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I am aware of the various relations with Dirac spinors and chirality but how does the fifth gamma matrix $\gamma^5$ behave with fermion fields, $\psi$?

Does the fifth gamma matrix have any particular commutation relation with $\psi$ that can be manipulated?

My issue has arisen out of the evaluation of the commutator, $[Q^5_a, Q^5_b]$ where, $$Q_a \equiv \int d^3x\ \psi^{\dagger}_r(x)(T_a)_{rs}\gamma^5\psi_s(x)$$

I have determined the algebra for $[Q_a, Q_b]$ (as above without $\gamma^5$ in) but I am unsure how to proceed with the inclusion of $\gamma^5$.

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  • $\begingroup$ To answer myself... $\frac{1}{2}(1\pm\gamma^5)\psi = \psi_{R,L}$ and so $\gamma^5\psi = \psi_R - \psi_L$ and the problem can be solved by realising that $[\gamma^5, T_a]_{rs}\psi_s(x)=0$ as I think I proved here math.stackexchange.com/a/1743587/272850 $\endgroup$ – Alexander McFarlane Apr 15 '16 at 12:31
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In spinor space, the fermionic fields $\psi$ are vectors, for example in the 4x4 representation of the $\gamma$-matrices:

$\psi^\dagger = \begin{pmatrix} \psi_1^* & \psi_2^* & \psi_3^* & \psi_4^* \end{pmatrix}^{T}$

and

$\psi=\begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3\\ \psi_4 \end{pmatrix}$

where $\psi_1$ to $\psi_4$ denote the spinor components. So the expression you gave above is of the form: row vector times matrix times column vector. Which overall is just a number. This also makes clear why the matrices can possibly commute with the field, the matrix multiplication would not make sense anymore.

To proceed in your question (The following are advices since I do not want to solve the problem for you and rather general. If you'd like more specific advice I'm happy to help). Usually one has scattering between particles of certain momentum. Therefore one would first want to use a Fourier transform (plane wave expansion) of the field to transform to momentum space.

After that a useful trick is to note that:

$\psi^\dagger \Gamma \phi = trace\left( \Gamma \phi \psi^\dagger \right)$

for any two spinors $\psi$, $\phi$ where $\Gamma$ is any combination of Dirac matrices. You can convince yourself of this identity by writing it out in components as I did above.

Then one can use completeness identities in Fourier space to get the required results in terms of momentum etc. Also note that usually the matrix element squared is computed, which is your expression squared (or something with similar structure). This simplifies calculations if the matrices are arranged in a convenient way (note that you can transpose Dirac numbers since they have no Dirac structure.).

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  • $\begingroup$ Thanks for not giving a spoiler :) it allowed me to delve quite deeply into QCD and understand the origins of this relation which it turns out is the Partially Conserved Axial Vector Current (PCAC) and to do with the QCD vacuum. This makes sense since the problem was posed by a QCD researcher! I am confused however, that you have said the matrices commute with the field. It appears to me from me most recent reading that they don't as $\frac{1}{2}(1\pm\gamma^5)\psi = \psi_{R,L}$ and so $\gamma^5\psi = \psi_R - \psi_L$ I think I may have resolved the issue though... $\endgroup$ – Alexander McFarlane Apr 15 '16 at 12:24
  • $\begingroup$ I am under the impression that $[\gamma^5,T_a]\psi(x) = 0$ and spent some time digging up papers on PCAC to try and prove it. In doing so I found an interesting question which seems to be posed by someone stuck on the same problem and I think I have a convincing proof in my answer math.stackexchange.com/a/1743587/272850 $\endgroup$ – Alexander McFarlane Apr 15 '16 at 12:28
  • $\begingroup$ Feel free to critique the above as I would welcome any issues with it btw $\endgroup$ – Alexander McFarlane Apr 15 '16 at 16:36
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    $\begingroup$ @AlexanderMcFarlane I'm afraid I might have completely misunderstood your question. I thought it was a standard problem about gamma matrices and didn't realise the SU(3) structure. The main point I was trying to get at is the role of the spinors as column and row vectors, since I thought that was what you were stuck on. But apparently it isn't, I apologize and will give it another thought. $\endgroup$ – Wolpertinger Apr 15 '16 at 17:30
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    $\begingroup$ No worries I appreciate the comments and it gave me some interesting things to think about nevertheless $\endgroup$ – Alexander McFarlane Apr 15 '16 at 17:54

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