I would like to know what is the contribution of the right source term to this fermion component equation of motion: \begin{equation} i(\partial_0+\partial_1)\psi_R= a \psi_R \end{equation} where $a$ is a real constant.

It's obvious that for $a=0$ we have a propagating wave to the right, since the imaginary factor doesn't contribute to anything. But on the other cases, this $i$ factor goes directly into the source as: \begin{equation} (\partial_0+\partial_1)\psi_R= ia' \psi_R \end{equation} so I do have two questions:

1) What effect would the imaginary source term make if it was only: $ia'$ (without field)

2) And the most important, what does it make when this term is also a reaction term, meaning it has the field there too: $ia' \psi_R$

Clarification: $\partial_0=\frac{\partial}{\partial t}$, $\partial_1=\frac{\partial}{\partial x}$ and a'=-a of course.

My try:

I've tried discretizing it (with same grid for space and time $\Delta$) to see the effect (i will be space index, j time index):

\begin{equation} \psi_R(i,j+1)-\psi_R(i,j)+\psi_R(i+1,j)-\psi_R(i,j)=ia' \Delta \psi_R(i,j) \end{equation} so: \begin{equation} \psi_R(i,j+1)=2\psi_R(i,j)-\psi_R(i+1,j)+ia' \Delta \psi_R(i,j) \end{equation} where we see that the imaginary source obviously gives imaginary components to the field, which in the next time evolution, will be negative (i*i=-1), and so on... giving rotations of the field $\psi$ in the complex plane proportional to $\Delta$ and to $\psi$ in the reactionary case 2), which just makes the rotations bigger or smaller depending on the previous quantity of $\psi$, which in the constant case 1) would just be a constant value of the field rotating in this complex plane always.

Of course all of this happens with dependence of the previous space coordinate, which means that matter is traveling, and so are the complex rotations??¿

Further context, not really necessary to answer the question (also check, comments for references if needed):

There are lots of things happening in this theory, such as the $\theta$ parameter, the Higgs mechanism, the anomaly on the currents, which are related to the Gauge fixing, but I don't really care about that here, only want the math interpretation of this concrete equation.

In case anyone wonders on the complete fermionic equations, they are: \begin{equation} \begin{cases} 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 \end{cases} \longrightarrow \begin{cases} i\partial_0 \psi_R = \left(-i\partial_1+e(A^0+A^1)\right) \psi_R \\ i\partial_0 \psi_R = \left(i\partial_1+e(A^0-A^1)\right) \psi_R \end{cases} \end{equation} which in a circle (of radius $r$) space-time (periodic boundaries for space) using linear combinations of eigenvalues of the Hamiltonian (i$\partial_t \psi_k= E_k \psi_k$), which have to fulfill the periodic boundary conditions have solutions like: \begin{equation} \psi(t,x)_{R/L}=\frac{1}{\sqrt{r}}\sum_k u(k)_{R/L} e^{-iE_kt} e^{i\left(k+\frac12\right)\frac{2\pi}{r}x} \ \text{ with } k \in \mathcal{Z} \end{equation} the equations become: \begin{equation} \begin{cases} E_k^R{\psi_k}_R = \left( \left(k+\frac12\right)\frac{2\pi}{r}+e(A^0+A^1)\right) {\psi_k}_R \\ E_k^L{\psi_k}_L = \left(-\left(k+\frac12\right)\frac{2\pi}{r} 1+e(A^0-A^1)\right) {\psi_k}_L \end{cases} \end{equation}which give us the spectrum: \begin{equation} \begin{cases} E_k^R = eA_0+\frac{2\pi}{r} \left(k+\frac12\right)+eA^1\\ E_k^L = eA_0-\frac{2\pi}{r} \left(k+\frac12\right)-eA^1 \end{cases} \end{equation} so we see that the $A_0$ gives energy linearly to both components ($\psi_R$ and $\psi_L$) creating or destroying both equally, which is related to the $j_V$, while the $A_1$ give energy also linearly to one and takes from the other, converting left to right and vice-versa, related with $j_A$. Where we see the relation of the gauge freedom with the anomaly of the vector and axial currents/charges ($Q_V=|\psi_R|^2+|\psi_L|^2$ and $Q_A=|\psi_R|^2-|\psi_L|^2$).

Which at the limit $r \xrightarrow[]{} \infty$, which would means going to the infinite normal Minkowsky 1+1d space-time again, doesn't make much sense to me.

So that is why I was asking for interpretation directly from the original equation, as a pure mathematical solution.

  • $\begingroup$ Discretizing the diferential equation I've come to the conclusion that the imaginary source terms, ends up adding rotations on the imaginary plane into the evolution of the field, which makes it not so trivial as I initially thought. Any one with some good ideas? :) $\endgroup$ Jul 6, 2022 at 19:17
  • 2
    $\begingroup$ To fix conventions, are you following a reference? Which page? $\endgroup$
    – Qmechanic
    Jul 8, 2022 at 10:39
  • 1
    $\begingroup$ Well I'm generalizing things, but some good convenction would be David Tong: Lectures on Gauge Theory, chapter 7: damtp.cam.ac.uk/user/tong/gaugetheory.html ,page 348. $\endgroup$ Jul 8, 2022 at 10:47
  • $\begingroup$ Or Peskin chapter 19, page 650. In both cases, they only present the free fermionic eq. of motion, and I would like to interpret the interaction one, which gives the equations above. ($\psi_R=\psi_+$) $\endgroup$ Jul 8, 2022 at 10:50

1 Answer 1


((2)) The solutions of your homogeneous linear partial differential equation $$ \left(i \partial_t + i\partial_x - a\right) \psi(x,t) = 0$$ are arbitrary linear combinations of $$\psi_p(x,t) = e^{i p x - i E t}$$ where $$E = p + a$$ and $p\in \mathbb R$. So what the $i a \psi $ term does is simply affecting the dispersion relation, i.e. the relation between the energy and momentum in your theory. Without it, you'd have $E=p$.
You might interpret it as an analogy of the usual Dirac equation with and without the mass term. What the mass term does is that it gives mass to the fermion, i.e., it affects the dispersion relation $E^2 = p^2 + m^2$.

The general solution can be cast in the form $$\psi(x,t) = \int \mathrm dp\, \tilde\psi(p) \,e^{ipx-i(p+a)t}$$ where $\tilde \psi(p)$ is an arbitrary function. It can be rewritten as $$ \psi(x,t) = e^{-iat} \int \mathrm dp\, \tilde\psi(p)\, e^{ip(x-t)} =e^{-iat} \Psi(x-t). $$ Any function of $x, t$ in the form $\Psi(x-t)$ is the solution of your equation for $a=0$. The term proportional to $a$ in your Eq. makes additional phase rotations with time evolution, as you found in the discretizing approximation.

((1)) The general solution to the non-homogeneous equation $$i\left(\partial_t + \partial_x\right) \psi(x,t) = a$$ reads $$\psi(x,t) = -iat + \Psi(x-t) $$ where, again, $\Psi$ is an arbitrary function. See this link, for example. Notice that choosing $\Psi(x-t) = -ia(x-t) + \tilde \Psi(x-t)$ would translate the form of the solution to $\psi(x,t) = -iax + \tilde\Psi(x-t) $, so both variables are indeed equivalent. I think now it's pretty obvious what the source term does.

  • $\begingroup$ Indeed I simulated the eq. of motion in C++, to visualize them, and they exactly do rotate in the complex plane with an angular velocity proportional to a, while also moving to the right or to the left (as the solution a=0 does). Also the a term is exactly the mass term yeah, how silly of me! The interactions with the gauge field (a term), mix the left and right spinor components exactly as the mass does! $\endgroup$ Jul 12, 2022 at 9:21
  • $\begingroup$ I have to say at this point I wasn't expecting an answer really, at least not one that made sense, you arrived just in time, thanks man! $\endgroup$ Jul 12, 2022 at 9:22

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