Is it possible that in 1+1d QED the negative charge of the $\psi^\dagger$ fields appears only in the quantum model when we introduce the commutator relation for quantization? And then in the classical electrodynamics model, we only have positive charges for $\psi$ and $\psi^*$?

Because the classic equation of motion for the space change in E is: $\partial_1 E=e(\psi^*_R \psi_R+ \psi^*_L \psi_L) = e(|\psi_R|^2 + |\psi_L|^2) \equiv e\rho $ where wee see that the electric field always increases going to the right, since everything that contributes are + modulus? Meaning every matter has positive charge?

Clearly in the quantum model we do have negative charges, since $Q\psi^\dagger= -\psi^\dagger$. So I don't really know how to compare both cases, if I got it wrong in the classic case, or if this abrupt change in the theory really appears in quantization (If so, then what would the classical limit of the theory do? Would charges of antifermions switch again?).


I would like to know which charge do particles and antiparticles have in QED in 1+1d, and its construction, like what is actually the electric charge operator like? And from where in the lagrangian it comes? \begin{equation} \mathcal{L}= \bar{\psi} i(D_0 \gamma^0 + D_1 \gamma^1) \psi + \frac12F_{01}^2= \psi_R^* i (D_0+D_1) \psi_R + \psi_L^* i(D_0-D_1)\psi_L + \frac12E^2 \end{equation}with D the covariant derivative including the coupling.

Because so far I've only been able to compute the vector ($j_V^\mu$) and axial ($j_A^\mu$) currents/charges on the model, which in the chiral states are:

\begin{equation} \begin{cases} j_V^0=\psi\rho_V= \vec{j}_A= \bar{\psi} \gamma^0 \psi = \begin{pmatrix} \psi_L^* & \psi^*_R \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1& 0 \end{pmatrix} \begin{pmatrix} \psi_R \\ \psi_L \end{pmatrix} = |\psi_R|^2 + |\psi_L|^2 \\ j_V^1=\vec{j}_V = \rho_A= \bar{\psi} \gamma^1 \psi = \begin{pmatrix} \psi_L^* & \psi^*_R \end{pmatrix} \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} \psi_R \\ \psi_L \end{pmatrix} = |\psi_R|^2 - |\psi_L|^2 \end{cases} \end{equation} and if we now define the Noether charges as the typical $ Q \equiv \int dx j^0$ with $\partial_t Q =0$ for conserved currents ($\partial_\mu j^\mu=0$). Then we will have: \begin{align} & \begin{cases} Q_R = \int \bar{\psi_R} \gamma^0 \psi_R \ dx= \int \psi_R^* \psi_R \ dx = \int |\psi_R|^2 \ dx \\ Q_L= \int \bar{\psi_L} \gamma^0 \psi_L \ dx= \int \psi_L^* \psi_L \ dx = \int |\psi_L|^2 \ dx \end{cases} \\ & \begin{cases} Q_V= \int \bar{\psi} \gamma^0 \psi \ dx= \int (|\psi_R|^2 + |\psi_L|^2) \ dx \\ Q_A= \int \bar{\psi} \gamma^1 \psi \ dx= \int (|\psi_R|^2 - |\psi_L|^2) \ dx \end{cases} \label{eq:Q_A} \end{align} but I kept reading that there are electric negative charged fermions, and I am super saturated right-now from not sleeping well, and can't think clearly, but I only see positive charged fermions for all the components, the right-handed fermions and it's antiparticle have both positive charge $\rho_V$, and the same happens for the left-handed fermions...

The problem I have is that the equations of motion that I obtain: \begin{equation} \begin{cases} \partial_1 F^{10}= e j_V^0 = e \bar{\psi}\gamma^0 \psi \ \xrightarrow[]{} \ \ \partial_1 E = e (|\psi_R|^2 + |\psi_L|^2) \\ \partial_0 F^{01}= e j_V^1 = e \bar{\psi}\gamma^1 \psi \ \xrightarrow[]{} \ \ -\partial_0 E = e (|\psi_R|^2 - |\psi_L|^2)\\ i(\partial_0+\partial_1)\psi_R = e(A^0-A^1) \psi_R \ \text{ and } \ i(\partial_0+\partial_1)\psi_R^* = -e(A^0-A^1) \psi_R^* \\ i(\partial_0-\partial_1)\psi_L=e(A^0 + A^1)\psi_L \ \text{ and } \ i(\partial_0-\partial_1)\psi_L^*=-e(A^0 + A^1)\psi_L^*\\ \end{cases} \end{equation} which from the first equation, means that the electric field increase is the same for any $|\psi|$ it doesn't matter if it is right-handed, left-handed or particle/antiparticle because everything that contributes to change the E in x, is a positive modulus! So it's like everything has a positive charge, don't it?

I think I'm missing some super important point, like confusing electric charge with vector charge or something, but I don't really get it right now.

  • 1
    $\begingroup$ Related post by OP: physics.stackexchange.com/q/717050/2451 $\endgroup$
    – Qmechanic
    Commented Jul 19, 2022 at 10:00
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    $\begingroup$ Hint: start with free fermions and write $Q$ in terms of creation and annihilation operators. $\endgroup$ Commented Jul 19, 2022 at 11:19
  • $\begingroup$ Yeah I am doing that now, but what about at the classical level? Because this problem of mine arises there directly, and the only solutions I found are in the quantized version.. :( $\endgroup$ Commented Jul 19, 2022 at 16:48
  • $\begingroup$ Is it posible, that the negative charges come from the anticomutators relations when you quantize? Which means that in the classical theory there are no such negative charges? $\endgroup$ Commented Jul 19, 2022 at 20:25
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    $\begingroup$ All positive charges at the classical level? That's wrong because $(\psi, \psi^*) \to e^{i\alpha} (\psi, \psi^*)$ is not a symmetry of the theory. $\endgroup$ Commented Jul 19, 2022 at 23:12

1 Answer 1


I just found the answer, I was describing the fermion fields in the classical limit with complex numbers as if they were scalar fields, which in the case of fermions because of they anti commutativity doesn't work, which gave results as positive charge for all matter or problems in the definitions of the energy of the states for electron fields and positron fields.

This is solved when quantizing, in the introduction of the anticomutations relations: \begin{equation} \{\psi_e,\psi_p\}=0 \ \longrightarrow \psi_e\psi_p=- \psi_p \psi_e \end{equation}

To do this in the classical limit, we would need to treat the electron $(\psi_e)$ and positron $(\psi_p)$ fields with Grasmann variables, and we would arrive into a different result for the charge that makes more sense. But in doing so, the intuition of $|\psi|^2$ propagating is lost, since grassmann numbers are not only complex number, telling us the intrinsic quantum behaviour of spin particles.

If we do such a thing from the start we could build a theory reassembling the quantum limit better, but that arises other problems, further references:


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