+ V_{SB}(v\,\cos\theta) \, .$$ We now consider several limits for the symmetry breaking piece of the potential.

Suppose that $\epsilon_1 = f_{\pi}m_{\pi}^2$ and all other $\epsilon$’s are zero. Minimizing the potential with respect to $v$ at fixed $\theta$ gives $$v(\theta) \, \approx \, f_{\pi}(1 + \Delta) \, ,$$ where $\Delta = - 2(m_{\pi}/m_{\sigma})^2 \sin^2(\theta/2)$, $m_{\sigma}^2 = 2\lambda f_{\pi}^2 + m_{\pi}^2$, and $$V_{\rm min}(\theta) - V_{\rm min}(0) \, \approx \,
2 f_{\pi}^2 m_{\pi}^2 \sin^2(\theta/2) \, .$$ Corrections to the approximate equalities are of relative order $\Delta << 1$. This is a tilted wine bottle bottom with only one minimum.

Now suppose that $\epsilon_2 = m_{\pi}^2$ and all other $\epsilon$’s are zero. Then $$v^2(\theta) \,=\, f_{\pi}^2 [1 - 2 m_{\pi}^2
 \sin^2\theta/m_{\sigma}^2] \, ,$$ where $m_{\sigma}^2 = 2\lambda f_{\pi}^2$, and $$V_{\rm min}(\theta) - V_{\rm min}(0) \, \approx \,
\frac{1}{2} f_{\pi}^2 m_{\pi}^2 \sin^2\theta \, .$$ We neglect a correction of relative order $(m_{\pi}/m_{\sigma})^2$. This potential has two degenerate minima located at $\theta = 0$ and $\pi$!

In general, one may expect that $V_{SB}$ allows for two minima, one at $\theta = 0$ and one at $\theta = \pi$. If they are not degenerate then by a simple redefinition of the fields we may, by 