# What is the Fermion's mass in this Lagrangian?

• LCSphysicist
LCSphysicist
Homework Statement
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Relevant Equations
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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:
vanhees71

## What is a fermion in the context of a Lagrangian?

A fermion is a particle that follows Fermi-Dirac statistics and is described by a Lagrangian in quantum field theory. Fermions include particles such as electrons, quarks, and neutrinos, which have half-integer spin (e.g., 1/2, -1/2). In a Lagrangian, fermions are typically represented by fields that obey the Dirac or Weyl equations.

## How is the mass of a fermion represented in a Lagrangian?

The mass of a fermion is represented in a Lagrangian by a term that couples the fermion field to itself. For a Dirac fermion, this term is usually of the form $$m\bar{\psi}\psi$$, where $$m$$ is the mass of the fermion, $$\psi$$ is the fermion field, and $$\bar{\psi}$$ is the Dirac adjoint of $$\psi$$. This term explicitly breaks chiral symmetry, indicating that the fermion has a non-zero mass.

## How do you extract the mass of a fermion from a given Lagrangian?

To extract the mass of a fermion from a given Lagrangian, identify the term that couples the fermion field to itself without any derivatives. For a Dirac fermion, this is typically the term $$m\bar{\psi}\psi$$. The coefficient $$m$$ in this term is the mass of the fermion. For example, in the Lagrangian $$\mathcal{L} = \bar{\psi}(i\gamma^\mu \partial_\mu - m)\psi$$, the mass of the fermion is $$m$$.

## What role does the fermion mass play in the dynamics described by the Lagrangian?

The fermion mass term in the Lagrangian affects the dynamics by giving the fermion a rest mass and influencing how it propagates and interacts with other fields. A non-zero mass term breaks chiral symmetry and affects the dispersion relation of the fermion, leading to different behavior compared to massless fermions. The mass also determines the energy-momentum relation for the fermion, influencing how it behaves under boosts and rotations.

## Can the mass of a fermion be zero in a Lagrangian, and what does that imply?

Yes, the mass of a fermion can be zero in a Lagrangian. When the mass term $$m\bar{\psi}\psi$$ is absent, the fermion is massless. This implies that the fermion obeys the Weyl equation rather than the Dirac equation, and it exhibits chiral symmetry. Massless ferm

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