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.

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    $\begingroup$ Inversion operator? Do you mean parity operator? $\endgroup$ – Qmechanic Jul 29 '16 at 22:57

the correct answer is \begin{equation} [Ĥ ,P̂ ]Ψ(r,t)=iĤγ̂_{0}Ψ(−r,t)−P̂ Ĥ Ψ(r,t)= \end{equation}

\begin{equation} i((α̂⋅p̂)+γ̂_{0}m)γ̂_{0}Ψ(−r,t)−iγ̂_{0}(−(α̂⋅p̂)+γ̂_{0}m)Ψ(−r,t)= \end{equation} \begin{equation} i[(α̂⋅p̂)γ̂_{0}+γ̂_{0}(α̂⋅p̂)]Ψ(−r,t)+i[γ̂_{0}mγ̂_{0}-γ̂_{0}mγ̂_{0}]Ψ(−r,t)= \end{equation} \begin{equation} i[(α̂⋅p̂)γ̂_{0}+γ̂_{0}(α̂⋅p̂)]Ψ(−r,t)= \end{equation} \begin{equation} =0\Rightarrow [Ĥ ,P̂ ]=0 \hspace{0.6cm} (1) \end{equation} To see that penultimate line is equal to zero, you have to use the explicit form of gamma Dirac's matrix, or use this: \begin{equation} γ̂_{0}(α̂⋅p̂)=γ̂_{0}\hat{{\alpha}_{i}}p^{i} \end{equation} but \begin{equation} γ̂_{0}\hat{{\alpha}_{i}}=-\hat{{\alpha}_{i}}γ̂_{0} \end{equation} Then

\begin{equation} i[(α̂⋅p̂)γ̂_{0}+γ̂_{0}(α̂⋅p̂)]Ψ(−r,t)= \end{equation} \begin{equation} i[\hat{{\alpha}_{i}}p^{i}γ̂_{0}-\hat{{\alpha}_{i}}γ̂_{0}p^{i}]Ψ(−r,t)= \hspace{0.6cm}(2) \end{equation}

And we can do \begin{equation} γ̂_{0}p^{i}=p^{i}γ̂_{0} \end{equation}

Thus the equation (2) is equal to zero and we proof the eq (1).


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