**Anticommutation Relation**

The first property of gamma matrices is the anticommutation relation, which states that the product of any two gamma matrices is equal to the negative of the product of the same two matrices in the opposite order:

${\gamma^{\mu},\gamma^{\nu}} = 2g^{\mu\nu}I_4$

where $g^{\mu\nu}$ is the metric tensor and $I_4$ is the 4×4 identity matrix.

**Proof**

To prove this property, we can start by expanding the left-hand side of the equation:

${\gamma^{\mu},\gamma^{\nu}} = \gamma^{\mu}\gamma^{\nu} + \gamma^{\nu}\gamma^{\mu}$

Next, we will use the specific form of the gamma matrices to calculate the product $\gamma^{\mu}\gamma^{\nu}$:

$\gamma^{\mu}\gamma^{\nu} = \sum_{\alpha}\sum_{\beta}(\gamma^{\mu}){\alpha\beta}(\gamma^{\nu}){\beta\alpha}$

Now, we use the fact that the gamma matrices are Hermitian and the fact that the indices $\alpha$ and $\beta$ are summed over to find the product of the gamma matrices and the product of the conjugate transpose of the gamma matrices :

$\gamma^{\mu}\gamma^{\nu} = \sum_{\alpha}\sum_{\beta}(\gamma^{\mu}){\alpha\beta}(\gamma^{\nu}){\beta\alpha} = \sum_{\alpha}\sum_{\beta}((\gamma^{\mu}){\alpha\beta})^*((\gamma^{\nu}){\beta\alpha})^*$

= $\sum_{\alpha}\sum_{\beta}((\gamma^{\mu})^{\dagger}){\beta\alpha}((\gamma^{\nu})^{\dagger}){\alpha\beta} = (\gamma^{\mu\dagger}\gamma^{\nu\dagger})_{\alpha\beta}$

Finally, we use the Hermiticity property of the gamma matrices and the fact that the indices are summed over to find the product of the conjugate transpose of the gamma matrices:

$(\gamma^{\mu\dagger}\gamma^{\nu\dagger}){\alpha\beta} = (\gamma^{\mu}\gamma^{\nu}){\beta\alpha} = -(\gamma^{\nu}\gamma^{\mu})_{\alpha\beta}$

Therefore, the anticommutation relation is true:

${\gamma^{\mu},\gamma^{\nu}} = \gamma^{\mu}\gamma^{\nu} + \gamma^{\nu}\gamma^{\mu} = -(\gamma^{\mu}\gamma^{\nu}) + (\gamma^{\mu}\gamma^{\nu}) = 2g^{\mu\nu}I_4$

This completes the proof.