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Is <p>_c = -<p>?

by qinglong.1397
Tags: <p>, <p>c
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qinglong.1397
#1
Oct31-12, 08:39 PM
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 [itex]C=i\gamma^0\gamma^2[/itex]. The spinor is [itex]\Psi[/itex] and its charge-conjugated spinor [itex]\Psi_C=-i\gamma^2\Psi^*[/itex].

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

[itex]<\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]^*[/itex]
[itex]=[\int d^3x\bar\Psi\vec p\Psi]^*=<\vec p>^*=<\vec p>[/itex]

where [itex]<\vec p>[/itex] is real.

Is there anything wrong with my calculation, because my teacher didn't give me the grade for this?
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qinglong.1397
#2
Nov7-12, 10:18 AM
P: 107
I guess my calculation is correct since nobody replies...
Bill_K
#3
Nov7-12, 12:10 PM
Sci Advisor
Thanks
Bill_K's Avatar
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?

qinglong.1397
#4
Nov7-12, 12:21 PM
P: 107
Is <p>_c = -<p>?

Quote Quote by Bill_K View Post
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
#5
Nov7-12, 12:25 PM
P: 3,014
What does [itex]\Psi^{\ast}[/itex] represent?
Bill_K
#6
Nov7-12, 12:34 PM
Sci Advisor
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Bill_K's Avatar
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.
qinglong.1397
#7
Nov7-12, 05:00 PM
P: 107
Quote Quote by Bill_K View Post
I'm just thinking that jμ does change sign while pμ does not, and the only difference between them is the derivative operator.
By [itex]j_\mu[/itex], do you mean electric current?

Actually, I found out that [itex]<\vec r>_C=-<\vec r>[/itex] and my teacher also got this.
cosmic dust
#8
Nov8-12, 09:23 AM
P: 123
I think there is a mistake in your calculation. If [itex] {{\Psi }_{C}}=-i{{\gamma }^{2}}{{\Psi }^{*}} [/itex] then [itex] \Psi _{C}^{\dagger }=i{{\Psi }^{T}}{{\gamma }^{2}}^{\dagger }=-i{{\Psi }^{T}}{{\gamma }^{2}} [/itex] since [itex] {{\gamma }^{2}}^{\dagger }=-{{\gamma }^{2}}
[/itex]. Then:
[itex] \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} [/itex]
since [itex]{{\gamma }^{2}}{{\gamma }^{2}}=-{{I}_{4}}[/itex] . Continuing the calculation we get:

[itex]{{\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[/itex]

since [itex]\left\langle \mathbf{p} \right\rangle[/itex] is real.
qinglong.1397
#9
Nov8-12, 09:29 AM
P: 107
Quote Quote by cosmic dust View Post
I think there is a mistake in your calculation. If [itex] {{\Psi }_{C}}=-i{{\gamma }^{2}}{{\Psi }^{*}} [/itex] then [itex] \Psi _{C}^{\dagger }=i{{\Psi }^{T}}{{\gamma }^{2}}^{\dagger }=-i{{\Psi }^{T}}{{\gamma }^{2}} [/itex] since [itex] {{\gamma }^{2}}^{\dagger }=-{{\gamma }^{2}}
[/itex]. Then:
[itex] \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} [/itex]
since [itex]{{\gamma }^{2}}{{\gamma }^{2}}=-{{I}_{4}}[/itex] . Continuing the calculation we get:

[itex]{{\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[/itex]

since [itex]\left\langle \mathbf{p} \right\rangle[/itex] is real.
Thanks for your reply! Although our results are the same, I still want to point out that you should've used [itex]\bar\Psi_C[/itex] instead of [itex]\Psi^\dagger_C[/itex], otherwise your [itex]<\vec p>_C[/itex] isn't Lorentz covariant.
cosmic dust
#10
Nov8-12, 10:22 AM
P: 123
Of course, my mistake... Let's take a look at this: [itex]\bar{\Psi } [/itex] is defined by [itex]\bar{\Psi }={{\Psi }^{\dagger }}{{\gamma }^{0}}[/itex] and so [itex]\overset{\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\, [/itex] will be:
[itex]\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 }[/itex]
Then the integrand of [itex]{{\left\langle \mathbf{p} \right\rangle }_{C}}[/itex] will be:

[itex]\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)}^{*}} [/itex]

and so we will get the same mean value. Is this OK ?
qinglong.1397
#11
Nov8-12, 12:30 PM
P: 107
Quote Quote by cosmic dust View Post
Of course, my mistake... Let's take a look at this: [itex]\bar{\Psi } [/itex] is defined by [itex]\bar{\Psi }={{\Psi }^{\dagger }}{{\gamma }^{0}}[/itex] and so [itex]\overset{\_\_}{\mathop{\left( {{\Psi }_{C}} \right)}}\, [/itex] will be:
[itex]\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 }[/itex]
Then the integrand of [itex]{{\left\langle \mathbf{p} \right\rangle }_{C}}[/itex] will be:

[itex]\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)}^{*}} [/itex]

and so we will get the same mean value. Is this OK ?
Good! I can now discuss it with my teacher. Thanks!


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