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Is <p>_c = -<p>?

  1. Oct 31, 2012 #1
    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?
     
  2. jcsd
  3. Nov 7, 2012 #2
    I guess my calculation is correct since nobody replies...
     
  4. Nov 7, 2012 #3

    Bill_K

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    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?
     
  5. Nov 7, 2012 #4
    I did use integration by parts.
     
    Last edited: Nov 7, 2012
  6. Nov 7, 2012 #5
    What does [itex]\Psi^{\ast}[/itex] represent?
     
    Last edited: Nov 7, 2012
  7. Nov 7, 2012 #6

    Bill_K

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    I'm just thinking that jμ does change sign while pμ does not, and the only difference between them is the derivative operator.
     
  8. Nov 7, 2012 #7
    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.
     
  9. Nov 8, 2012 #8
    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.
     
  10. Nov 8, 2012 #9
    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.
     
    Last edited: Nov 8, 2012
  11. Nov 8, 2012 #10
    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 ?
     
  12. Nov 8, 2012 #11
    Good! I can now discuss it with my teacher. Thanks!
     
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