Is there an error in the Feynman rules for QFT in the Mandl & Shaw book?

[Vic Sandler]

I have found an error in the QFT book by Mandl & Shaw, 2nd ed. In eqn (8.53), there is a minus sign in front of $e^4$. However, this cannot be justified on the basis of how the equation is derived. In the unnumbered eqn below eqn (8.52) there is a minus sign, and the trace in this eqn is to be replaced by $-32(p_2p_1')(p_2'p_1)$. This should cancel the minus signs.

There is no simple way to fix this error. The eqn (8.54) is correct and carries the minus sign. This tells me that the minus sign in eqn (8.53) is correct. The minus sign in $-32(p_2P_1')(p_2'p_1)$ is also correct and thus I am led to believe that the minus sign in the unnumbered eqn is incorrect. This opens a can of worms.

In passing from the first line of that unnumbered eqn to the second line, it is necessary to commute the $u(p_2')$ from the extreme right hand side past 7 other fermion operators to the extreme left hand side. According to Feynman rule 8 on page 120, this should cause a sign change in the second line of the eqn and fix the problem.

If rule 8 is not supposed to be applied here, then why not?

If rule 8 is supposed to be applied, then this opens up other problems. For instance, on page 136, starting with eqn (8.41a), there are various places where minus signs are needed, but missing. Basically, A and B should have them, but the product X does not need one. A worse problem shows up in eqns (8.78) on page 146. Each of the four right hand sides would require a minus sign and thus $|\mathcal{M}|^2$ would be negative.

A possible way out of this dilemna would be to notice that in taking the complex conjugate of $\mathcal{M}_a$ it is necessary to commute two fermion operators and this would provide the extra minus sign necessary to make the square of the amplitude positive.

If rule 8 is not supposed to be applied here, then why not?

If rule 8 is supposed to be applied, then this opens up other problems. On page 132, in passing from eqn (8.23) to the unnumbered eqn below it, a fermion operator must be commuted past three other fermion operators and this would introduce a minus sign. I'm not sure where things stand any more.

Can someone please straighten me out?

 

[andrien]

Oh,common man,calm down!There is nothing wrong with it.What you are understanding as fermion interchange is only the result of conjugation.The element is like M_aM_b*,right.So when you take the conjugate of M_b,the matrix element is reversed.Just see 8.28 for example and you will understand.There is no fierz reshuffling or interchange of fermions.

 

[Vic Sandler]

Thanks andrien, I will do my best to remain calm. I seriously doubt that I got it right and the book got it wrong. Please keep that in mind. However ...

A little later in equation (8.29), he slings the rightmost fermion operator around to the leftmost position, passing 3 fermion operators in the process. If Feynman rule 8 is applied to both (8.28) and (8.29), then the final result comes out correct even if the intermediate step were wrong in lacking a minus sign.

I haven't done a thorough search through the book yet, but in the few examples I did look at, with but a single exception, such minus signs always paired off so that the final results came out correctly even if intermediate steps were wrong. The one exception is the derivation of eqn (8.54) which seems to require that rule 8 be applied in the way I have suggested. Barring that, how do you get from eqn (8.48a,b) to eqn (8.54), both of which seem to be correct?

 

[andrien]

You are just doing mistake for the conjugate with interchange of fermion operators.If you take for example the moller scattering and you interchange the final state,you get the crossed diagram which differs from the direct one by a minus sign.When you write |M|²,then for an element like u^-₂γ_μu₁ for M,the value of |M|² will come out as (u^-₂γ_μu₁)(u^-₁γ_μu₂).There is no use of rule 8 or so far anything like it here.Moreover,if you are writting a matrix element as (u^-₄γ_μu₃)(u^-₂γ_μu₁) and you want to interchange 4 and 1,then you can not just swap it with a minus sign.If you want to do it then you have to use Fierz reshuffling.

 

[Vic Sandler]

If that is the case, then what solution do you propose for the problem of the derivation of eqn (8.54)? I have worked out eqns (8.48a,b) as carefully as I could starting from the Feynman diagrams and they seem correct to me. However, if there is a sign error in one of them, then that would solve the whole problem. I can find no error in the derivation up until eqn (8.43) for which I get a sign change. Eqn (8.54) can be found on the net in the same form, so what gives?

In the following link, eqn (3.4) matches M & S eqn (8.54) except for a factor of 4 which might be attributable to a different definition of E.

http://physics.weber.edu/schroeder/feynman/feynman3.pdf

 

[HallsofIvy]

"Common man"? Why Mr Sandler is clearly not common at all!

 

[andrien]

I have worked out eqns (8.48a,b) as carefully as I could starting from the Feynman diagrams and they seem correct to me. However, if there is a sign error in one of them, then that would solve the whole problem.

I think the problem you are trying to state that one of 8.48a or b has wrong sign.Then you can there apply the exchange principle to see why there should be a sign difference.However if that is not convincing,then you will have to learn the operator product expansion in which you will have to sandwich the e² order interaction between the initial and final states and break field operators into positive and negative frequency part.This would lead to 16 possibilities out of which 4 will survive because of inlet and outlet condition.You will also have to use anticommutation relation between field operators.Finally after all that,you will see that the amplitude for the annihilation diagram must have a different sign from direct diagram.

If that is the case, then what solution do you propose for the problem of the derivation of eqn (8.54)?

There is nothing wrong with the derivation given in the book,it is fine.

 

[Vic Sandler]

There is an explicit minus sign in the unnumbered eqn below eqn (8.52). Do you see it? It is just to the left of $e^4$. Also, there is an explicit minus sign in the unnumbered eqn below eqn (A,14a) on page 140. Do you see it? It is just to the left of the 32. If you use that eqn to simplify the trace, then the two minus signs should cancel each other. Is this correct? The result should be that there is no minus sign to the left of $e^4$ in eqn (8.53). However, there is a minus sign there. Do you see it? I can't explain it. If you can, I wish you would do so. How do you multiply $-e^4$ by $-32$ and get $-32e^4$ for the result?

 

[Vic Sandler]

I have worked out the Moller scattering computation and it shows that my suggestion for applying Feynman rule 8 does not work so I am no longer interested in that track. However, the problem of the minus sign in eqn (8.53) remains. There must be an extra minus sign lurking about somewhere. Can anyone help me find it?

 

[Vic Sandler]

I have found the problem, it is a simple typo in the book. The minus sign in $-e^4$ on the first line of eqn (8.53) is wrong and should not be there. That fixes everything. The reason I could not figure this out is because I was under the mistaken impression that $(p_1 - p_1')^2 = 4E^2sin^2(\theta/2)$. Actually, $(p_1 - p_1')^2 = -4E^2sin^2(\theta/2)$. Therefor, the minus sign in $-e^4$ on the second line of eqn (8.53) is correct.

 

[andrien]

The reason I could not figure this out is because I was under the mistaken impression that $(p_1 - p_1')^2 = 4E^2sin^2(\theta/2)$. Actually, $(p_1 - p_1')^2 = -4E^2sin^2(\theta/2)$. Therefor, the minus sign in $-e^4$ on the second line of eqn (8.53) is correct.

I was checking the results and all eqn seemed correct but if you will do such mistakes then there is no way for anyone to know what is it which is really wrong.So in future try to post your solution so that it can be easy to figure out what the problem is,rather than jumbling around guessing to find out where the problem lies.

 

[Vic Sandler]

Of course you are correct andrien. However, in my defense, in the second sentence of the OP, I did point directly to the problem.