- #1

LCSphysicist

- 646

- 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}##?

$$

\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}##?

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