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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.

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1 Answer 1

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(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.

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  • $\begingroup$ I made those edits you suggested $\endgroup$
    – user261609
    May 30, 2021 at 17:46
  • $\begingroup$ Why does setting p to 0 give me the trace formula? $\endgroup$
    – user261609
    May 30, 2021 at 17:50
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    $\begingroup$ 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, 2021 at 4:12
  • $\begingroup$ Thanks! This was very helpful and insightful! $\endgroup$
    – user261609
    May 31, 2021 at 13:05

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