# Non-hermiticity of Dirac Lagrangian: null momentum?

The usual Dirac Lagrangian is $L(\psi,\bar\psi)=\bar\psi(i\not\partial-m)\psi$. The canonical momenta are $$\pi=\frac{\partial L}{\partial \psi_{,0}}=i\psi^\dagger \\ \bar \pi=\frac{\partial L}{\partial \bar\psi_{,0}}=0$$

The fact that $\bar\pi=0$ is nowhere discussed in the books I've read. If we ignore this and move on, we can write $\{\psi(t,x),\pi(t,y)\}=i\delta(x-y)$, which in fact is equivalent to $\{\psi(t,x),\psi^\dagger(t,y)\}=\delta(x-y)$.

We can write the solution of the EOM as $$\psi(x)=\sum_s \int \frac{d^3 p}{(2\pi)^3\sqrt{2E}}\ b^s_p u^s(p) e^{ip\cdot x}+c^{s\dagger}_p v^s(p) e^{-ip\cdot x}$$ where $b,c$ are creation-annihilation operators.

Furthermore, we prove that, for example, $\{b^s(p),b^t(q)\}=(2\pi)^3\delta_{st}\delta(p-q)$ and a similar relation for $c_p$. After that, we easily prove that $\{H,c_p^\dagger\}=\omega_p c_p^\dagger$, which means that $|p\rangle=c_p^\dagger|0\rangle$, and $H|p\rangle=\omega_p|p\rangle$, i.e., $|p\rangle$ has energy $\omega_p$.

So far so good. If we were to study this theory from first principles, though, we should also impose $\{\bar\psi(t,x),\bar\pi(t,y)\}=i\delta(x-y)$, which is impossible as $\bar\pi=0$ (we should impose this commutator because the fields are supposed to be $\psi$ and $\bar\psi$: they are to be thought of as independent variables)

This is easily seen to be related to the fact that the lagrangian is non-hermitian. We can fix this by subtracting the total derivative $\frac{i}{2}\partial_\mu(\bar\psi\gamma^\mu\psi)$; we end up with $$L(\psi,\bar\psi)=\frac{i}{2}\bar\psi \not\partial\psi-\frac{i}{2}\partial_\mu\bar\psi\gamma^\mu\psi-\bar\psi m\psi$$

From this new lagrangian we get the same PDE (Dirac equation), as we should expect. Also, the conjugate momenta are both non-zero: $$\pi=\frac{\partial L}{\partial \psi_{,0}}=\frac{i}{2}\psi^\dagger \\ \bar \pi=\frac{\partial L}{\partial \bar\psi_{,0}}=\frac{i}{2}\psi$$

It is not hard to see that these gives rise to the same hamiltonian as before. The problem comes from the $\frac{1}{2}$ factor in $\pi,\bar\pi$: they pop up here and there. For example, the canonical commutation relations are changed to $\{\psi(t,x),\psi^\dagger(t,y)\}=\boldsymbol{2}\delta(x-y)$, which, in turns, are equivalent to $\{b^s(p),b^t(q)\}=\boldsymbol{2}(2\pi)^3\delta_{st}\delta(p-q)$ and $\{H,c_p^\dagger\}=\boldsymbol{2}\omega_p c_p^\dagger$.

This last relation is awful: a state $|p\rangle$ has energy equal to $\boldsymbol{2}\omega_p$. I feel that both approaches are unsatisfying, up to a certain point at least. The first one is the usual approach, but the lagrangian is non-hermitian and the momentum conjugate to $\bar\psi$ is null, so a systematic use of the CCR is not possible. The second one fixes both these problems, has the same equation of motion and the same hamiltonian, but the CCR are wrong by a factor of two, causing particles energy to be $2\omega_p$ instead of $\omega_p$.

I know that there are many questions in SE similar to this one, such as

Wrong sign anticommutation relation for the Dirac field?

Dirac equation as Hamiltonian system

etc.

but these people were asking about signs or other issues. My question is about the factors of $2$ that appears every now and then, changing the energy spectrum of the theory.

Finally, there is this qustion From Lagrangian to Hamiltonian in Fermionic Model which hits pretty close, but the question is not clear (some free indices in the lagrangian) and OP says that the hamiltonian changes with the new lagrangian. It can easly be seen that this is not true: the hamiltonian is the same, but the CCR are different (and this is not addressed). Also, the answers claim that the momenta don't change, which I believe is not true. They also say that the fields are such that $\partial \psi/\partial\bar\psi_{,0}\neq 0$, and this is not explained any further (and I don't even know if its true or not).

So, can anyone please shed any light on this? Why do we accept the first approach irrespective of its flaws? Shouldn't the fact that $\bar\pi=0$ arouse any suspicion? Why is the second approach not working properly?

OP is asking about how to perform the singular Legendre transformation for a Lagrangian theory of Dirac fermions. This was already done in my Phys.SE answer here using the Faddeev-Jackiw method. However, OP wants to consider the traditional Dirac-Bergmann$^1$ analysis. Potential complication were listed in my Phys.SE answer here.

We start from the manifestly real$^2$ Lagrangian density

$${\cal L}~=~\frac{i}{2}\sum_{\mu=0}^3\left(\bar\psi\gamma^{\mu}\partial_{\mu}\psi -\partial_{\mu}\bar\psi~\gamma^{\mu}\psi\right)-m\bar\psi \psi ~=~{\cal L}_0-{\cal H}, \tag{1}$$

where

$${\cal L}_0~:=~\frac{i}{2}\left(\psi^{\dagger}\dot{\psi} -\dot{\psi}^{\dagger}\psi\right), \qquad {\cal H}~:=~\frac{i}{2}\sum_{j=1}^3\left(\bar\psi\gamma^j\partial_j\psi -\partial_j\bar\psi~\gamma^j\psi\right)+m\bar\psi \psi.\tag{2}$$

The first issue is that the complex fields

$$\psi_{\alpha}~\equiv~(\psi^1_{\alpha}+i\psi^2_{\alpha})/\sqrt{2} \quad\text{and}\quad \psi^{\dagger}_{\alpha}~\equiv~(\psi^1_{\alpha}-i\psi^2_{\alpha})/\sqrt{2}, \tag{3}$$

are not independent fields. It is probably most convincing if we rewrite the theory (1) using the real and imaginary parts, $\psi^1$ and $\psi^2$, which are two independent real-valued fields. Then

$${\cal L}_0~=~ \frac{i}{2}\sum_{a=1}^2 (\psi^a)^{T}\dot{\psi}^a .\tag{4}$$

Now define imaginary-valued canonical momenta

$$\pi_a^{\alpha}~:= ~\frac{\partial_R {\cal L}}{\partial\dot{\psi}_{\alpha}^a }~=~\frac{i}{2} \psi^a_{\alpha} \tag{5}$$

as the right$^3$ derivative of ${\cal L}$ wrt. $\dot{\psi}_{\alpha}^a$. The non-zero canonical equal-time super-Poisson brackets read

$$\{ \psi^a_{\alpha}({\bf x},t),\pi_b^{\beta}({\bf y},t) \}_{PB}~=~\delta^a_b \delta_{\alpha}^{\beta} ~\delta^3({\bf x}-{\bf y}). \tag{6}$$

[Curiously, the canonical Poisson bracket $\{\cdot,\cdot\}$ itself is intrinsically imaginary in the fermionic sector.] Eq. (5) yields two primary constraints

$$\chi_a~:=~\pi_a - \frac{i}{2} \psi^a~\approx~0 ,\tag{7}$$

which are second-class constraints

$$\Delta^{\alpha\beta}_{ab}({\bf x}-{\bf y})~:=~\{\chi^{\alpha}_a({\bf x},t),\chi_b^{\beta}({\bf y},t)\}_{PB}~=~-i\delta_{ab}~\delta^{\alpha\beta} ~\delta^3({\bf x}-{\bf y}),\tag{8}$$

with inverse matrix

$$(\Delta^{-1})_{\alpha\beta}^{ab}({\bf x}-{\bf y})~=~i\delta^{ab}~\delta_{\alpha\beta} ~\delta^3({\bf x}-{\bf y}).\tag{9}$$

The Poisson bracket (6) should be replaced with the Dirac bracket. The fundamental Dirac brackets read

$$\{ \psi^a_{\alpha}({\bf x},t),\psi^b_{\beta}({\bf y},t) \}_{DB} ~=~-i\delta^{ab}~\delta_{\alpha\beta} ~\delta^3({\bf x}-{\bf y}).\tag{10}$$

The result (10) is in agreement with the Faddeev-Jackiw method, see e.g. eq. (5') in my Phys.SE answer here, which also list the corresponding canonical anticommutation relations (CARs).

References:

1. A. Das, Lectures on QFT, (2008); chapter 10.

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$^1$ Dirac-Bergmann analysis of Grassmann-odd fields is also considered in my Phys.SE answer here.

$^2$ One may show that the Lagrangian density (1) is real using

$$(\gamma^{\mu})^{\dagger}~=~ \gamma^0\gamma^{\mu}\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{11}$$

Conventions: In this answer, we will use $(+,-,-,-)$ Minkowski sign convention, and Clifford algebra

$$\{\gamma^{\mu},\gamma^{\nu}\}_+~=~2\eta^{\mu\nu}{\bf 1}_{4 \times 4}.\tag{12}$$

$^3$ Left and right derivatives of Grassmann-odd variables are also discussed in my Phys.SE answer here.