3
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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 '18 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 '18 at 16:05
  • $\begingroup$ I think you misinterpreted the statement of the exercise. @Javier is correct. $\endgroup$ – Jon Feb 23 '18 at 16:06
  • $\begingroup$ Where? In what context? Where they perhaps talking about gauge invariance? $\endgroup$ – Javier Feb 23 '18 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 '18 at 16:07
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '18 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 '18 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 '18 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 '18 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 '18 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.