Spinor indices manipulation in the Noether's current for free fermions

I can't solve this apparent paradox; I have the free lagrangian for massive fermions $\mathscr L = i\bar\Psi\gamma^\mu\partial_\mu\Psi - m\bar\Psi\Psi$ which is invariant under the global phase transformation

$$\Psi \rightarrow e^{i\lambda}\Psi \hspace{40pt} \bar\Psi \rightarrow e^{-i\lambda}\bar\Psi$$

Therefore I can compute the Noether conserved current

$$J^\mu = \frac{\partial\mathscr L}{\partial(\partial_\mu\Psi)}\frac{\delta\Psi}{\delta\lambda} +\frac{\partial\mathscr L}{\partial(\partial_\mu\bar\Psi)}\frac{\delta\bar\Psi}{\delta\lambda} = \bar\Psi\gamma^\mu\Psi$$

Now I see in the notes of my Professor the following calculation:

Since usually we want a real current, we can rewrite $J^\mu = i\bar\Psi\gamma^\mu\Psi$ or with explicit spinor indices $J^\mu = i\bar\Psi^\alpha(\gamma^\mu)_\alpha^{\ \ \ \beta}\Psi_\beta = i\Psi^{\dagger\sigma}(\gamma^0)_\sigma^{\ \ \ \alpha}(\gamma^\mu)_\alpha^{\ \ \ \beta}\Psi_\beta$ where $\alpha, \beta, \sigma = 1, 2, 3, 4$. We can prove this is real in fact

$$(J^\mu)^\dagger = -i(\Psi^\dagger)_\beta(\gamma^{\mu\dagger})^\beta_{\ \ \ \alpha}(\gamma^{0\dagger})^\alpha_{\ \ \ \sigma}(\Psi^\sigma)$$

but we know that for any $\psi^\alpha$ and $\chi^\alpha$ we can write

$$\psi^\alpha\chi_\alpha = \varepsilon^{\alpha\beta}\psi_\beta\chi_\alpha = - \psi_\beta\varepsilon^{\beta\alpha}\chi_\alpha = -\psi_\beta\chi^\beta$$

this means that when we raise and lower a contracted spinor index, we get a minus sign in front of our expression, hence

$$(J^\mu)^\dagger = -i(\Psi^\dagger)_\beta(\gamma^{\mu\dagger})^\beta_{\ \ \ \alpha}(\gamma^{0\dagger})^\alpha_{\ \ \ \sigma}(\Psi^\sigma)$$

$\color{red}{\mathrm{raise \ and \ lower} \ \beta}$

$$(J^\mu)^\dagger = -(\color{red}-) i(\Psi^\dagger)^\beta(\gamma^{\mu\dagger})_{\beta\alpha}(\gamma^{0\dagger})^\alpha_{\ \ \ \sigma}(\Psi^\sigma)$$

raise and lower $\color{green}\alpha$ then $\color{blue}\sigma$

$$(J^\mu)^\dagger = -(\color{red}-)(\color{green}-)(\color{blue}-) i(\Psi^\dagger)^\beta(\gamma^{\mu\dagger})_\beta^{\ \ \ \alpha}(\gamma^{0\dagger})_\alpha^{\ \ \ \sigma}(\Psi_\sigma) = + i\Psi^\dagger\gamma^{\mu\dagger}\gamma^{0\dagger}\Psi$$

now we can insert the identity $\gamma^0\gamma^0 = \Bbb I_4$ and since $\gamma^{0\dagger} = \gamma^0$ and $\gamma^0\gamma^{\mu\dagger}\gamma^0 = \bar\gamma^\mu = \gamma^\mu$ we eventually get

$$(J^\mu)^\dagger = i\color{red}{\Psi^\dagger\gamma^0}\color{\green}{\gamma^0\gamma^{\mu\dagger}\gamma^0}\Psi = i\color{red}{\bar\Psi}\color{green}{\gamma^\mu}\Psi = J^\mu$$

we conclude that $J^\mu$ is real.

Well I can't understand this procedure since, without adding that $i$, I would have simply done in this way

\begin{aligned} (J^\mu)^\dagger & = \Psi^\dagger\gamma^{\mu\dagger}\gamma^{0\dagger}\Psi = \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{inserting \ identity} \gamma^0\gamma^0 = \Bbb I_4 \\ & = \color{red}{\Psi^\dagger\gamma^0}\color{green}{\gamma^0\gamma^{\mu\dagger}\gamma^0}\Psi = \\ & = \color{red}{\bar\Psi}\color{green}{\gamma^\mu}\Psi = J^\mu \end{aligned}

and I would have done in this way also because if I were to apply the same procedure used by my professor to the kinetic term of the lagrangian I would get

$$[i\bar\Psi\gamma^\mu\partial_\mu\Psi]^\dagger = [i\Psi^{\dagger\sigma}(\gamma^0)_\sigma^{\ \ \ \alpha}(\gamma^\mu)_\alpha^{\ \ \ \beta}(\partial_\mu\Psi)_\beta]^\dagger = - i(\partial_\mu\Psi^\dagger)_\beta(\gamma^{\mu\dagger})^\beta_{\ \ \ \alpha}(\gamma^{0\dagger})^\alpha_{\ \ \ \sigma}\Psi^\sigma$$

now I raise and lower $\color{red}\beta$, $\color{green}\alpha$ and $\color{blue}\sigma$ and i get

$$-(\color{red}-)(\color{green}-)(\color{blue}-) i(\partial_\mu\Psi^\dagger)^\beta(\gamma^{\mu\dagger})_\beta^{\ \ \ \alpha}(\gamma^{0\dagger})_\alpha^{\ \ \ \sigma}\Psi_\sigma = +i\partial_\mu\Psi^\dagger\gamma^{\mu\dagger}\gamma^{0\dagger}\Psi$$

Now I can insert the identity $\gamma^0\gamma^0 = \Bbb I_4$ and then integrate by parts

$$i\partial_\mu\Psi^\dagger\gamma^{\mu\dagger}\gamma^{0\dagger}\Psi = -i\Psi^\dagger\gamma^0\gamma^0\gamma^{\mu\dagger}\gamma^0\partial_\mu\Psi + i\partial_\mu(\Psi^\dagger\gamma^0\gamma^0\gamma^{\mu\dagger}\gamma^0\Psi) = -i\bar\Psi\gamma^\mu\partial_\mu\Psi + i\partial_\mu(\bar\Psi\gamma^\mu\Psi)$$

i.e. up to a total derivative I get

$$(i\bar\Psi\gamma^\mu\partial_\mu\Psi)^\dagger = - i\bar\Psi\gamma^\mu\partial_\mu\Psi$$

Therefore with this procedure the lagrangian (i.e. the action) would no longer be real. What is wrong with this?

• $J^0 = \bar\psi\gamma^0\psi =\psi^\dagger\gamma^0\gamma^0\psi$, which is non-negative. Hence $i$ times that would be imaginary. – Phoenix87 Jan 2 '17 at 17:44

Your current is right (without the $i$). Your professor's proof is wrong when he/she obtains a minus sign when raising/lowering indices in the gamma matrices: these are $c$-numbers, and therefore you don't get a minus sing when you move its indices around.
In the expression $-i(\Psi^\dagger)_\beta(\gamma^{\mu\dagger})^\beta_{\ \ \ \alpha}(\gamma^{0\dagger})^\alpha_{\ \ \ \sigma}(\Psi^\sigma)$ there is no negative sign associated to raising/lowering the $\alpha$ indices, and therefore the $\color{green}{-}$ is wrong.
Your proof of the hermiticity of $J^\mu$ (without the $i$) is correct, and so is your analysis of the hermiticity of the Lagrangian ($\mathcal L$ is indeed hermitian; again, the faulty logic lies in the $\color{green}{-}$ sign).
• Thank you very much, i was going crazy! Can I ask just one more thing? to raise and lower a spinor index I can use the antisymmetric symbol $\varepsilon^{\alpha\beta}$, what "object" should I use to raise/lower that $\alpha$ index? – M. M. R. Jan 2 '17 at 18:41
• @M.M.Riva short answer: don't raise/lower $\alpha$. Your proof of the hermiticity of $J$ without indices is the best proof you can write. Don't use index notation. Somewhat longer answer: you can't raise/lower $\alpha$ because it is a Dirac index, not a Weyl index. Dirac indices cannot be raised/lowered ($\varepsilon$ is only defined for Weyl spinors). If you want to use Weyl indices, I recommend you to read chapters 33 to 36 of Srednicki's book (there is a free copy here, in his webpage) – AccidentalFourierTransform Jan 3 '17 at 16:12