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Show that the imaginary part of the Dirac Lagrangian [density] is a total derivative.

My attempt:

The Dirac Lagrangian density is given by, $$ \mathcal{L} \ = \ -\bar{\psi}(\gamma^{\mu}\partial_{\mu} + m)\psi $$ Now, as far as I am aware, the hermiticity conditions of the gamma matrices depend on which representation one uses - we used the Weyl representation in class but with the $(-, +, +, +)$ metric, so there was a factor of $-i$ in every matrix - essentially meaning that the matrices would be anti-hermitian (and therefore, their complex conjugate should furnish a negative sign). Including this is not necessary for my solution, but I will still do it.

The spinors themselves are complex (as far as I know) and all I can say is that the product $\psi^{\dagger}\psi$ should be real. Thus, $$ \mathcal{L}^{*} \ = \ -\bar{\psi}^{*}(-\gamma^{\mu}\partial_{\mu} + m)\psi^{*}. $$ This implies, $$ \mathcal{L} - \mathcal{L}^{*} = -\bar{\psi}(\gamma^{\mu}\partial_{\mu} + m)\psi - \bar{\psi}^{*}(\gamma^{\mu}\partial_{\mu} - m)\psi^{*}$$ $$ = -\partial_{\mu}(\bar{\psi}\gamma^{\mu}\psi + \bar{\psi}^{*}\gamma^{\mu}\psi^{*}) + (\partial_{\mu}\bar{\psi} \gamma^{\mu} - m\bar{\psi})\psi + (\partial_{\mu}\bar{\psi}^{*} \gamma^{\mu} + m\bar{\psi}^{*})\psi^{*} .$$ Here, the expressions in the brackets in the second and third terms vanish if I impose the equations derived from Dirac Lagrangian itself. This means, $$ \frac{1}{2i}(\mathcal{L} - \mathcal{L}^{*}) = \mathrm{Im}(\mathcal{L}) = \partial_{\mu}f^{\mu} $$ This is a total derivative as I am supposed to prove.

My issues with my solution:

In order to derive this, I impose the dynamical equations derived from the starting Lagrangian - is it at all valid? The Dirac equation says, $$ (\gamma^{\mu}\partial_{\mu} + m)\psi = 0.$$ This clearly says that if I impose this equation on the starting density, I get $\mathcal{L} = 0$ - this does not seem to be different from what I have done in my own solution, except that I imposed the equation later instead of at the beginning. The imaginary part being $0$ is of course a valid solution (since any constant field will have a vanishing divergence) but it is a trivial solution.

At the same time, the only other option I see is to use both the idea that the chosen representations of the gamma matrices are anti-hermitian and the plane wave solutions of the Dirac field to derive something - but there are similar problems there too (since then I would need to look into all the momentum states, $u(p), \ v(p)$) and the entire procedure would be tedious and I am not even sure if I will get an answer. The second issue to that is that then the proof is not be general - I will have assumed a form of both the matrices and the spinors.

Any help will be appreciated. Thanks!

EDIT:

Due to the comments by @MadMax, I feel like I should provide some other information that I skipped (because I felt it was not needed). One of them is the negative sign in front of the Lagrangian that I included in this edit.

In our course, we defined the gamma matrices as, $$ \gamma^{0} = -i\begin{pmatrix} 0 & 1\\1&0\end{pmatrix},\ \gamma^{i} = -i\begin{pmatrix} 0 & \sigma^{i}\\-\sigma^{i}&0\end{pmatrix}$$ The Dirac adjoint was defined as, $$ \bar{\psi} = i\psi^{\dagger}\gamma^{0}$$ And the metric, as I mentioned earlier, is $(-,+,+,+)$. Finally, the Lagrangian is defined as, $$ \mathcal{L} = -\bar{\psi}(\gamma^{\mu}\partial_{\mu} + m)\psi $$

As @MadMax mentions, the mass term would have picked up a factor of $i$, but since our definition of the Dirac adjoint was different, it doesn't.

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  • $\begingroup$ For metric $(-, +, +, +)$, instead of $\mathcal{L} \ = \ \bar{\psi}(\gamma^{\mu}\partial_{\mu} + m)\psi$, should your Lagrangian be $\mathcal{L} \ = \ i\bar{\psi}(\gamma^{\mu}\partial_{\mu} + m)\psi$? Assuming that we are talking about Grassmann-odd $\psi$ field, there is a missing i, right? $\endgroup$
    – MadMax
    Sep 25, 2023 at 18:30
  • $\begingroup$ We didn't have an i in class; but we defined the gamma matrices with an extra factor of i. $\endgroup$
    – ShKol
    Sep 25, 2023 at 19:25
  • $\begingroup$ "We didn't have an i in class", then you might consider asking your professor for refund of your tuition :) $\endgroup$
    – MadMax
    Sep 25, 2023 at 19:30
  • $\begingroup$ "but we defined the gamma matrices with an extra factor of i", because of this extra i factor in the gamma matrices, the mass term picks up an extra i since $\bar{\psi} = \psi^{\dagger}\gamma_0$ and there is an extra i coming from $\gamma_0$. As for the kinetic term, the extra i from $\gamma_0$ cancels out the extra i from $\gamma_\mu$, therefore it stays the same as the metric (+,-,-,-) format. $\endgroup$
    – MadMax
    Sep 25, 2023 at 19:36
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    $\begingroup$ For the Lagrangian, you can take a look at Edward Witten's paper referenced in this post: physics.stackexchange.com/questions/513768/… $\endgroup$
    – MadMax
    Sep 25, 2023 at 21:03

1 Answer 1

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Give a representation of the Gamma matrices, there exists some matrix $\beta$ which satisfies $$ (\gamma^\mu)^\dagger = \beta \gamma^\mu \beta^{-1} , \qquad \beta^\dagger = - \beta . $$ The Dirac conjugate is defined by $$ {\bar \psi} \equiv i \psi^\dagger \beta . $$ Now, let's consider the Dirac Lagrangian, $$ {\cal L} = {\bar \psi} ( \gamma^\mu \partial_\mu + m ) \psi $$ Taking a complex conjugating and manipulating further, we find \begin{align} {\cal L}^* &= [ {\bar \psi} ( \gamma^\mu \partial_\mu + m ) \psi ]^* \\ &= [ i (\psi^\dagger \beta ) ( \gamma^\mu \partial_\mu + m ) \psi ]^\dagger \\ &= - i \psi^\dagger ( ( \gamma^\mu )^\dagger \overleftarrow{\partial_\mu} + m ) (\psi^\dagger \beta )^\dagger \\ &= - i \psi^\dagger ( \beta \gamma^\mu \beta^{-1}\overleftarrow{\partial_\mu} + m )\beta^\dagger \psi \\ &= - {\bar \psi} ( \gamma^\mu \overleftarrow{\partial_\mu} + m ) \psi \\ &= {\bar \psi} ( \gamma^\mu \partial_\mu + m ) \psi - \partial_\mu [ {\bar \psi} \gamma^\mu \psi ] \\ &= {\cal L} - \partial_\mu [ {\bar \psi} \gamma^\mu \psi ] . \end{align}

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  • $\begingroup$ I understand the steps here - but I don't understand why taking a complex conjugate is akin to taking the Hermitian. Is it because the Lagrangian is a scalar and taking the conjugate transpose just means taking a conjugate? Also, why does there always exist such a matrix like $\beta$? In the usual conventions for the gamma matrices and such, it always seems like the relation holds coincidentally. Thanks! $\endgroup$
    – ShKol
    Sep 26, 2023 at 4:04
  • $\begingroup$ Your calculation is wrong. By $- {\bar \psi} ( \gamma^\mu \overleftarrow{\partial_\mu} + m ) \psi = {\bar \psi} ( \gamma^\mu \partial_\mu + m ) \psi - \partial_\mu [ {\bar \psi} \gamma^\mu \psi ]$, you changed the sign of the mass term. Please double check your math. $\endgroup$
    – MadMax
    Sep 26, 2023 at 15:05

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