I am self studying QFT from the textbook An Introduction of Quantum Field Theory and the corresponding solutions from Zhong-Zhi Xianyu. The generalized Fierz Transformation is derived in problem 3.6. I was able to do the basis normalization is part (a) and was able to do the computation in part (c); now I am working on the main part of the proof in part (b) and I am having some troubles understanding the proof outlined in the solution.

(1) In equation 3.54 of the solutions, why can one say that $\bar{u}_4 \Gamma^E \Gamma^C u_4 = Trace(\Gamma^E\Gamma^C)$? I tried proving this by example by choosing two specific examples of $\Gamma^E$ and $\Gamma^C$ so that I can get a specific product $\Gamma^E \Gamma^C$. Now for $u_4$ I chose $u_4 = (\sqrt{p\cdot \sigma}\xi \hspace{0.5cm} \sqrt{p\cdot \bar{\sigma}}\xi)^T$ (there is a bar over the second sigma), but I could not get something that resembled a trace. Can anyone show me why the statement $\bar{u}_4 \Gamma^E \Gamma^C u_4 = Trace(\Gamma^E\Gamma^C)$ is true?

(2) In the math line below equation equation 3.54 we have $(\bar{u}_2 \Gamma^F u_5)(\bar{u}_4 \Gamma^E u_1)(\bar{u}_1\Gamma^A u_2)(\bar{u}_3\Gamma^B u_4) = Trace(\Gamma^E \Gamma^A \Gamma^F \Gamma^B)$. It appears that the author believed $(\bar{u}_2 \Gamma^F u_5)(\bar{u}_4 \Gamma^E u_1)(\bar{u}_1\Gamma^A u_2)(\bar{u}_3\Gamma^B u_4) = (\bar{u}_4 \Gamma^E u_1)(\bar{u}_1\Gamma^A u_2)(\bar{u}_2 \Gamma^F u_5)(\bar{u}_3\Gamma^B u_4)$. Why can we change the order of these products? Aren't these matrix multiplications and doesn't order matter in multiplication? I am confused.

(3) Finally, I noticed to go from 3.53 to 3.54 in the solutions the author basically implies $(\bar{u}_4\Gamma^E u_1)(\bar{u}_1\Gamma^C u_4) = \bar{u}_4 \Gamma^E \Gamma^C u_4 = Trace(\Gamma^E\Gamma^C)$. This implies, to me, that $u_1 \bar{u}_1 = identity$. Why is this true? In order to answer this question I first noted that this product means $u_1 \bar{u}_1 = u_1u_1^\dagger \gamma^0$. Next I wrote $u_1$ as $(\sqrt{p\cdot \sigma}\xi \hspace{0.5cm} \sqrt{p\cdot \bar{\sigma}}\xi)^T$. I carried out the matrix multiplication and used the identity $(p\cdot \sigma)(p \cdot \bar{\sigma}) = p^2$. My final result consisted of a 4 by 4 matrix with 4 non zero components. Elements had the form $p \cdot \sigma \xi\xi^\dagger, \pm p \xi\xi^\dagger, p \cdot \bar{\sigma} \xi\xi^\dagger$. So I am stuck.


(2) Each $(\bar{u}\Gamma u)$ is a C-number, so feel free to change the orders.

(1) & (3) I think the author is trying to find out the value of $C^{AB}_{\phantom{AB}CD}$ given the general Fierz identity $$(\bar{u}_1Au_2)(\bar{u}_3Bu_4)=C^{AB}_{\phantom{AB}CD}(\bar{u}_1Cu_4)(\bar{u}_3Du_2)$$ is correct. In other words, he was substituting in some special $u$'s. Recall that $$\sum_{spin}u^s(p)\bar{u}^s(p)=\gamma\cdot p+m\\ \sum_{spin}v^s(p)\bar{v}^s(p)=\gamma\cdot p-m.$$ So $u^s(p)\bar{u}^s(p)-v^s(p)\bar{v}^s(p)$ gives you an identity matrix.

It remains to argue that the general Fierz identity is correct. Or to say, if I get the $C$'s for a special set of $u$'s, then it will still be true for other $u$'s. To this end, you can view $(\bar{*}_1A*_2)(\bar{*}_3B*_4)$ as a linear function from $(R^4)^4$ to $R$. So Fierz identity is merely a transformation between different bases for this linear function. And this finishes the argument.

I have a different understanding for this question. Notice that the trace in question (a) is very similar to the Killing form of Lie algebra. So I tend to understand it in the Lie algebra way. We can think $(\bar{*}'A*)(\bar{*}'B*)$ as a linear function from two $*$ to two $*'$. But the order of the two $*$ is changed on the left-hand side. Then we compute the Killing form (trace in the matrix representation) with $(\bar{*}'E*)(\bar{*}'F*)$ of both sides. We can find that on one side it is $Tr(EA)Tr(FB)$ while on the other side it is $Tr(ECFD)$.

The sketch for this computation is as follows. On the right hand side $(\bar{*}'A*_2)(\bar{*}'B*_4)$ maps $(u_2)_i(u_4)_j$ to $A_{li}(u_2)_iB_{kj}(u_4)_j$. So a matrix form of this map can be written as $(AB)_{lk;ij}=A_{li}B_{kj}$. While the matrix form of $(\bar{*}'C*_4)(\bar{*}'D*_2)$ in the same bases is $(CD)_{lk;ij}=C_{lj}B_{ki}$. To compute the Killing form with $(\bar{*}'E*_2)(\bar{*}'F*_4)$, we multiply their matrix forms and take the trace. For RHS $$(EF)_{pq;lk}(AB)_{lk;pq}=E_{pl}F_{qk}A_{lp}B_{kq}=Tr(EA)Tr(FB).$$ For LHS $$(EF)_{pq;lk}(CD)_{lk;pq}=E_{pl}F_{qk}C_{lq}D_{kp}=Tr(ECFD).$$

In this method, we summed over many $u$'s implicitly in the trace. So I believed that my construction of four $u$ and $v$ to get an identity matrix in (1) & (3) is the correct way.

  • $\begingroup$ I made those edits you suggested $\endgroup$
    – user758469
    May 30 at 17:46
  • $\begingroup$ Why does setting p to 0 give me the trace formula? $\endgroup$
    – user758469
    May 30 at 17:50
  • 1
    $\begingroup$ Sorry I made a stupid mistake. The spirit is that find some combinations in $(R^4)^2$ which give you an identity matrix. I have corrected them in the answer and provide a Lie algebra point of view (although the computation is almost the same as Xianyu's). $\endgroup$
    – Youran
    May 31 at 4:12
  • $\begingroup$ Thanks! This was very helpful and insightful! $\endgroup$
    – user758469
    May 31 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.