# Computing a trace containing $\gamma$-matrices

I want to compute the following trace

$$Tr \Big( Y(\not{\!p_1'}+m) \Big) \ \ (1)$$

Where

$$\not{\!A} := \gamma^{\alpha} A_{\alpha} \ \ (2)$$

$$Y:= 4 \not{\!f_1} \not{\!p} \not{\!f_1} + m[-16(pf_1)+16 f_1^2] + m^2 ( 4 \not{\!p} - 16 \not{\!f_1})+16 m^3 \ \ (3)$$

The answer is given to be:

$$Tr \Big( Y(\not{\!p_1'}+m) \Big)=16 \Big( 2(f_1p)(f_1p') -f_1^2(pp')+m^2[-4(pf_1)+4f_1^2]+m^2 [(pp')-4(f_1 p')] +4m^4 \Big) \ \ (4)$$

We need to use the following properties

1) If $$(\gamma^{\alpha}\gamma^{\beta}...\gamma^{\mu}\gamma^{\nu})$$ contains an odd number of $$\gamma$$-matrices

$$Tr(\gamma^{\alpha}\gamma^{\beta}...\gamma^{\mu}\gamma^{\nu})=0$$

2)

$$Tr(\not{\!A}\not{\!B})=4AB$$

3)

$$Tr(\not{\!A}\not{\!B}\not{\!C}\not{\!D})=4\Big( (AB)(CD)-(AC)(BD)+(AD)(BC) \Big)$$

4) Given $$A, B$$ to be square matrices

$$Tr(A+B)=Tr(A) + Tr(B)$$

Besides: I've assumed that the trace of a scalar is equal to itself. (i.e. $$Tr(m)=m$$)

Applying such properties I get

$$Tr \Big( Y(\not{\!p_1'}+m) \Big) = 16 \Big( (f_1 p)(f_1 p') - f_1^2(pp') + f_1p'p p' \Big) + 16m^2p^2-64 m^2 f_1 p + 16m^2[-pf_1+f_1^2]+16m^4 \ \ (5)$$

Which is not the provided solution $$(4)$$. What am I missing?

Source: Second Edition QFT, Mandl & Shaw page 144.

Any help is appreciated.

The trace of a scalar is not the scalar itself. In this context, the scalar $$m$$ represents $$m\mathbb1$$, where $$\mathbb1$$ is the $$4\times4$$ identity matrix. Thus $$\operatorname{Tr}m=4m$$.
• joriki thank you. But only the last term of my answer changes $Tr \Big( Y(\not{\!p_1'}+m) \Big) = 16 \Big( (f_1 p)(f_1 p') - f_1^2(pp') + f_1p'p p' \Big) + 16m^2p^2-64 m^2 f_1 p + 16m^2[-pf_1+f_1^2]+32m^4$; $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$so I am still missing something... May 31, 2020 at 14:33
• @JD_PM: Sorry, I confused the Dirac matrices and the Pauli matrices (it was some decades ago that I studied physics :-) – they're actually $4\times4$ matrices, so $\operatorname{Tr}m=4m$. This affects not only your last term but all scalars, including the ones in square brackets, and appears to bring all of them in agreement with $(4)$. Another mistake is that you replaced $p'$ by $p$ in the terms $16m^2p^2-64m^2f_1p$; fixing that brings those in agreement with $(4)$ as well. Let me know if any discrepancies remain after that. May 31, 2020 at 14:53
• joriki I appreciate your help, I am closer to get the answer! Now I have $Tr \Big( Y(\not{\!p_1'}+m) \Big) = 16 \Big( (f_1 p)(f_1 p') - f_1^2(pp') + f_1p'p p' + m^2[pp'-4f_1p'] + m^2[-4p_1f_1+4f_1^2]+4m^4\Big)$ May 31, 2020 at 15:22
• Note this is the same as $(4)$ but I've got $f_1p'p p'$ as an extra term and I do not have the factor of 2 in front of $(f_1 p)(f_1 p')$... May 31, 2020 at 15:23
• joriki I've just seen I applied property 3) incorrectly! It is not $f_1p'p p'$ but $f_1p'pf_1$, which leads to the desired answer :) $Tr \Big( Y(\not{\!p_1'}+m) \Big) = 16 \Big( 2(f_1 p)(f_1 p') - f_1^2(pp') + m^2[pp'-4f_1p'] + m^2[-4p_1f_1+4f_1^2]+4m^4\Big)$ May 31, 2020 at 15:32