Consider the Dirac Lagrangian,


and take the Dirac spinor chiral decomposition with $\psi_{L}=\frac{1}{2}\left(1-\gamma^{5}\right)\psi$ and $\psi_{R}=\frac{1}{2}\left(1+\gamma^{5}\right)\psi$. How can I obtain the Lagrangian in the form,

$$ L=\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}-m\left(\psi_{L}\psi_{R}+\psi_{R}\psi_{L}\right)\;? $$

Plugging $\psi=\psi_{L}+\psi_{R}$ in the first expression doesn't seem to work, as I get crossed terms (I guess). Could you advise?


I mean, $$ \begin{array}{cl} L & =\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi\\ & =\left(\overline{\psi}_{L}+\overline{\psi}_{R}\right)\left(i\gamma^{\mu}\partial_{\mu}-m\right)\left(\psi_{L}+\psi_{R}\right)\\ & =\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}+\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}-m\left(\overline{\psi}_{L}\psi_{R}+\overline{\psi}_{R}\psi_{L}+\overline{\psi}_{L}\psi_{L}+\overline{\psi}_{R}\psi_{R}\right)\\ & =???\\ & =\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}-m\left(\overline{\psi}_{L}\psi_{R}+\overline{\psi}_{R}\psi_{L}\right) \end{array} $$

But why do we have $\overline{\psi}_{L}\psi_{L}+\overline{\psi}_{R}\psi_{R}=0$ and $\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}=0$? For example, let's just look at the mass term for now. We have $$ \overline{\psi}_{L}=\psi_{L}^{\dagger}\gamma^{0}=\frac{1}{2}\psi^{\dagger}\left(1-\left(\gamma^{5}\right)^{\dagger}\right)\gamma^{0} $$


$$ \begin{array}{cl} \overline{\psi}_{L}\psi_{L} & =\frac{1}{2}\psi^{\dagger}\left(1-\left(\gamma^{5}\right)^{\dagger}\right)\gamma^{0}\frac{1}{2}\left(1-\gamma^{5}\right)\psi\\ & =\frac{1}{4}\psi^{\dagger}\left(1-\gamma^{0}\gamma^{5}\gamma^{0}\right)\gamma^{0}\left(1-\gamma^{5}\right)\psi\\ & =\frac{1}{4}\psi^{\dagger}\left(\gamma^{0}-\gamma^{0}\gamma^{5}\gamma^{0}\gamma^{0}\right)\left(1-\gamma^{5}\right)\psi\\ & =\frac{1}{4}\psi^{\dagger}\left(\gamma^{0}-\gamma^{0}\gamma^{5}1\right)\left(1-\gamma^{5}\right)\psi\\ & =\frac{1}{4}\psi^{\dagger}\gamma^{0}\left(1-\gamma^{5}\right)\left(1-\gamma^{5}\right)\psi\\ & =\frac{1}{4}\overline{\psi}\left(1-\gamma^{5}-\gamma^{5}-\gamma^{5}\left(-\gamma^{5}\right)\right)\psi\\ & =\frac{1}{4}\overline{\psi}\left(1-2\gamma^{5}+1\right)\psi\\ & =\frac{1}{2}\overline{\psi}\left(1-\gamma^{5}\right)\psi \end{array} $$

and similarly,

$$ \overline{\psi}_{R}\psi_{R}=\frac{1}{2}\overline{\psi}\left(1+\gamma^{5}\right)\psi $$

hence, $$ \begin{array}{cl} \overline{\psi}_{L}\psi_{L}+\overline{\psi}_{R}\psi_{R} & =\frac{1}{2}\overline{\psi}\left(1-\gamma^{5}\right)\psi+\frac{1}{2}\overline{\psi}\left(1+\gamma^{5}\right)\psi\\ & =\overline{\psi}-\frac{1}{2}\overline{\psi}\gamma^{5}\psi+\frac{1}{2}\overline{\psi}\gamma^{5}\psi\\ & =\overline{\psi}\psi. \end{array} $$

I didn't get zero. What am I missing?

  • $\begingroup$ The cross terms vanish in the kinetic term involving γ, and are the only ones surviving in the mass term lacking γ. Can you see that with your chiral projectors? It should be illustrated in any creditable text you learned about such projectors from. $\endgroup$ Jun 6, 2018 at 12:55
  • $\begingroup$ @CosmasZachos I don't see it yet. Can you look through my "edit"? $\endgroup$
    – Igaturtle
    Jun 6, 2018 at 14:49
  • $\begingroup$ I can't see what you are doing, but you must conclude that $\overline{\psi_L}=(\overline{\psi})_R$. $\endgroup$ Jun 6, 2018 at 15:05
  • 2
    $\begingroup$ $\gamma_5$ is hermitian, no? $\endgroup$ Jun 6, 2018 at 15:26
  • $\begingroup$ Thanks! That was the missing link in my reasoning. Now it checks out. $\endgroup$
    – Igaturtle
    Jun 6, 2018 at 16:17

2 Answers 2


The $\psi_L$ spinor transforms in the $(\frac12,0)$ representation of $SL(2,\mathbb C)$, and $\psi_R$ under the conjugate representation $(0,\frac12)$, so we can construct a Dirac spinor as the column,

$$\psi = \begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix}$$

and it suffices to treat $\psi$ as a column vector of two elements, with $\bar \psi = \psi^\dagger \gamma^0$. You simply need to apply basic linear algebra to work out the Lagrangian now. Recall that,

$$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu\\ \tilde \sigma^\mu & 0 \end{pmatrix}$$

where $\sigma^\mu = (\mathbb I_2, \vec \sigma)$ and $\tilde \sigma^\mu = (\mathbb I_2,-\vec \sigma)$ where $\vec \sigma$ is the vector whose components are the Pauli matrices. Carrying out this calculation is a matter of multiplying columns, matrices and rows.

  • $\begingroup$ Sure, I've seen it done that way, which I believe is valid only in the chiral representation. But I'm trying to follow a different and most general approach here - one using a definition of the chiral spinors in an arbitrary representation of the Clifford algebra, by means of the projection operators $P_{\pm}=\frac{1}{2}\left(1\pm\gamma^{5}\right)$ only. I believe it should be possible. $\endgroup$
    – Igaturtle
    Jun 6, 2018 at 12:27
  • $\begingroup$ @Igaturtle Yes, you just need to use some gamma matrix identities. $\endgroup$
    – JamalS
    Jun 6, 2018 at 12:52

You got the incorrect terms since your definition of hermitian $(\gamma^5)^\dagger = \gamma^0 \gamma^5 \gamma^0 = - \gamma^5$ is simply wrong. One should have $(\gamma^5)^\dagger = \gamma^5$ as:

$$ (\gamma^5)^\dagger = (i\gamma^0\gamma^1\gamma^2\gamma^3)^\dagger = (i)^\dagger(\gamma^3)^\dagger(\gamma^2)^\dagger(\gamma^1)^\dagger(\gamma^0)^\dagger = (-i)(-\gamma^3)(-\gamma^2)(-\gamma^1)\gamma^0 = i\gamma^3\gamma^2\gamma^1\gamma^0 = i\gamma^0\gamma^1\gamma^2\gamma^3 = \gamma^5, $$


$$ (\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0, $$

for $\mu=0, 1, 2, 3$.

  • $\begingroup$ Thanks, that's right, it is simply wrong, as Cosmas Zachos pointed out in the comment section 20 minutes before this answer. I think I should give credit to him, do you agree? $\endgroup$
    – Igaturtle
    Jun 6, 2018 at 16:22

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