A classical example of spontaneous symmetry breaking

[spaghetti3451]

Homework Statement

A simple classical example that demonstrates spontaneous symmetry breaking is described by the Lagrangian for a scalar with a negative mass term:

$\mathcal{L}=-\frac{1}{2}\phi\Box\phi + \frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4}$.

(a) How many constants $c$ can you find for which $\phi(x)=c$ is a solution to the equations of motion? Which solution has the lowest energy (the ground state)?

(b) The Lagrangian has a symmetry under $\phi \rightarrow -\phi$. Show that this symmetry is not respected by the ground state. We say the vacuum expectation value of $\phi$ is $c$, and write $\langle\phi\rangle=c$. In this vacuum, the $\mathbb{Z}_2$ symmetry $\phi \rightarrow -\phi$ is spontaneously broken.

(c) Write $\phi(x)=c+\pi(x)$ and substitute back into the Lagrangian. Show that now $\pi = 0$ is a solution to the equations of motion. How does $\pi$ transform under the $\mathbb{Z}_2$ symmetry $\phi \rightarrow -\phi$ ? Show that this is a symmetry of $\pi$'s Lagrangian.

Homework Equations

The Attempt at a Solution

How does the Lagrangian $\mathcal{L}=-\frac{1}{2}\phi\Box\phi + \frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4}$ for the scalar field $\phi(x)$ have a negative mass term?

As far as I can see, the mass term $\frac{1}{2}m^{2}\phi^{2}$ is positive!

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[Fred Wright]

In the Klein-Gordon eqn. the mass term is the same sign as the kinetic term. If you consider the Lagrangian as the sum of a kinetic term and a potential term you see that the potential term
½m²Φ² - (λ/4!)Φ⁴ triggers a broken symmetry. This is the famous Goldstone's Mexican hat. Varying the Lagrangian w.r.t. Φ gives the minima
Φ = m√(3!/λ)e^iθ, where θ varies between 0 and 2π (i.e. an infinite number of ground states).

 

[spaghetti3451]

In the Klein-Gordon eqn. the mass term is the same sign as the kinetic term.

The Klein-Gordon equation has two terms $\Box\phi$ and $m^{2}\phi$. Are you referring to $\Box\phi$ as the kinetic term?

If you consider the Lagrangian as the sum of a kinetic term and a potential term you see that the potential term ½m²Φ² - (λ/4!)Φ⁴ triggers a broken symmetry.

Would you be able to demonstrate the math that explains the meaning of a broken symmetry?

Varying the Lagrangian w.r.t. Φ gives the minima Φ = m√(3!/λ)e^iθ, where θ varies between 0 and 2π (i.e. an infinite number of ground states).

Why do you parameterise the ground states by $\theta$? In other words, why does the variation of the Lagrangian with respect to $\phi$ give the ground states of the field theory?Thank you so much for the answer! But, it's best if you only explain the query in my first post as too much of this information from you is only making me confused.