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

$<\vec p>_C=\int d^3x\bar\Psi_C\vec p\Psi_C=\int d^3x\Psi^T\gamma^0\gamma^2\vec p\gamma^2\Psi^*=-[\int d^3x\bar\Psi\vec p^*\Psi]^*$
$=[\int d^3x\bar\Psi\vec p\Psi]^*=<\vec p>^*=<\vec p>$

where $<\vec p>$ 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 $<\vec r>_C=-<\vec r>$ 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 ${{\gamma }^{2}}^{\dagger }=-{{\gamma }^{2}}$. Then:
$\begin{align}<br /> & {{\left\langle \mathbf{p} \right\rangle }_{C}}=\int{{{d}^{3}}\mathbf{x}\Psi _{C}^{\dagger }\mathbf{p}{{\Psi }_{C}}}=\int{{{d}^{3}}\mathbf{x}\left( -i{{\Psi }^{T}}{{\gamma }^{2}} \right)\mathbf{p}\left( -i{{\gamma }^{2}}{{\Psi }^{*}} \right)}=-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}{{\gamma }^{2}}\mathbf{p}{{\gamma }^{2}}{{\Psi }^{*}}}= \\ <br /> & \quad \quad =-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\gamma }^{2}}{{\gamma }^{2}}{{\Psi }^{*}}}=\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\Psi }^{*}}} \\ <br /> \end{align}$
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 ${{\gamma }^{2}}^{\dagger }=-{{\gamma }^{2}}$. Then:
$\begin{align}<br /> & {{\left\langle \mathbf{p} \right\rangle }_{C}}=\int{{{d}^{3}}\mathbf{x}\Psi _{C}^{\dagger }\mathbf{p}{{\Psi }_{C}}}=\int{{{d}^{3}}\mathbf{x}\left( -i{{\Psi }^{T}}{{\gamma }^{2}} \right)\mathbf{p}\left( -i{{\gamma }^{2}}{{\Psi }^{*}} \right)}=-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}{{\gamma }^{2}}\mathbf{p}{{\gamma }^{2}}{{\Psi }^{*}}}= \\ <br /> & \quad \quad =-\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\gamma }^{2}}{{\gamma }^{2}}{{\Psi }^{*}}}=\int{{{d}^{3}}\mathbf{x}{{\Psi }^{T}}\mathbf{p}{{\Psi }^{*}}} \\ <br /> \end{align}$
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 $<\vec p>_C$ 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!