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## 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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