What is the Fermion's mass in this Lagrangian?

LCSphysicist
Messages
644
Reaction score
162
Homework Statement
.
Relevant Equations
.
We have a Lagrangian of the form:
$$

\mathcal{L} = \overline{\psi} i \gamma_{\mu} \partial^{\mu} \psi - g \left( \overline{\psi}_L \psi_R \phi + \overline{\psi}_R \psi_L \phi^* \right) + \mathcal{L}_{\phi} - V(|\phi|^2)

$$
Essentially, what we are studying is spontaneous symmetry breaking. First, we must find the minimum of $$V(|\phi|^2)$$ to determine the vacuum state. We obtain:
$$

\langle \phi \rangle = v = \sqrt{\frac{m^2}{\lambda}}

$$
Now, let's perform the following expansion:
$$

\phi = (v + h(r, t)) e^{-\frac{i \pi(r, t)}{f}}

$$
Now, the question arises: How do we find the mass of the "new particles," ##\pi## and ##h##? This part is straightforward. However, the challenge lies in determining the fermion mass, denoted as ##m_{\psi}##, and its coupling to ##\pi## and ##h##.

I assume that the only terms that matter in answering this question are:

$$

\overline{\psi} i \gamma_{\mu} \partial^{\mu} \psi - g \left( \overline{\psi}_L \psi_R \phi + \overline{\psi}_R \psi_L \phi^* \right)

$$
Now, let's expand this term as follows:
$$

\overline{\psi} i \gamma_{\mu} \partial^{\mu} \psi - g \left( \overline{\psi}_L \psi_R \left( (v + h) e^{i \frac{\pi}{f}} \right) + \overline{\psi}_R \psi_L \left( (v + h) e^{-i \frac{\pi}{f}} \right) \right)

$$
The challenge here is to determine the fermion mass. My idea is to write a Lagrangian equivalent to the Dirac Lagrangian, where the constant ##c## that should appear in the Lagrangian, i.e., ##c \overline{\psi} \psi##, represents the mass. However, I can't find such a term in the Lagrangian we have. To proceed, I first rewrite ##\psi_{L,R}## in terms of ##\psi## itself, resulting in:

$$

- g (v+h) \overline{\psi} \left( \cos\left(\frac{\pi}{f}\right) + i \gamma^5 \sin\left(\frac{\pi}{f}\right) \right) \psi

$$

Next, I expand the trigonometric expressions to obtain:

$$

g (v+h) \overline{\psi} \left( 1 - \frac{1}{2} \left(\frac{\pi}{f}\right)^2 + i \gamma^5 \frac{\pi}{f} \right) \psi

$$
This expansion results in terms such as:

$$

- g v \overline{\psi} \psi - g h \overline{\psi} \psi - \frac{i g v \gamma^5}{f} \overline{\psi} \pi \psi + \frac{g v}{2 f^2} \overline{\psi} \pi \pi \psi + O(\ldots)

$$

So, the fermion mass would be ##g v##, the coupling ##h \psi \overline{\psi}## would be ##g##, and the ##\overline{\psi} \pi \psi## coupling would be ##\frac{i g v \gamma^5}{f}##?
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top