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$ \newcommand{\p}{\partial} $ I am trying to show that the current $J^{\mu} = (\gamma_{\nu}\partial^{\nu} \phi - m\phi)\gamma^{\mu}\psi$ is conserved for all fields that satisfy the Klein-Gordon and Dirac equations: $$(\partial_{\mu}\partial^{\mu} - m^2)\phi=0 \tag{KG}$$ $$(\gamma_{\mu}\partial^{\mu}-m)\psi=0 \tag{Drc}$$ So I begin by taking the derivative since we know that a conserved current satisfies $\partial_{\mu}J^{\mu}=0$. For convenience, since I do not know how to write slashed quantities here I define $D = \gamma \cdot \partial $. Therefore I have

$$\begin{aligned} \p_{\mu}J^{\mu} &= \p_{\mu}(D\phi \gamma^{\mu}\psi) + D\phi \p_{\mu}\gamma^{\mu}\psi - m(\p_{\mu}\phi) \gamma^{\mu}\psi - m\phi(\p_{\mu}\gamma^{\mu}\psi) \\ &= (DD\phi)\psi + D\phi D\psi - m(D\phi) \psi - m\phi (D\psi) \\ &= (DD\phi)\psi + D\phi D\psi - D(m\phi \psi) \\ &= D(D\phi \, \psi) - D\phi D\psi +D\phi D\psi - D(m\phi \psi) \\ &= D[(D\phi-m\phi)\psi] \end{aligned}$$ but as you can see the scalar is in the place of the spinor and thus I cannot use (Drc). Any thoughts/help on this?

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  • $\begingroup$ Are you sure it's not the massless Dirac equation? $\endgroup$
    – Prahar
    Commented Dec 27, 2014 at 15:48
  • $\begingroup$ Well, I am taking it from a book so I have copied the question exactly like I see it. In specific the SUPERGRAVITY textbook, page 109. $\endgroup$
    – Marion
    Commented Dec 27, 2014 at 15:49
  • $\begingroup$ Sorry I miscalculated. It wouldn't work for massless. $\endgroup$
    – Prahar
    Commented Dec 27, 2014 at 15:52

1 Answer 1

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$$ \newcommand{\p}{\partial} \begin{aligned} \p_{\mu}J^{\mu} &= \gamma_{\nu}(\p_{\mu}\p^{\nu}\phi)\gamma^{\mu}\psi+\gamma_{\nu}\p^{\nu}\phi \gamma^{\mu}\p_{\mu}\psi-m\p_{\mu}\phi\gamma^{\mu}\psi - m\phi\gamma^{\mu}\p_{\mu}\psi\\ \end{aligned} $$ Now the first piece can be written as $$\gamma_{\nu}(\p_{\mu}\p^{\nu}\phi)\gamma^{\mu} \psi= \gamma_{\nu}\gamma_{\mu}(\p^{\mu}\p^{\nu}\phi)\psi=(D D\phi)\psi$$By means of the property (using your notation, $A = \gamma_{\mu}a^{\mu}=\gamma^{\mu}a_{\mu}$): $$AB=2(a\dot{}b)-BA$$and you have $DD = 2 (\p_{\mu}\p^{\mu}) - DD$ that leads to $DD = \p_{\mu}\p^{\mu}$.

So you have: $$(\p_{\mu}\p^{\mu}\phi)\psi$$ which cancels out the last piece whence you use (Drc) obtaining: $$-m\phi\ m\psi = -m^2\phi\psi$$ and then (KG).

It is left: \begin{aligned} &= \gamma_{\nu}\p^{\nu}\phi \gamma^{\mu}\p_{\mu}\psi-m\p_{\mu}\phi\gamma^{\mu}\psi\\ \end{aligned} By using (Drc) in the first part and changing the mute index of the summation: \begin{aligned} &= \gamma_{\nu}\p^{\nu}\phi m \psi-m\p_{\mu}\phi\gamma^{\mu}\psi\\ &= m D\phi \psi - m D \phi\psi\\ &=0 \end{aligned} I should not have done mistakes. Fell free to ask or correct me.

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  • $\begingroup$ Hi and thanks for your answer. I will go through it in detail. $\endgroup$
    – Marion
    Commented Dec 27, 2014 at 16:42
  • $\begingroup$ Let me know if it works for you or if you need references for what I wrote. $\endgroup$
    – Gian Abr
    Commented Dec 27, 2014 at 16:49
  • $\begingroup$ References would be useful anyway for me and other readers. Would you be kind enough to provide them? $\endgroup$
    – Marion
    Commented Dec 27, 2014 at 21:50

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