# The proof that Dirac's hamiltonian commutes with inversion operator

I tried to check the statement that Dirac free Hamiltonian commutes with inversion operator. For $$\hat {P}\Psi(\mathbf r , t) = i\hat {\gamma}_{0}\Psi (-\mathbf r , t), \quad \hat {H} = (\hat {\alpha} \cdot \hat {\mathbf p}) + \hat {\gamma}_{0}m$$ I got

# $$[\hat {H}, \hat {P}]\Psi (\mathbf r, t) = i\hat {H}\hat {\gamma}_{0} \Psi (-\mathbf r , t ) - \hat {P} \hat {H}\Psi (\mathbf r, t) =$$

$$i ((\hat {\alpha} \cdot \hat {\mathbf p }) + \hat {\gamma}_{0} m)\hat {\gamma}_{0}\Psi (-\mathbf r , t) - i (-(\hat {\alpha} \cdot \hat {\mathbf p }) + \hat {\gamma}_{0} m)\hat {\gamma}_{0}\Psi (-\mathbf r , t) =$$ $$=2i(\hat {\alpha} \cdot \hat {\mathbf p} )\hat {\gamma}_{0}\Psi (-\mathbf r , t).$$ Where is the mistake?

Maybe, my mistake is in the following: $$\hat {P} \hat {H}(\mathbf p) \Psi (\mathbf r , t) \neq i\hat {H} (-\mathbf p )\hat {\gamma}_{0}\Psi (-\mathbf r , t),$$ the correct one is $$\hat {P} \hat {H}(\mathbf p) \Psi (\mathbf r , t) = i\hat {\gamma}_{0}\hat {H} (-\mathbf p )\Psi (-\mathbf r , t).$$ But I don't understand, why. For example, when I acting by $\hat {P}$ on energy expression $$\hat {P} E = \hat {P} \int \Psi^{+}((\hat {p} \cdot \hat {\alpha}) + \hat {\gamma}_{0} m)\Psi d^{3}\mathbf r = \int \hat {P}(\Psi^{+}) (-(\hat {p} \cdot \hat {\alpha}) + \hat {\gamma}_{0} m)\hat {P} (\Psi )d^{3}\mathbf r,$$ the $\hat {P}$ only changes the sign in $\hat {\mathbf p}$ summand of hamiltonian, while the $\hat {\gamma}_{0}$ doesn't act on it.

• Inversion operator? Do you mean parity operator? – Qmechanic Jul 29 '16 at 22:57