- #1

JD_PM

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- Homework Statement
- 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)##

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

- Relevant Equations
- Please see properties below

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

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

The provided solution is

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

My solution is not equivalent to ##(5)##. What am I missing?Any help is appreciated.

Thank you

EDIT: I was wrong assuming that ##Tr(m)=m##; Of course that traces aren't defined for scalars! . It should be ##m \times Tr (1_n) = n m##, where ##1_n## is the n-identity matrix

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

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

The provided solution is

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

My solution is not equivalent to ##(5)##. What am I missing?Any help is appreciated.

Thank you

EDIT: I was wrong assuming that ##Tr(m)=m##; Of course that traces aren't defined for scalars! . It should be ##m \times Tr (1_n) = n m##, where ##1_n## is the n-identity matrix

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