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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?

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    $\begingroup$ 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 $\Psi\propto\delta(\phi_{2}-v)\delta(\phi_{1})$. $\endgroup$
    – Buzz
    Aug 27, 2023 at 3:33

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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 $$ e^{i\theta \int\! dx~ (\phi_1 \partial_0 \phi_2 -\phi_2 \partial_0 \phi_1 ) } \phi_2 e^{-i\theta \int\! dx~ (\phi_1 \partial_0 \phi_2 -\phi_2 \partial_0 \phi_1 ) }= \phi_1\sin\theta + \phi_2\cos\theta. $$

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