. Otherwise, the 3 momenta could be quite unrelated...which is why I'm thinking now that there may probably be no "identity" that I'm seeking. (It's a bit tempting to assume that there's something to simplify just about everything a novice can possibly encounter in QFT ;-)).

Tom, I'll have a look at the Fierz identity too.

 

[Avodyne]

Avodyne, your idea assumes that the 3-momenta arguments of both spinors are identical

No, it doesn't. You take your expression for $G^{\mu\nu}$, and substitute my expression in two places (with different momenta in each place, and with m->-m where you have v's instead of u's).

 

[maverick280857]

No, it doesn't. You take your expression for $G^{\mu\nu}$, and substitute my expression in two places (with different momenta in each place, and with m->-m where you have v's instead of u's).

Ah, neat. That's because I'm interested in $G^{\mu\nu}$ and not the expression I wrote in my first post. Nice.