# Calculating the vacuum in an $O(2)$ spontaneous symmetry breaking

Consider the Lagrangian (we ignore the quantum correction here) $$\mathcal{L}=-\frac{1}{2} \partial_\mu \phi_1 \partial^\mu \phi_1-\frac{1}{2} \partial_\mu \phi_2 \partial^\mu \phi_2-\frac{\mu^2}{2} \left(\phi_1 \phi_1+\phi_2 \phi_2\right)-\frac{\lambda}{4} \left(\phi_1 \phi_1+\phi_2 \phi_2\right)^2$$ where $$\phi_1,\phi_2$$ are two real scalar fields.

It is obvious that this theory has a global $$O(2)$$ symmetry. In the case where $$\mu^2<0$$, any $$\phi$$ such that $$\phi_1^2+\phi_2^2=v^2\equiv\frac{|\mu^2|}{\lambda}$$ corresponds to a global minimum of potential $$V(\phi):=\frac{\mu^2}{2} \left(\phi_1 \phi_1+\phi_2 \phi_2\right)+\frac{\lambda}{4} \left(\phi_1 \phi_1+\phi_2 \phi_2\right)^2$$.

Therefore, this corresponds to an infinite number of degenerate vacuums $$\vert \Omega_\theta\rangle$$ in the corresponding system, such that $$\langle\Omega_\theta\vert\phi_1\vert \Omega_\theta\rangle=v\sin(\theta),\langle\Omega_\theta\vert\phi_2\vert \Omega_\theta\rangle=v\cos(\theta)$$, these vacuums can be connected by $$O(2)$$ transformations, thus in the $$\mu^2<0$$ case we have a spontaneous symmetry breaking in $$O(2)$$ symmetry.

So here's my question: Is there any way to calculate the explicit expression of $$\vert \Omega_{\theta=0}\rangle$$? How?

• The only explicit representation I know for a state like this would be as a wave function$al$, but then the answer is rather trivial, just $\Psi\propto\delta(\phi_{2}-v)\delta(\phi_{1})$.
– Buzz
Aug 27, 2023 at 3:33

There are dozens of questions on this site constructing field eigenstates and field coherent states in terms of real field operators and the "empty" Fock state $$|0\rangle$$ to your satisfaction. Something like $$|\Omega_0\rangle\propto \exp(v\int\!dx ~(\phi_2)_+)|0\rangle$$.
You only need construct this way the state $$|\Omega_0\rangle$$ s.t. $$\langle \Omega_0| \phi_2|\Omega_0\rangle=v$$, $$\langle \Omega_0| \phi_1|\Omega_0\rangle=0$$, and then rotate around the circle of minima, $$|\Omega_\theta\rangle= e^{i\theta \int\! dx~ (\phi_1 \partial_0 \phi_2 -\phi_2 \partial_0 \phi_1 ) }|\Omega_0\rangle,$$ or something of the sort given less cavalier normalizations, since $$e^{i\theta \int\! dx~ (\phi_1 \partial_0 \phi_2 -\phi_2 \partial_0 \phi_1 ) } \phi_2 e^{-i\theta \int\! dx~ (\phi_1 \partial_0 \phi_2 -\phi_2 \partial_0 \phi_1 ) }= \phi_1\sin\theta + \phi_2\cos\theta.$$