# Deriving the Generalized Fierz Transformation from Schroeder's Textbook

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.

• I made those edits you suggested May 30 at 17:46
• Why does setting p to 0 give me the trace formula? May 30 at 17:50
• 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). May 31 at 4:12
• Thanks! This was very helpful and insightful! May 31 at 13:05