Is it possible to retrieve the matrix elements of the $\gamma$s by simply knowing their anti-commutation relation:
$$
\{\gamma^\mu, \gamma^\nu\}=2\,g^{\mu\nu}\,\mathbb{I}_{4}
$$
I'm just trying to reconstruct Dirac's reasoning when he first encountered these relations (previous to his equation). The analogy with the Pauli matrices is evident, which at that time were newly introduced, so it's reasonable to assume some kind of analogy. But it's still too early to guess correctly.
He might have recognized the Clifford algebra… but again I think it's not enough.
Dirac Matrices – On the Derivation of Dirac Matrices in Quantum Mechanics
clifford-algebradirac-matricesquantum mechanicsspecial-relativity
Related Solutions
Although the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}$ is the most famous, there is an expression for the commutator:
$$[\gamma^\mu,\gamma^\nu] = 2\gamma^\mu \gamma^\nu - 2 \eta^{\mu\nu}$$
The matrix defined by $[\gamma^\mu,\gamma^\nu]$ actually has a purpose: it forms a representation of the Lorentz algebra. If we define $S^{\mu\nu}$ as $1/4$ the commutator, then we have,
$$[S^{\mu\nu},S^{\rho\sigma}] = \eta^{\nu\rho}S^{\mu\sigma} - \eta^{\mu\rho}S^{\nu\sigma} + \eta^{\mu\sigma}S^{\nu\rho} - \eta^{\nu\sigma}S^{\mu\rho}= \eta^{\rho[\nu}S^{\mu]\sigma} + \eta^{\sigma [\mu}S^{\nu]\rho}$$
which is the Lorentz algebra. One can verify this by simply using the first commutator, and the rule for the commutator involving a product.
There is a particularly important use for the commutator, namely defining $\sigma^{\mu\nu} = \frac{i}{2} [\gamma^\mu,\gamma^\nu]$, the action of a spin-$\frac32$ particle is given by,
$$\mathcal L = -\frac{1}{2}\bar{\psi}_\mu \left( \varepsilon^{\mu\lambda \sigma \nu} \gamma_5 \gamma_\lambda \partial_\sigma -im\sigma^{\mu\nu}\right)\psi_v,$$
which can be used to describe the superpartner to the graviton, namely the gravitino, thus making it necessary for supergravity theories.
It's pretty annoying that P&S just give you $$S^{\mu \nu} = \frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]$$ from thin air, here is a way to derive it similar to Bjorken-Drell's derivation (who start from the Dirac equation) but from the Clifford algebra directly, assuming that products of the gamma matrices form a basis. Given a Clifford algebra of $\gamma^{\mu}$'s satisfying \begin{align} \{ \gamma^{\mu} , \gamma^{\mu} \} = 2 \eta^{\mu \nu} I \end{align} we note that for an invertible transformation $S$ we have \begin{align} 2 \eta^{\mu \nu} I &= 2 \eta^{\mu \nu} S^{-1} S \\ &= S^{-1}(2 \eta^{\mu \nu}) S \\ &= S^{-1}\{ \gamma^{\mu} , \gamma^{\mu} \} S \\ &= \{ S^{-1} \gamma^{\mu} S, S^{-1}\gamma^{\mu} S \} \\ &= \{ \gamma'^{\mu} , \gamma'^{\mu} \} \end{align} showing us that the Clifford algebra of matrices $$\gamma'^{\mu} = S^{-1} \gamma^{\mu} S$$ also satisfies the Clifford algebra, hence any set of matrices satisfying the Clifford algebra can be obtained from a given set $\gamma^{\mu}$ using a non-singular transformation $S$. Since the anti-commutation relations involve the metric $\eta_{\mu \nu}$, and we know the metric is left invariant under Lorentz transformations $$\eta^{\mu \nu} = \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \eta^{\rho \sigma} $$ this immediately implies \begin{align} 2 \eta^{\mu \nu} I &= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} 2 \eta^{\rho \sigma} I \\ &= \Lambda^{\mu} \, _{\rho} \Lambda^{\nu} \, _{\sigma} \{ \gamma^{\rho} , \gamma^{\sigma} \} \\ &= \{ \Lambda^{\mu} \, _{\rho} \gamma^{\rho} , \Lambda^{\nu} \, _{\sigma} \gamma^{\sigma} \} \\ &= \{ \gamma'^{\mu} , \gamma'^{\mu} \} \end{align} which shows that the Lorentz transformation of a gamma matrix also satisfies the Clifford algebra, and so is itself a gamma matrix, and hence can be expressed in terms of some non-singular transformation $S$ \begin{align} \gamma'^{\mu} &= \Lambda^{\mu} \, _{\nu} \gamma^{\nu} \\ &= S^{-1} \gamma^{\mu} S \end{align} where $S$ is to be determined. Since the operators $S$ represent performing a Lorentz transformation on $\gamma^{\mu}$, and Lorentz transformations on fields expand as $I - \frac{i}{2}\omega_{\mu \nu} M^{\mu \nu}$, we expand $\Lambda$ and $S$ as \begin{align} \Lambda^{\mu} \, _{\nu} &= \delta ^{\mu} \, _{\nu} + \omega^{\mu} \, _{\nu} \\ S &= I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu} \end{align} where $\Sigma^{\mu \nu}$ must be anti-symmetric and constructed from a basis of gamma matrices, hence from \begin{align} \gamma'^a &= \Lambda^a \, _{\mu} \gamma^{\mu} \\ &= (\delta^a \, _{\mu} + \omega^a \, _{\mu})\gamma^{\mu} \\ &= \gamma^a + \omega^a \, _{\mu} \gamma^{\mu} \\ &= \gamma^a + \omega_{b \mu} \eta^{a b} \gamma^{\mu} \\ &= \gamma^a + \omega_{b \mu} \eta^{a [b} \gamma^{\mu]} \\ &= \gamma^a + \frac{1}{2} \omega_{b \mu} (\eta^{a b} \gamma^{\mu} - \eta^{a \mu} \gamma^b) \\ &= \gamma^a + \frac{1}{2} \omega_{\nu} (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) \\ &= S^{-1} \gamma^a S \\ &= (I - \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \gamma^a (I + \frac{i}{2} \omega_{\mu \nu} \Sigma^{\mu \nu}) \\ &= \gamma^a - \frac{i}{2} \omega_{\mu \nu} [\gamma^a, \Sigma^{\mu \nu}] \end{align} we have the relation (which can be interpreted as saying that $\gamma^a$ transforms as a vector under spinor representations of Lorentz transformations, as e.g. in Tong's QFT notes) \begin{align} i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) = [\gamma^a, \Sigma^{\mu \nu}] \end{align} and we know $\Sigma^{\mu \nu}$, since it is anti-symmetric, must involve a product's of $\gamma$ matrices (because of the 16-dimensional basis formed from Clifford algebra elements), only two by the left-hand side, and from \begin{align} \gamma^{\mu} \gamma^{\nu} &= - \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu \neq \nu, \\ \gamma^{\mu} \gamma^{\mu} &= \gamma^{\nu} \gamma^{\mu} , \ \ \ \mu = \nu, \end{align} we expect that \begin{align} \Sigma^{\mu \nu} &= c [\gamma^{\mu},\gamma^{\nu}] \\ &= c (\gamma^{\mu} \gamma^{\nu} - \gamma^{\nu} \gamma^{\mu}) \\ &= 2 c ( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu}) \end{align} for some $c$ which we constrain by the (vector) relation above \begin{align} i (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}) &= [\gamma^a, \Sigma^{\mu \nu}] \\ &= c [\gamma^a, 2( \gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu})] \\ &= 2 c [\gamma^a, \gamma^{\mu} \gamma^{\nu}] \\ &= 2 c ( \gamma^{\mu} [\gamma^a,\gamma^{\nu}] + [\gamma^a, \gamma^{\mu}] \gamma^{\nu}) \\ &= 2 c [ \gamma^{\mu} 2( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + 2( \gamma^a \gamma^{\mu} - \eta^{a \mu}) \gamma^{\nu}] \\ &= 4 c [ \gamma^{\mu} ( \gamma^a \gamma^{\nu} - \eta^{a \nu}) + ( \gamma^{\mu} \gamma^{a} + 2 \eta^{a \mu} - \eta^{a \mu}) \gamma^{\nu}] \\ &= 4 c (\eta^{a \mu} \gamma^{\nu} - \eta^{a \nu} \gamma^{\mu}). \end{align} This gives the result $c = i/4$. The generator of Lorentz transformations of gamma matrices is \begin{align} \Sigma^{\mu \nu} &= \dfrac{i}{4} [\gamma^{\mu},\gamma^{\nu}] \\ &= \dfrac{i}{2}(\gamma^{\mu} \gamma^{\nu} - \eta^{\mu \nu}) \ \ \text{i.e.} \\ S &= I - \frac{i}{2} \omega_{\mu \nu} (\frac{i}{4} [\gamma^{\mu},\gamma^{\nu}]) \\ &= I + \dfrac{1}{8} \omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu}]. \end{align} Using the fact that the gamma matrices transform as a vector under the spinor representation of an infinitesimal Lorentz transformation, \begin{align} [\Sigma^{\mu \nu}, \gamma^{\rho}] = i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho}) \end{align} we can show the spinor representation of a Lorentz transformation satisfies the Lorentz algebra commutation relations, since for $\rho \neq \sigma$ \begin{align} [\Sigma^{\mu \nu},\Sigma^{\rho \sigma}] &= \frac{i}{2}[\Sigma^{\mu \nu},\gamma^{\rho} \gamma^{\sigma}] \\ &= \frac{i}{2}([\Sigma^{\mu \nu},\gamma^{\rho} ] \gamma^{\sigma} + \gamma^{\rho} [\Sigma^{\mu \nu}, \gamma^{\sigma}]) \\ &= \frac{i}{2}\{ i (\gamma^{\mu} \eta^{\nu \rho} - \gamma^{\nu} \eta^{\mu \rho}) \gamma^{\sigma} + \gamma^{\rho} i (\gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\nu} \eta^{\mu \sigma}) \} \\ &= - \frac{1}{2}\{ \gamma^{\mu} \eta^{\nu \rho} \gamma^{\sigma} - \gamma^{\nu} \eta^{\mu \rho} \gamma^{\sigma} + \gamma^{\rho} \gamma^{\mu} \eta^{\nu \sigma} - \gamma^{\rho} \gamma^{\nu} \eta^{\mu \sigma} \} \\ &= \frac{i}{2}\{ \eta^{\nu \rho} (2 \Sigma^{\mu \sigma} + \eta^{\mu \sigma}) - \eta^{\mu \rho} (2 \Sigma^{\nu \sigma} - \eta^{\nu \sigma}) + (2 \Sigma^{\rho \mu} - \eta^{\rho \mu}) \eta^{\nu \sigma} - (2 \Sigma^{\rho \nu}) - \eta^{\rho \nu}) \eta^{\mu \sigma} \} \\ &= i ( \eta^{\nu \rho} \Sigma^{\mu \sigma} - \eta^{\mu \rho} \Sigma^{\nu \sigma} + \Sigma^{\rho \mu} \eta^{\nu \sigma} - \Sigma^{\rho \nu} \eta^{\mu \sigma} ). \end{align} This method generalizes from $SO(3,1)$ to $SO(N)$, see e.g. Kaku QFT Sec. 2.6, and the underlying reason for doing any of this in the first place is that one seeks to find projective representations which arise due to the non-simple-connectedness of these orthogonal groups. Regarding your question about arbitrary metrics $g_{\mu \nu}$, this method applies to, and arises due to the non-simple-connectedness of, special orthogonal groups, you can't generalize to arbitrary metrics, this is a problem which can be circumvented in supergravity and superstring theory using veilbein's.
References:
- Bjorken, J.D. and Drell, S.D., 1964. Relativistic quantum mechanics; Ch. 2.
- Kaku, M., 1993. Quantum field theory: a modern introduction. Oxford Univ. Press; Sec. 2.6.
- Tong, Quantum Field Theory Notes http://www.damtp.cam.ac.uk/user/tong/qft.html.
- http://www.damtp.cam.ac.uk/user/examples/D18S.pdf
- Does $GL(N,\mathbb{R})$ own spinor representation? Which group is its covering group? (Kaku's QFT textbook)
Best Answer
Yes, you can find a set of $\gamma$ matrices just by knowing their commutation relations $$\{\gamma^\mu, \gamma^\nu\}=2\,g^{\mu\nu}\,\mathbb{I}_{4}$$
One such set is the Dirac basis: $$\gamma^0=\begin{pmatrix}I_2&0\\0&-I_2\end{pmatrix},\quad \gamma^k=\begin{pmatrix}0&\sigma^k\\-\sigma^k&0\end{pmatrix} \text{ with }k=1,2,3$$ But this solution is not unique. Another widely used set is the Weyl basis: $$\gamma^0=\begin{pmatrix}0&I_2\\I_2&0\end{pmatrix},\quad \gamma^k=\begin{pmatrix}0&\sigma^k\\-\sigma^k&0\end{pmatrix} \text{ with }k=1,2,3$$
Furthermore, from a set of $\gamma$ matrices you can (with any unitary $4\times 4$ matrix $U$) construct many more sets of $\gamma$ matrices by $$\gamma'^\mu = U^\dagger\gamma^\mu U$$ which will automatically satisfy the required commutation relations as well.
And for the $\gamma^\mu$ you are not restricted to $4\times 4$ matrices. You can also find $8\times 8$ matrices or even bigger matrices. But, since the physical results got from these will be the same, there is no need to bother with them.