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Is <p>_c = <p>? 
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#1
Oct3112, 08:39 PM

P: 107

I calculated the expectation value of the momentum of the chargeconjugated 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 chargeconjugated 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? 


#2
Nov712, 10:18 AM

P: 107

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



#3
Nov712, 12:10 PM

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?



#4
Nov712, 12:21 PM

P: 107

Is <p>_c = <p>?



#5
Nov712, 12:25 PM

P: 3,014

What does [itex]\Psi^{\ast}[/itex] represent?



#6
Nov712, 12:34 PM

Sci Advisor
Thanks
P: 4,160




#7
Nov712, 05:00 PM

P: 107

Actually, I found out that [itex]<\vec r>_C=<\vec r>[/itex] and my teacher also got this. 


#8
Nov812, 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. 


#9
Nov812, 09:29 AM

P: 107




#10
Nov812, 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 ? 


#11
Nov812, 12:30 PM

P: 107




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