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I want to understand clearly why $ P \gamma^{\mu} P = \gamma^{\mu} $, where $ P $ is the parity operator.

This result follow for example from pag. 66 of Peskin-Schroeder. The parity operator acts on Dirac fields in this way: $ P \psi (t, \textbf{x}) P = \gamma^0 \psi (t, -\textbf{x}) $, assuming no other phase factors. On the Dirac bilinear $ \bar{\psi} \gamma^{\mu} \psi (t, \textbf{x}) $, using the fact that $ P^2=1 $, you have $P \bar{\psi} \gamma^{\mu} \psi (t, \textbf{x})P = P \bar{\psi} P P \gamma^{\mu} P P \psi (t, \textbf{x}) P = \bar{\psi}\gamma^0 P \gamma^{\mu} P \gamma^0 \psi (t,- \textbf{x}) = \bar{\psi}\gamma^0 \gamma^{\mu} \gamma^0 \psi (t,- \textbf{x}) = (-1)^{\mu}\bar{\psi} \gamma^{\mu} \psi (t, -\textbf{x}) $

With $(-1)^{\mu} =1 $ if $ \mu=0 $ and $(-1)^{\mu} =-1 $ if $ \mu=1,2,3 $. So $ P \gamma^{\mu} P = \gamma^{\mu} $ follows because $ P$ and $\gamma^{\mu} $ act on different spaces? Or there are other explanations? (Please, do all the steps in the answer)

To be clear: I use $ \gamma^0= \bigl(\begin{smallmatrix} 0&1\\ 1&0 \end{smallmatrix} \bigr)$ and $ \gamma^{i}= \bigl(\begin{smallmatrix} 0&\sigma^i\\ -\sigma^i&0 \end{smallmatrix} \bigr)$.

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    $\begingroup$ You have to invert the logic. $v^\mu= {\psi(t, \textbf{x})} \gamma^{\mu} \psi(t, \textbf{x}) $ is a Lorentz vector. By definition of the parity operator, you have $Pv^\mu P=(-1)^{\mu}v'^\mu$; with $v'^\mu={\psi(t, -\textbf{x})} \gamma^{\mu} \psi(t, -\textbf{x})$. To do this, the equalities that you have written in your question mean that you must have $P\gamma^{\mu}P = \gamma^{\mu}$ $\endgroup$ – Trimok Jun 18 '14 at 9:09
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One of the ways for getting this result is the trivial consequence that parity operator must change the sign of integral values of 3-momentum and current while it leaves invariant full energy and charge values. For example, for energy density we have with $\Psi ' = \hat {P}\Psi$ ($\hat {P} = \hat {U}P_{\mathbf x \to -\mathbf x}$) the following result:

$$ E \to E' = \Psi{'}^{\dagger}(-\mathbf x, t) \left((\hat {\mathbf p} \cdot \hat {\alpha}) + \gamma_{0}m\right)\Psi{'}(-\mathbf x , t) = $$ $$ =\Psi^{\dagger}(\mathbf x , t)\hat {P}^{\dagger}\left((\hat {\mathbf p} \cdot\hat {\alpha} ) + \gamma_{0}m\right)\hat {P}\Psi(\mathbf x, t) = $$ $$ =\Psi^{\dagger}(\mathbf x , t)\left(-\left(\hat {\mathbf p} \cdot \hat {P}^{+}\hat {\alpha}\hat {P} \right) + \hat {P}^{\dagger}\gamma_{0}\hat {P}m\right)\Psi(\mathbf x, t) = $$ $$ =\Psi^{\dagger}(\mathbf x, t) \left((\hat {\mathbf p} \cdot\hat {\alpha} ) + \gamma_{0}m\right)\Psi(\mathbf x , t) \Rightarrow $$ $$ \hat {U}^{\dagger}\hat {\alpha}\hat {U} = -\hat {\alpha}, \quad \hat {U}^{\dagger}\gamma_{0}\hat {U} = \gamma_{0} \Rightarrow \hat {U}^{\dagger}\gamma^{\mu}\hat {U} = \gamma_{\mu}, \qquad (1) $$

so your equality is possible only in a form of $(1)$ ($\hat{P}^{\dagger}$ formally coincide with $\hat{P}$).

In the first line I write the expression for the energy density after transformation of the function of the state ($\Psi \to \Psi {'}$). In the second one I have used the expressions $\Psi{'} = \hat {P}\Psi , \Psi{'}^{+} = \Psi^{+}\hat {P}^{+}$; in the third one I have moved $\hat {P}^{\dagger}$ right and $\hat {P}$ left on ($\hat {P}^{\dagger}$ acts on $\hat {\mathbf p}$ by only changing its sign) and after that I have got bilinear forms $\hat {P}^{+}\hat {\alpha}\hat {P}, \hat {P}^{+}\gamma_{0}\hat {P}$. Finally, I have equated this result to the energy without inversion, because energy doesn't change (by its physical meaning) under spatial inversion and I have got $(2)$ by equating corresponding expressions from the last line and from the previous one: $-\left(\hat {\mathbf p} \cdot \hat {P}^{+}\hat {\alpha}\hat {P} \right)$ to $(\hat {\mathbf p} \cdot\hat {\alpha} )$ etc.

The last consequence can be obtained by the following way: $$ \hat {U}^{\dagger}\gamma_{0}\hat {U} = \gamma_{0} \Rightarrow \hat {U}^{\dagger}\hat{\alpha}\hat {U} = \hat {U}^{\dagger}\gamma_{0}\mathbf {\gamma}\hat {U} = \gamma_{0}\hat {U}^{\dagger}\mathbf {\gamma}\hat {U} = -\gamma_{0} \gamma \Rightarrow $$ $$ \hat {U}^{\dagger}\gamma_{0}\hat {U} = \gamma_{0}, \quad \hat {U}^{\dagger}\gamma \hat {U} = -\gamma . $$

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  • $\begingroup$ Can you do in detail all the steps? Also, I don't understand the last line. $\endgroup$ – Rexcirus Jun 17 '14 at 19:26
  • $\begingroup$ @Rexcirus : I have fixed my answer. $\endgroup$ – Andrew McAddams Jun 18 '14 at 4:27
  • $\begingroup$ Thanks for the edit. Anyway, why in the first step is $ \textbf{p} $ and not $ -\textbf{p}$? Why only the spinors have changed? If $ -\textbf{p}$ is correct, then $ U \alpha U= \alpha $ and finally $ U \gamma^{\mu} U= \gamma^{\mu} $. From the argument of the Dirac bilinear I expect this formula to be valid, and not the one with $ \gamma_{\mu} $ on the right side. $\endgroup$ – Rexcirus Jun 18 '14 at 8:04
  • $\begingroup$ @Rexcirus : let's assume your version. You wrote that for some operator's density $\Psi^{\dagger}\hat {A}\Psi$ there must be simultaneously replacements $\hat {A} \to \hat {A}{'} = \hat {P}\hat {A}\hat {P}^{\dagger}$ and $\Psi \to \Psi{'} = \hat {P}\Psi$, $\Psi^{\dagger} \to \Psi^{\dagger}{'} = \Psi^{\dagger}\hat {P}^{\dagger}$. But after these replacements the density will be unchnanged. $\endgroup$ – Andrew McAddams Jun 18 '14 at 16:18
  • $\begingroup$ There must be one of two transformations: 1) you transform the state funstions $\Psi $ (so the value $\Psi^{\dagger}\hat {A}\Psi$ refers to the density of $\hat {A}$ in the NEW system which is given by $\Psi{'}$), and 2) you transform the operator $\hat {A}$ expression. $\endgroup$ – Andrew McAddams Jun 18 '14 at 16:21

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