# Is <p>_c = -<p>?

by qinglong.1397
Tags: <p>, <p>c
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 P: 107 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?
 P: 107 I guess my calculation is correct since nobody replies...
 Sci Advisor Thanks P: 4,160 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?
P: 107
Is <p>_c = -<p>?

 Quote by 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?
I did use integration by parts.
 P: 3,014 What does $\Psi^{\ast}$ represent?
Sci Advisor
Thanks
P: 4,160
 I did use integration by parts.
I'm just thinking that jμ does change sign while pμ does not, and the only difference between them is the derivative operator.
P: 107
 Quote by Bill_K I'm just thinking that jμ does change sign while pμ 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.
 P: 123 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} & {{\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 }^{*}}}= \\ & \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 }^{*}}} \\ \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.
P: 107
 Quote by 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} & {{\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 }^{*}}}= \\ & \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 }^{*}}} \\ \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.
 P: 123 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 ?
P: 107
 Quote by 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 ?
Good! I can now discuss it with my teacher. Thanks!