I'm currently trying to learn about the Dirac equation in curved spacetime and have come across an odd remark in Nakahara's well-known textbook "Geometry, Topology and Physics" that I would like to understand.

In section 7.10.2, Nakahara writes the Dirac in two-dimensional spacetime as

$$S[\bar{ψ,ψ}] = ∫_M d^2x\sqrt{-g}\, \bar{ψ}\,iγ^a e_a^µ\left[∂_µ+\frac12iΓ^{b\ c}_{\ µ}Σ_{bc}\right]ψ$$


  • $Γ^{b\ c}_{\ µ}$ are the coefficients for the spin connection, $\nabla_µ e_c = Γ^{b\ }_{\ µc} e_b$
  • $e_a = e^µ_a ∂_µ$ are the vielbein vectors
  • and $Σ_{bc}=\frac14 i[γ_b,γ_c]$ are the spinor representations of the generators for the Lorentz Lie algebra $so(1,1)$.

(I have dropped the mass and $U(1)$ gauge field terms that are also present in the book.)

Then, he writes

It is interesting to note that the spin connection term vanishes if $\dim M=2$.

and argues that the Lagrangian has to be made hermitian first, so he ends up with

$$S[\bar{ψ,ψ}] = ∫_M d^2x\sqrt{-g}\, \bar{ψ}\,ie_a^µ\left[γ^a \overleftrightarrow{∂}_µ + \frac12iΓ^{b\ c}_{\ µ}\{γ^a,Σ_{bc}\}\right]ψ$$

In two dimensions, there is only one generator $Σ_{01}=γ_0γ_1$, and it anticommutes with both gamma matrices, $\{γ^0,Σ_{01}\}=\{γ^1,Σ_{01}\}=0$, so the spin connection term vanishes.

However, I do not understand why the Lagrangian has to be made hermitian first, so my questions are:

  1. Is the Dirac operator in curved spacetime hermitian?

  2. If not, why is making it hermitian like Nakahara does a good idea?

For the first question, I am currently trying to calculate the adjoint, but did not get the terms to cancel yet.

  • $\begingroup$ I might be wrong, but I think that, conceptually, you allways need a real valued function as action because you can't talk of maximum and minimum with complex numbers, even though you can use the complex expression, feed the EL equations and get the right answer. $\endgroup$
    – Hydro Guy
    Commented Feb 24, 2015 at 13:13
  • $\begingroup$ @HydroGuy: Ooh, I like that way of putting it. It's not entirely obvious that being real on all fields $ψ$ is equivalent to the operator being hermitian, but if I remember correctly, that is true in complex Hilbert spaces. $\endgroup$ Commented Feb 28, 2015 at 20:20

1 Answer 1


1) Dirac operator Pseudo-Hermitian in flat and not pseudo-hermitian in curved space if you define it as in your first equation (without symmetrizing).

because $(i\gamma^a e^\mu_a\partial_\mu)^\dagger = \gamma^0 (i\gamma^a e^\mu_a\partial_\mu)\gamma^0$, whereas $(\gamma^a \left[ \gamma^b,\gamma^c\right])^\dagger =\gamma^0 ( \left[ \gamma^b,\gamma^c\right]\gamma^a)\gamma^0 $

2) Lagrangian must always be hermitian so that the S-matrix is unitary (QM).

In flat space the kinetic term in the Lagrangian is not hermitian, so in principle we should add the hermitian conjugate $\mathcal{L} = \bar{\psi}i\gamma\partial \psi + h.c.$, however if we do not its ok because the lagrangian will still be hermitian up to a total derivative.. This statement was actually redundant because hermiticity means $(\psi,A\psi) = (A^\dagger \psi, \psi) = (A\psi,\psi)$ so total derivatives don't matter for the definition.

This does not work anymore for a nonzero spin connection as demonstrated above. So you should explicitly add the hermitian conjugate.

  • $\begingroup$ Thanks, but could you expand on why the Lagrangian has to be hermitian? For instance, I get that the Hamiltonian has to be hermitian, so the potential energy of the Lagrangian should be of the form $-\psi^{\dagger}H\psi$ with $H$ a hermitian operator. What about the time derivatives, though? Also, you say "pseudo-hermitian" due to the fact that $\overline{\psi}$ differs from the adjoint $\psi^\dagger$ by the matrix $\gamma^0$? $\endgroup$ Commented Feb 27, 2015 at 10:46
  • $\begingroup$ There are many different ways to demonstrate the necessity of a hermitian Lagrangian in QFT. I'll mention one: Non-Hermitian kinetic term in Lagrangian $\Rightarrow$ complex momenta $\Rightarrow$ an imaginary non infinitesimal pole in the free propagator $\Rightarrow$ Finite lifetime for stable particles $\Rightarrow$ particles vanishing into nothing. This example is also a result of the complex energy from the expectation value of the hamiltonian for any excited state $E^*_n =< n|H|n>^\dagger \neq E_n$ $\endgroup$
    – Ali Moh
    Commented Feb 27, 2015 at 12:10
  • $\begingroup$ Note that even for unstable particles to begin with, this gives a higher probability to decay than given by the sum of legitimate channels... Probablilty > 100% $\Rightarrow$ non-unitary $\endgroup$
    – Ali Moh
    Commented Feb 27, 2015 at 12:13
  • $\begingroup$ For your second question, yes. the definition of pseudo-hermitian is just that $A^\dagger = \gamma^0 A \gamma^0$... Notice that related to this statement is the dirac representation of the lorentz group is pseudo-Unitary Finally notice that if an operator $A$ is pseudo hermitian, then $\gamma^0 A$ is hermitian. So if you define the dirac operator with a $\gamma^0$ in front, and symmetrize in nonfat background you end up with a full fledged hermitian operator.. $\endgroup$
    – Ali Moh
    Commented Feb 27, 2015 at 12:15
  • $\begingroup$ $\gamma^0 A$ is hermitian if your metric's signature is $\eta^{00}=+1$ otherwise it is $i\gamma^0 A$ $\endgroup$
    – Ali Moh
    Commented Feb 27, 2015 at 15:27

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