Calculating <p>_c for Charge-Conjugated Dirac Spinor

[qinglong.1397]

I calculated the expectation value of the momentum of the charge-conjugated Dirac spinor and found that it was the negative of that of the Dirac spinor. Here is the calculation.

Charge conjugation operator is chosen to be C=i\gamma^0\gamma^2. The spinor is \Psi and its charge-conjugated spinor \Psi_C=-i\gamma^2\Psi^*.

The expectation value of the momentum of \Psi_C=-i\gamma^2\Psi^* is given by

&lt;\vec p&gt;_C=\int d^3x\bar\Psi_C\vec p\Psi_C=\int d^3x\Psi^T\gamma^0\gamma^2\vec p\gamma^2\Psi^*=-[\int d^3x\bar\Psi\vec p^*\Psi]^*
=[\int d^3x\bar\Psi\vec p\Psi]^*=&lt;\vec p&gt;^*=&lt;\vec p&gt;

where &lt;\vec p&gt; is real.

Is there anything wrong with my calculation, because my teacher didn't give me the grade for this?

 

[qinglong.1397]

I guess my calculation is correct since nobody replies...

 

[Bill_K]

It's hard to tell from what you have written. Did you remember that p acts to the right, and when you switch it from acting on one ψ to the other, you have to integrate by parts?

 

[qinglong.1397]

It's hard to tell from what you have written. Did you remember that p acts to the right, and when you switch it from acting on one ψ to the other, you have to integrate by parts?

I did use integration by parts.

 

[Dickfore]

What does \Psi^{\ast} represent?

 

[Bill_K]

I did use integration by parts.

I'm just thinking that j_μ does change sign while p_μ does not, and the only difference between them is the derivative operator.

 

[qinglong.1397]

I'm just thinking that j_μ does change sign while p_μ does not, and the only difference between them is the derivative operator.

By j_\mu, do you mean electric current?

Actually, I found out that &lt;\vec r&gt;_C=-&lt;\vec r&gt; and my teacher also got this.

 

[cosmic dust]

I think there is a mistake in your calculation. If {{\Psi }_{C}}=-i{{\gamma }^{2}}{{\Psi }^{*}} then \Psi _{C}^{\dagger }=i{{\Psi }^{T}}{{\gamma }^{2}}^{\dagger }=-i{{\Psi }^{T}}{{\gamma }^{2}} since {{\gamma }^{2}}^{\dagger }=-{{\gamma }^{2}}<br />. Then:
\begin{align}<br /> &amp; {{\left\langle \mathbf{p} \right\rangle }_{C}}=\int{{{d}^{3}}\mathbf{x}\Psi _{C}^{\dagger }\mathbf{p}{{\Psi }_{C}}}=\int{{{d}^{3}}\mathbf{x}\left( -i{{\Psi }^{T}}{{\gamma }^{2}} \right)\mathbf{p}\left( -i{{\gamma }^{2}}{{\Psi }^{*}} \right)}=-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}{{\gamma }^{2}}\mathbf{p}{{\gamma }^{2}}{{\Psi }^{*}}}= \\ <br /> &amp; \quad \quad =-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\gamma }^{2}}{{\gamma }^{2}}{{\Psi }^{*}}}=\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\Psi }^{*}}} \\ <br /> \end{align}
since {{\gamma }^{2}}{{\gamma }^{2}}=-{{I}_{4}} . Continuing the calculation we get:

{{\left\langle \mathbf{p} \right\rangle }_{C}}=\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\Psi }^{*}}}={{\left( \int{{{d}^{3}}\mathbf{x}{{\Psi }^{\dagger }}\mathbf{p}\Psi } \right)}^{*}}={{\left\langle \mathbf{p} \right\rangle }^{*}}=\left\langle \mathbf{p} \right\rangle

since \left\langle \mathbf{p} \right\rangle is real.

 

[qinglong.1397]

I think there is a mistake in your calculation. If {{\Psi }_{C}}=-i{{\gamma }^{2}}{{\Psi }^{*}} then \Psi _{C}^{\dagger }=i{{\Psi }^{T}}{{\gamma }^{2}}^{\dagger }=-i{{\Psi }^{T}}{{\gamma }^{2}} since {{\gamma }^{2}}^{\dagger }=-{{\gamma }^{2}}<br />. Then:
\begin{align}<br /> &amp; {{\left\langle \mathbf{p} \right\rangle }_{C}}=\int{{{d}^{3}}\mathbf{x}\Psi _{C}^{\dagger }\mathbf{p}{{\Psi }_{C}}}=\int{{{d}^{3}}\mathbf{x}\left( -i{{\Psi }^{T}}{{\gamma }^{2}} \right)\mathbf{p}\left( -i{{\gamma }^{2}}{{\Psi }^{*}} \right)}=-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}{{\gamma }^{2}}\mathbf{p}{{\gamma }^{2}}{{\Psi }^{*}}}= \\ <br /> &amp; \quad \quad =-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\gamma }^{2}}{{\gamma }^{2}}{{\Psi }^{*}}}=\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\Psi }^{*}}} \\ <br /> \end{align}
since {{\gamma }^{2}}{{\gamma }^{2}}=-{{I}_{4}} . Continuing the calculation we get:

{{\left\langle \mathbf{p} \right\rangle }_{C}}=\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\Psi }^{*}}}={{\left( \int{{{d}^{3}}\mathbf{x}{{\Psi }^{\dagger }}\mathbf{p}\Psi } \right)}^{*}}={{\left\langle \mathbf{p} \right\rangle }^{*}}=\left\langle \mathbf{p} \right\rangle

since \left\langle \mathbf{p} \right\rangle is real.

Thanks for your reply! Although our results are the same, I still want to point out that you should've used \bar\Psi_C instead of \Psi^\dagger_C, otherwise your &lt;\vec p&gt;_C isn't Lorentz covariant.

 

[cosmic dust]

Of course, my mistake... Let's take a look at this: \bar{\Psi } is defined by \bar{\Psi }={{\Psi }^{\dagger }}{{\gamma }^{0}} and so \overset{\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\, will be:
\overset{\_\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\,={{\left( -i{{\gamma }^{2}}{{\Psi }^{*}} \right)}^{\dagger }}{{\gamma }^{0}}=-i{{\Psi }^{T}}{{\gamma }^{2}}{{\gamma }^{0}}=-i{{\Psi }^{T}}{{\gamma }^{0}}{{\gamma }^{0}}{{\gamma }^{2}}{{\gamma }^{0}}=-i{{\left( {{\Psi }^{\dagger }}{{\gamma }^{0}} \right)}^{*}}{{\gamma }^{2}}^{\dagger }=-i{{\bar{\Psi }}^{*}}{{\gamma }^{2}}^{\dagger }
Then the integrand of {{\left\langle \mathbf{p} \right\rangle }_{C}} will be:

\overset{\_\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\,\mathbf{p}{{\Psi }_{C}}=-{{\bar{\Psi }}^{*}}{{\gamma }^{2}}^{\dagger }\mathbf{p}{{\gamma }^{2}}{{\Psi }^{*}}=-{{\bar{\Psi }}^{*}}\mathbf{p}{{\Psi }^{*}}=-{{\left( \bar{\Psi }{{\mathbf{p}}^{*}}\Psi \right)}^{*}}={{\left( \bar{\Psi }\mathbf{p}\Psi \right)}^{*}}

and so we will get the same mean value. Is this OK ?

 

[qinglong.1397]

Of course, my mistake... Let's take a look at this: \bar{\Psi } is defined by \bar{\Psi }={{\Psi }^{\dagger }}{{\gamma }^{0}} and so \overset{\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\, will be:
\overset{\_\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\,={{\left( -i{{\gamma }^{2}}{{\Psi }^{*}} \right)}^{\dagger }}{{\gamma }^{0}}=-i{{\Psi }^{T}}{{\gamma }^{2}}{{\gamma }^{0}}=-i{{\Psi }^{T}}{{\gamma }^{0}}{{\gamma }^{0}}{{\gamma }^{2}}{{\gamma }^{0}}=-i{{\left( {{\Psi }^{\dagger }}{{\gamma }^{0}} \right)}^{*}}{{\gamma }^{2}}^{\dagger }=-i{{\bar{\Psi }}^{*}}{{\gamma }^{2}}^{\dagger }
Then the integrand of {{\left\langle \mathbf{p} \right\rangle }_{C}} will be:

\overset{\_\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\,\mathbf{p}{{\Psi }_{C}}=-{{\bar{\Psi }}^{*}}{{\gamma }^{2}}^{\dagger }\mathbf{p}{{\gamma }^{2}}{{\Psi }^{*}}=-{{\bar{\Psi }}^{*}}\mathbf{p}{{\Psi }^{*}}=-{{\left( \bar{\Psi }{{\mathbf{p}}^{*}}\Psi \right)}^{*}}={{\left( \bar{\Psi }\mathbf{p}\Psi \right)}^{*}}

and so we will get the same mean value. Is this OK ?

Good! I can now discuss it with my teacher. Thanks!