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I asked this question as a subquestion in another thread, but got the answer below and thought it deserved a thread of its own.

Two well-known representation of the gamma matrices are the Weyl and Dirac-Pauli reps. The Weyl rep is often used when dealing with ultra-relativistic systems (or massless ones) and the Dirac for the other case. But I've never come across in any QFT book how to derive these reps, they just give it to you. So what is the procedure to derive a rep for the gamma matrices?

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    $\begingroup$ Hi mate - unsure if you are notified when I edit my answers but I added another section to the answer below due to the slightly different phrasing of your question. Took me a bit of time to figure out - sorry for the delay :). $\endgroup$ – AltLHC Jan 29 '14 at 12:10
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One option is to start out with the matrix representation for two sets of conjugate Grassmann numbers (see previous thread), $\theta_i, \pi_i$ with $i=1,...,N$, such that

$\{\theta_i,\theta_j\}=0,\quad\{\pi_{i},\pi_{j}\} = 0, \quad \{\theta_i,\pi_j\} = \delta_{ij}$

Then a $2N$-dimensional Clifford algebra can be built by

$\gamma_{i}=\theta_{i}+\pi_{i}\\ \gamma_{N+i}=i(\theta_{i}-\pi_{i})$

Given the above anti-commutation relations it is straightforward to verify that $\{\gamma_{i},\gamma_{j}\}=2\delta_{ij}\mathbf{1}$. For a odd number of dimensions the last $\gamma$-matrix can be found by considering the product

$\gamma_{2N+1} = i^N\prod_{i=1}^{2N}\gamma_{i} = i^N\gamma_{1}\gamma_2...\gamma_{2N}$

To get a representation of the Dirac algebra $\{\gamma_{\mu},\gamma_\nu\}=2g_{\mu\nu}\mathbf{1}$ with signature (+,-,-,...,-) simply rotate all but one of the matrices in the representation such that $\gamma_i\to i\gamma_i$ (and relabel a bit).

This approach enables one to derive general representations of the gamma matrices from Grassmann numbers.

However, another option exists, namely to start out with a lower dimensional representation of the Clifford algebra (which can be computed by the method described above). A well-known case of a lower-dimensional representation, which was also known to Weyl & Dirac, would be the Pauli matrices:

$\sigma_{1} = \left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right], \quad \sigma_{2} = \left[\begin{matrix} 0 & -i \\ i & 0 \end{matrix}\right], \quad \sigma_{3} = \left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right] $

From these matrices outer products, $\rho_i = \mathbf{1}\otimes \sigma_i$ and $\eta_i = \sigma_i \otimes \mathbf{1}$, can be formed. It is then clear that $[\rho_i,\eta_j]=0$ which makes it possible to choose five matrices from the set $\{\rho_i,\eta_j,\rho_i\eta_j\}$ which fulfill the Clifford algebra.

To make this approach a bit more explicit, consider starting with a diagonal matrix from the initial set for simplicity - let us choose $\rho_3$ ($\eta_3$ would have been another option). This leaves us with two potential sets of matrices, namely $\{\rho_1,\rho_2\eta_1,\rho_2\eta_2,\rho_2\eta_3\}$ and $\{\rho_{2},\rho_{1}\eta_{1},\rho_{1}\eta_{2},\rho_{1}\eta_{3}\}$. Since $\rho_1$ is real I choose the first set, making the matrices:

\begin{align} \gamma_0 &= \left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}\right] = \rho_3, &&\gamma_1 = \left[\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{matrix}\right] = i\rho_2\eta_1 \\ \gamma_2 &= \left[\begin{matrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{matrix}\right] = i\rho_2\eta_2, &&\gamma_3 = \left[\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{matrix}\right] = i\rho_2\eta_3\\ \gamma_5 &= \left[\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{matrix}\right] = \rho_1 \end{align}

where I have taken the liberty to rotate three of them as described above. In this way the Dirac representation is found. Notice that a few choices were made along the way but that several of them can be motivated by the search for a simple representation (choosing diagonal and/or real when possible).

This approach can naturally be generalized to generate higher dimensional representations as well.

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  • $\begingroup$ The following question might be related to this question: physics.stackexchange.com/questions/103529/… Could you provide a possible answer? $\endgroup$ – linuxfreebird Mar 16 '14 at 15:14
  • $\begingroup$ This answer contains wrong information. First, @AltLHC did not distinguish anticommutation relation and Poisson bracket. I have to point it out that Grassmann numbers are conjugate with themselves. The quantization of these anti-commutative numbers give rise to the Clifford algebra. For the Clifford algebra, we find that their anti-commutators correspond to the canonical quantization of fermions. By taking semi-classical limit $\hbar\rightarrow 0$, we recover the Grassmann valued numbers. $\endgroup$ – Xiaoyi Jing Mar 2 '16 at 7:21
  • $\begingroup$ This is a wrong answer, which is misleading. $\endgroup$ – Xiaoyi Jing Mar 2 '16 at 7:33
  • $\begingroup$ @AltLHC To be more precise, from the pseudo-classical mechanics of the 'classical fermions', we start from a bunch of anti-commuting numbers $\theta_{i}\theta_{j}+\theta_{j}\theta_{i}=0$, forming the Grassmann algebra $\Lambda$. Remark: At this stage all Grassmann generators anticommute. It does not make any sense to have a relation $ab+ba=1$ within the Grassmann algebra. On this algebra, we can construct its symplectic structure, after which we can find the 'graded-Poisson bracket' $\left\{\theta_{i},\theta_{j}\right\}_{PB}=\delta_{ij}$. $\endgroup$ – Xiaoyi Jing Mar 2 '16 at 7:38
  • $\begingroup$ @AltLHC Then, the canonical quantization is straightforward. We do it by replacing $\left\{\theta_{i},\theta_{i}\right\}_{PB}=\delta_{ij}$ by $[\hat{\theta}_{i},\hat{\theta}_{j}]_{+}=\hbar\delta_{ij}$. Remark, although the Grassmann numbers $\theta_{i}$ and $\theta_{j}$ anti-commute, i.e. $\theta_{i}\theta_{j}+\theta_{j}\theta_{i}=0$, their classical graded Poisson bracket $\left\{\theta_{i},\theta_{i}\right\}_{PB}$ does not vanish. $\endgroup$ – Xiaoyi Jing Mar 2 '16 at 7:42
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Here's a general solution to the problem of defining the representations of gamma matrices.

A representation of gamma matrices is completely defined by defining the following five state vectors:

$$ D_0 = D_{++} = \left[\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}\right],\;\;\; D_1 = D_{+-} = \left[\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix}\right],\;\;\; D_2 = D_{-+} = \left[\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}\right],\\ D_3 = D_{--} = \left[\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}\right],\;\;\;\;\; V = \left[\begin{matrix} 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \end{matrix}\right].\\ $$

The reason for calling these the "D" and "V" states is that the density matrices of the first four are diagonal while the last one is an example of what Schwinger called the "vacuum" in one of his papers on the Schwinger Measurement Algebra. I've given two notations for the D_xx in order to simplify the derivation.

The four diagonal states are defined by a "Complete Set of Commuting Observables". This is just enough information to determine a quantum state. For the case of Dirac gamma matrices, a complete set of commuting observables will have two (distinct) observables that happen to be "roots of unity". That is, the two observables $P$ and $Q$ have to satisfy:

$$PP = QQ = 1,\;\;P\;\textrm{not equal to}\;Q ,\;\;\textrm{and}\;\; PQ = QP,$$

with neither equal to 1 or -1. An example of two gamma algebra elements that satisfy these requirements are

$$P = \gamma^3\gamma^0,\;\;Q = i\gamma^1\gamma^2$$

Ignoring signs, there are sixteen possible products of gamma matrices:

$$\{1,\gamma^1,\gamma^2,\gamma^1\gamma^2, ... \gamma^1\gamma^2\gamma^3\gamma^0\}$$

To use one of these as a root of unity we have to make it square to +1. So if it squares instead to -1, simply multiply it by i. We can't use 1 as $P$, so after converting the above to roots of unity, we have 15 choices for $P$.

Once we choose $P$, there are 14 roots of unity left. Of these, just six will commute with $Q$ so we can choose one of those six. And having chosen $P$ and $Q$, their product $PQ=QP$ will also be a root of unity that commutes with $P$ and $Q$. So in addition to $\{P,Q\}$, we could as well have chosen $\{P,PQ\}$ or $\{Q,PQ\}$ and ended up with the same representation.

Next we have to choose $V$. This is a single state that is a "complementary state" to our Complete Set of Commuting Observables. This is a basic concept in quantum mechanics; the usual example is position and momentum. The requirement is that the transition probabilities between the state $V$ and the $D_n$ states all need to be the same.

In fact, the $V$ state is easy to choose. Of the 16 products of gamma matrices, one was trivial (i.e. 1), and three are the $\{P,Q,PQ\}$. The remaining 12 products are complementary observables. All we need to do is to choose two of them that are distinct and that commute. Call them $R$ and $S$. We can choose $R$ completely freely from those 12 products. After choosing $R$, there will be four choices for $S$ and, as before, they will come in pairs so there are really only two choices. Then the $V$ state will take +1 or -1 as its eigenvalues with respect to $R$ and $S$ so with $R$ and $S$ in hand, there will be four choices for $V$.

Things are easier if we write them in pure density matrix form. If we use the +1 eigenvalues for the state $V$, its pure density matrix state will be:

$$\rho_V = (1+R)(1+S)/4 =\frac{1}{4}\left[\begin{matrix} 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \\ 1&1&1&1 \end{matrix}\right]$$

Now we can write the four (pure) density matrices corresponding to $D_n$ as:

$$ \rho_{D++} = (1+P)(1+Q)/4\\ \rho_{D+-} = (1+P)(1-Q)/4\\ \rho_{D-+} = (1-P)(1+Q)/4\\ \rho_{D--} = (1-P)(1-Q)/4\\$$

When written in the representation, the above are to be diagonal matrices, each with a single 1 on the diagonal and the rest zero.

Similarly, we can pick out the nm matrix element of the representation by multiplications of density matrices of the form:

$$M_{nm} = \rho_{Dn}\;\rho_V\;\rho_{Dm}$$

For example: $$M_{02} = \rho_{D0}\;\rho_V\;\rho_{D2}$$ which in the representation is explicitly: $$\frac{1}{4}\left[\begin{matrix}0&0&1&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{matrix}\right]= \frac{1}{4}\left[\begin{matrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{matrix}\right] \left[\begin{matrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{matrix}\right] \left[\begin{matrix}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0\end{matrix}\right]$$

Since we now know which algebra elements represent a particular spot in the matrix, we can read off the representation by seeing which algebra elements contribute to which matrix elements. For example, if $\gamma_0$ shows up in $M_{02}$, then the representation of $\gamma_0$ will have a 1 in its (0,2) position.

As it turns out, I've been doing research on gamma matrices recently and have written an application using Java that allows the user to work through the above steps by clicking buttons in a graphical user interface. It shows resulting representation and you can also show the results for other products of gamma matrices such as $i\gamma^1\gamma^2$. I put in some presets so you can get standard gamma matrix representations in a single key click and found this question while using a search engine to find more cases. I'll link it in when it's done.

Also, it should be noted that the above explicitly describes only the representations that happen to represent the products of gamma matrices as matrices whose 16 elements include four complex phases and 12 zeroes. By rotations and boosts, you can get more general representations.

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