I am having some basic questions about how to interpret Lagrangians, lets start with Dirac:

$L = \bar{\Psi} (i \gamma^{\mu} \partial_{\mu} -m) \Psi$,

where $\Psi$ is a Dirac-Spinor, $m$ is the mass, $\gamma^{\mu}$ is a gamma matrix and $\partial_{\mu}$ is the derivative.

1) Is $m$ a vector or a matrix or a scalar? I always thought that it is a scalar, but for some reason $\bar{\Psi} m \Psi$ is allowed in the Lagrangian, but $m \bar{\Psi} \Psi$ isn't, so it can't be a scalar! EDIT m is a scalar and can be anywhere in the Dirac equation, but these terms are not valid for the Standard model!

2) What exactly is violated so that $m \bar{\Psi} \Psi$ is not allowed? Is it not invariant under something? EDIT these terms are both not valid for the Standard model, because they are not gauge invariant!

3) I always thought that a Dirac spinor contains all the possible for a fermion, the two spin states for a particle and the two spin states for the antiparticle. Is this assumption correct?

4) Why do we need $\bar{\Psi}$? What does it represent? Do the particles and anti-particles trade places? Or is my interpretation in 3) wrong and $\Psi$ represents the particle and $\bar{\Psi}$ and antiparticle.

5) The Dirac equation describes a free massive fermion moving through space and time, does the $\bar{\Psi}$ indicate an interaction?

I was trying to understand it from wikipedia, but I failed. Any answer to any of the questions above, will be appreciated.

  • $\begingroup$ Who says that $\bar \Psi \Psi$ isn't allowed? $m$ is a scalar so you can put it wherever you want. $\endgroup$
    – Javier
    Feb 23, 2018 at 16:02
  • $\begingroup$ An exercise in quantum-field theory states the following: "The following terms are not allowed in the Standard Model Lagrangian. For each term, explain briefly why. 1) $m \bar{\Psi}\Psi$" and then some more Lagrangian terms... $\endgroup$
    – Alex
    Feb 23, 2018 at 16:05
  • $\begingroup$ I think you misinterpreted the statement of the exercise. @Javier is correct. $\endgroup$
    – Jon
    Feb 23, 2018 at 16:06
  • $\begingroup$ Where? In what context? Where they perhaps talking about gauge invariance? $\endgroup$
    – Javier
    Feb 23, 2018 at 16:07
  • 1
    $\begingroup$ @Alex That's something completely different. For a generic Dirac fermion you can write $m \bar{\Psi} \Psi$. In the standard model you can't because it violates gauge invariance. Note that you can't write $\bar{\Psi} m \Psi$ either for the same reason. $\endgroup$
    – knzhou
    Feb 23, 2018 at 16:07

2 Answers 2


By the way that is not the Dirac equation, but the Dirac Lagrangian/action.

1) $m$ is a scalar. Mass terms for fermionic fields are allowed in the Standard Model, you are confusing mass terms for gauge fields which are not allowed on their own, but come in through spontaneous symmetry breaking (Higgs mechanism).

2) $m\Psi \bar{\Psi}$ is allowed, as any phase change (local of global) will cancel out. Again, you are confusing a mass term for a gauge field $mA^\mu A_\mu$. This would violate gauge invariance, $A_\mu \rightarrow A_\mu + \partial_\mu\Lambda$.

EDIT The above two answers are true for Dirac Lagrangians and EM interactions, as stated in the question. In the presence of a weak interaction, fermions are affected differently depending on their chirality. This then introduces a gauge-dependent mass term, only saved by the Higgs mechanism.

3) No, it's only the equation of motion for a spin-1/2 fermion. If you construct the spin operator $\mathbf{S}^2$, you'll find that the eigenvalues are $\frac{3}{4}\hbar^2$, corresponding to $S(S+1)$ with $S = 1/2$.

For spin 3/2 fermions, the equation is this, etc.

4) What is the interpretaion of the complex conjugate of a number? Really, you just make up whatever term in the Lagrangian gives you the correct Dirac equation (when applying the Euler-Lagrange equations), which you know is correct from experiment.
You can always justify the form of the Lagrangian, for example having $\Psi \bar{\Psi}$ means that you have local and global phase invariance, and that the resulting potential $\propto \Psi^2$ has a minimum, thereby leading to a stable field theory.

5) $\bar{\Psi}$ is not an interaction. The Dirac equation is the equation obeyed by the a free massive spin-1/2 fermion. Or, more correctly, by its field operator (whereby I am making the distinction between relativistic quantum mechanism and quantum field theory).

NB though that you can just set the mass to $0$, and you get the so called Weyl fermions.

To get interacations, you need non-linear terms.

The one that usually comes up is $\propto J^\mu A_\mu = \Psi \gamma^\mu \bar{\Psi}A_\mu$, where $A_\mu$ is the electro-magnetic gauge potential. This term is not linear, and it represents the interaction between a spin-1/2 fermion $\Psi$ and a spin-1 vector boson $A_\mu$.

You can also make two different fermions interact by having a term that goes like $\Psi_1 \cdot \Psi_2$, where both obey their individual Dirac equation.

Wikipedia is really bad for this stuff unless you already know roughly what is going on, I would recomend looking any undergraduate lecture series on gauge field theories. The Cambridge one is quite good.

  • $\begingroup$ Thank you for your answer! I was reading more up, and I'm a little bit confused by some statements: 1) + 2) This was a discussion in the comments, but I actually think that mass terms are indeed not allowed in the SM, unless they contain the Higgs scalar. This term does not contain it, furthermore it is not gauge invariant, which is a requirement for SM. $\endgroup$
    – Alex
    Feb 24, 2018 at 20:00
  • $\begingroup$ 3) I think you answered a different question here. $\Psi$ is a spinor, not an equation. My questions was regarding the entries of the spinor. 4) $\bar{\Psi}$ is not the complex conjugate, it's the hermitian conjugate times the $\gamma^0$ matrix $\bar{\Psi}=\Psi^{\dagger} \gamma^0$, and I don't understand this object, but I do understand the argument for a make up object to make the equation work... In this field charge conjugate seems to have the notation $\Psi^*$. 5) So although we have two $\Psi$s in the Lagrangian, it is only one fermion interacting with the photon $A_{\mu}$? $\endgroup$
    – Alex
    Feb 24, 2018 at 20:22
  • $\begingroup$ 1) + 2), I am sorry - but no. Any mass term for $\psi$ would not violate gauge invariance. I don't know your level and how much you already know, but "gauge" invariance here just means "phase" invariance, and it is trivial to see that $m\bar{\Psi} \Psi$ is left invariant by any global or local phase transormation, in terms of any $U(N)$. Now, if you started off with the kinetic-only Dirac lagrangian + the Higgs potential, then sure the mass term would drop out naturally, with the $m$ coefficient being in terms of the Higgs coupling $g$ and the VEV. $\endgroup$
    – SuperCiocia
    Feb 24, 2018 at 22:44
  • $\begingroup$ But nothing prevents you from inclduing the $m\bar{\Psi}\Psi$ term to begin with, it's just nicer to get it from the Higgs potential. What you are not allowed to do, however, is putting a mass term for a spin-1 gauge field $A_\mu$, and $mA_{\mu}A^{}\mu$ is not invariant under the gauge transformation $A_\mu \rightarrow A_\mu + \partial_\mu \Lambda$ as you can trivially show. The fact that weak gauge bosons have mass though, means that we need the Higgs mechanism here to get the mass term. $\endgroup$
    – SuperCiocia
    Feb 24, 2018 at 22:47
  • $\begingroup$ 3) I misunderstood your question then, I apologise. It has 4 degrees of freedom, interpreted as spin UP & spin DOWN, electron & positron. $\endgroup$
    – SuperCiocia
    Feb 24, 2018 at 22:48

A scalar $m$ breaks the gauge symmetry for any gauge field whose left versus right behaviors are different (i.e. chiral fields). If you break $Ψ$ into left and right helicity components $L$ and $R$, then $\bar{Ψ} m Ψ$ is actually $\bar{L} m R + \bar{R} m L$. For dynamics, this entails an oscillation between left and right helicity components.

You can view it as follows. Generically, helicity refers to the component of angular momentum along an axis collinear with the momentum. For bodies moving at under light speed, the direction of motion is relative. To an observer moving the same way as the body is, but at a faster speed, the body appears to be moving the other way and its helicity has the opposite sign; so that helicity is relative in both left/right'edness and magnitude.

For light-speed bodies, there is no overtaking them and no rest frame. So, helicity can be an invariant. In the classification into "tardion", "luxon", "tachyon" (respectively for sub-light, light-speed and super-light systems), luxons has a further sub-classification into generic versus the (unnamed) "helical" subclass. In the latter, angular momentum is entirely on an axis collinear with momentum and helicity is an invariant - both its sign and magnitude. Photons fall in this class. In particular, photons have helicity, not spin.

The only other subclass that has helicity as an invariant are the ones where it is trivially so: "spin zero" bodies, that possess no intrinsic angular momentum and therefore zero helicity.

Here's where the problem occurs: for the weak force, the charge is left-helicity! Up to proportion. For anti-matter, it is right helicity. Charge is an invariant property - which forces helicity to be an invariant property, which forces the bodies to be helical luxons, since that's the only (non-trivial) case where helicity is an invariant.

In the 1950's and before that, they would have summed it up as: "WHOOPS!" and pulled out the red "REJECT!" stamp to plaster on top of the published papers.

The $Ψ$ in the Dirac equation gives the appearance of two light-speed helical states, $L$ and $R$, with the equation giving you an oscillation between the two states, the oscillation taking place at a rate proportional to $m$. In effect, it's like a light speed body that is constantly and randomly zig-zagging between left and right states, whose time-averaged speed and path are that of a tardion with a positive rest mass $m$.

To make this co-habit consistently with a chiral field, like the weak force, you have effectively turn on $m$ as a field by treating it as $m = g φ$, where $g$ is the $φ$-charge of the body and the field $φ$ is permanently and everywhere switched on. So, now the Dirac term $\bar{Ψ} m Ψ$ becomes the "Yukawa term" $\bar{L} g φ R + \bar{R} φ^† g L$. What does $φ$ do? It drives the oscillation between the left and right helicity states $L$ and $R$. The $g$'s can be skewed, so that "mass" eigenstates need not line up with "charge" eigenstates. So, $g$ could actually be a matrix as well: $\bar{L} g φ R$ and $\bar{R} φ^† g^† L$.

So, all the tardions that interact with the chiral field are explained away as actually being helical luxons that only have the appearance of being of positive mass $m$ and slower than light, thanks to the zig-zagging action driven by the $φ$ field. Fundamentally, everything that interacts non-trivially with the weak force is a de facto helical luxon of zero mass. Any appearance of being otherwise is just an illusion brought about by the action of the $φ$ field.

These terms can be made to respect gauge symmetry by endowing $φ$ with suitable transformation properties under a gauge transform. That means the "on-state" for $φ$ can't be unique. Under a gauge transform, it transforms to a different on-state. But only one is selected out as the actual on-state - everywhere in the Universe. Hence, the term "broken symmetry".

For the weak force (which has 3 modes) - and the more comprehensive electro-weak force (which has 4 modes) it is a part of (the other part being the "hypercharge") - one residual mode of gauge symmetry remains; the other three are frozen by the broken symmetry. The forces exhibited by the three frozen modes are short range, because the vacuum is opaque with respect to them, thanks to the interference brought about by the on-state of the $φ$ field.

The fourth mode is long range and the vacuum is transparent with respect to it, since the on-state of the $φ$ also has the gauge symmetry. It also happens to be the one mode that is left-right symmetric. That didn't actually have to be the case. There's nothing fundamental in the model for the electroweak theory that required it. It just is, and that's an unexplained feature.

The long-range force for the fourth mode is the electromagnetic force. It is a combination of the hypercharge and one of the three modes of the weak force. The hypercharge is what actually satisfies the Maxwell equations (and its quantum, the $B$, has zero weak nuclear charge), while the electromagnetic field does not(!) - but rather a non-linear set of field equations that include also the effect of the weak force. Electroweak technically falsifies Maxwell's theory for electromagnetism. The electromagnetic field interacts with the weak nuclear force. In particular, the photon interacts with the weak nuclear force. It's not neutral.


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