# A question about trace technology

## Main Question or Discussion Point

Hi everybody!

While reading Peskin-Schroeder, i stuck in the in equation 5.4 about the unpolarized cross section of the $$e^- e^+ \rightarrow \mu^- \mu^+$$ annihilation. i didn't understand how this relationof the electron trace came to be, and where the indices came from?

$$\operatorname{tr}[(p\!\!/^\prime-m_e) \gamma^\mu (p\!\!/ + m_e) \gamma^\nu] = 4[p^{\prime \mu} p^\nu +p'^\nu p^\mu - g^{\mu \nu} (p \cdot p' + m^2_e)]$$

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malawi_glenn
Homework Helper
what have you tried? Do you know how traces with gamma matrices works at all?

if not, then I can give some relations that are useful

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Hi,

I remember when I first encountered with those... very confused I was. First time, I did it "brute force". Calculated all the components on one side, check they were the same on the other. Of course it is stupid, and not the right way to do it. But by doing so, I understood how to work it out neatly.

Everybody has his own methods...

malawi_glenn
Homework Helper
This example can be calculated on 1/2 page, if one does all calculations. But one have to know all trace identies and so on.

$$\text{Tr}(\gamma ^{\alpha}\gamma ^{\beta}) = g^{\alpha \beta}$$

nrqed
Homework Helper
Gold Member
Hi everybody!

While reading Peskin-Schroeder, i stuck in the in equation 5.4 about the unpolarized cross section of the $$e^- e^+ \rightarrow \mu^- \mu^+$$ annihilation. i didn't understand how this relationof the electron trace came to be, and where the indices came from?

$$\operatorname{tr}[(p\!\!/^\prime-m_e) \gamma^\mu (p\!\!/ + m_e) \gamma^\nu] = 4[p^{\prime \mu} p^\nu +p'^\nu p^\mu - g^{\mu \nu} (p \cdot p' + m^2_e)]$$

Are you asking about how to get from the trace (on the left) to the final result or are you asking how one gets from a Feynman amplitude to a trace expression (in other words how the left hand side comes from)?

To get from the left side to the right side is a simple application of trace formula. Look up the equations 5.5, 5.8, 5.9 etc. Or are you asking how where those trace formula come from? (using the cyclicity of the trace operation and using identities for the product of gamma matrices does the trick)

yes, i just want to know how to get from the left hand side to the right hand side. the problem mabye is thas i am not familiar with traces of gamma matrices. i'll try to learn it then come with a more concrete question..
but if anybody can give some hints that helps i'll be grateful.

nrqed
Homework Helper
Gold Member
yes, i just want to know how to get from the left hand side to the right hand side. the problem mabye is thas i am not familiar with traces of gamma matrices. i'll try to learn it then come with a more concrete question..
but if anybody can give some hints that helps i'll be grateful.
But then it's just a matter of applying the trace formula given in equation 5.5. Just expand the expression you gave in your first post (keeping in mind that that masses simplys multiply identity matrices) and use directly the formula of 5.5. That's all there is to it.

malawi_glenn
Homework Helper
Then if you want a lot of rules and properties of gamma matrices and dirac-slash notation, you can find it on many places on the internet. And their derivation. Do you know how the minkowski metric tensor works?

I am actually trying to make my own collection of formulas, and a collection of proofs. Just to have it all collected in a nice reference and TeX - practicing. If you are interessted you can send me a PM.

ok tell me please how to trace this thing $$\operatorname{tr}(\gamma^\nu\gamma^\mu p'_\mu\gamma^\mu\gamma^\nu p_\nu)$$? why it can't show $$\operatorname{tr}(\gamma^\nu\gamma^\mu p'_\mu\gamma^\mu\gamma^\nu p_\nu)$$? the last is what i want to ask

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malawi_glenn
Homework Helper
ok tell me please how to trace this thing $$\operatorname{tr}(\gamma^\nu\gamma^\mu p'_\mu\gamma^\mu\gamma^\nu p_\nu)$$? why it can't show $$\operatorname{tr}(\gamma^\nu\gamma^\mu p'_\mu\gamma^\mu\gamma^\nu p_\nu)$$? the last is what i want to ask
how did you get the same indicies on two of them? you must as general as possible, i.e:

$$\operatorname{tr}[(p\!\!/^\prime-m_e) \gamma^\mu (p\!\!/ + m_e) \gamma^\nu] = \operatorname{tr}[(p'_{\alpha}\gamma^{\alpha}-m_e) \gamma^\mu (p_{\beta}\gamma^{\beta}+ m_e) \gamma^\nu]$$

Then use the trace formula for 4 gamma matrices which I mailed you a link to.

$$\operatorname{tr}(8p'^\nu p^\mu) - \operatorname{tr}(2mp'^\mu \gamma^\nu) + \operatorname{tr}(2m\gamma^\mu p^\nu) + \operatorname{tr}(2m^2g^{\mu\nu})$$
am i on the right way?

malawi_glenn
Homework Helper
no, lets take this one for instance:

$$\text{Tr}(p'_{\alpha}\gamma^{\alpha}\gamma^\mu p_{\beta}\gamma^{\beta} \gamma^\nu) = \text{Tr}(p'_{\alpha}p_{\beta}\gamma^{\alpha}\gamma^\mu \gamma^{\beta} \gamma^\nu) = 4p'_{\alpha}p_{\beta}(g^{\alpha\mu}g^{\beta \nu} - g^{\alpha\beta}g^{\mu\nu}+g^{\alpha\nu}g^{\mu\beta}) = ...$$

i didn't know that we can commute p with gamma and put it out of the trace!! would you tell me why we can do this?

malawi_glenn
Homework Helper
p_a is just a number, an element in a 4-vector.

$$p_a\gamma ^a = p_0\gamma ^0 + p_1\gamma ^1 + p_2\gamma ^2 +p_3\gamma ^3$$

$$p_0 = E$$

$$p_1 = p_x$$ etc.

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thanx for help Glenn! i'll give it another shot

malawi_glenn
Homework Helper
thanx for help Glenn! i'll give it another shot
np, just knock yourself out hehe, cheers

ok, i have another question that should have came earlier
$$(p'\!\!\!\!\!/ \ -m)_{da}\gamma^\mu_{ab}(p\!\!\!\!\!/ \ +m)_{bc}\gamma^\nu_{cd} =\operatorname{Tr}[(p'\!\!\!\!\!/ \ -m)\gamma^\mu(p\!\!\!\!\!/ \ +m)\gamma^\nu]$$
where the trace came from here?

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nrqed
Homework Helper
Gold Member
ok, i have another question that should have came earlier
$$(p'\!\!\!\!\!/ \ -m)_{da}\gamma^\mu_{ab}(p\!\!\!\!\!/ \ +m)_{bc}\gamma^\nu_{cd} =\operatorname{Tr}[(p'\!\!\!\!\!/ \ -m)\gamma^\mu(p\!\!\!\!\!/ \ +m)\gamma^\nu]$$
where the trace came from here?

Let's say you have three matrices A, B and C. If you are calculating the following sum:

$$A_{cd} B_{de} C_{ec}$$

this is the same as

$$tr(ABC)$$

The fact that the last index of the third matrix is the same as the first index of the first matrix means that if you work with the actual matrices, you have to multiply tham and trace the product. That's all there is to it.

i'll try to prove it to make sure i got it..
thanks nrqed!

malawi_glenn