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  1. Consider the Spontaneous symmetry breaking in the theory $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{\mu^2}{2}\phi^2+\frac{\lambda}{4!}\phi^4.$$ By the ground state of a classical field theory we mean the minimum value of the full Hamiltonian. Then in studying spontaneous symmetry breaking why do we only minimize the potential $V(\phi)=\frac{\mu^2}{2}\phi^2-\frac{\lambda}{4!}\phi^4$ and do not pay attention to the energy contributed by the gradient term $\partial_\mu\phi\partial^\mu\phi$? Certainly, if in the ground state, $\phi(x,t)$ has a spatial variation then $(\nabla\phi)^2$ term would contribute to the energy density. So why do we only minimize $V(\phi)$?

  2. Is it assumed that $\phi(x,t)=\phi_0=\text{a constant independent of space-time}$ in the ground state? Is it necessary in all theories that $\phi(x,t)=\text{a constant}$ in ground state of the system?

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  • $\begingroup$ Comment to the post (v2): $V$ should be $-V$ to have a minimum in the first place. $\endgroup$ – Qmechanic Jul 29 '15 at 19:41
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With a Lagrangian like: $\mathcal{L} = \partial_\mu \phi^\dagger \, \partial^\mu \phi - V(\phi) = \mathring{\phi^\dagger} \mathring{\phi} + \partial_i \phi^\dagger \, \partial^i\phi - V(\phi) $, the Hamiltonian is: \begin{equation*} \mathcal{H} = \frac{\partial \mathcal{L}}{\partial \mathring{\phi}} \mathring{\phi} + \mathring{\phi^\dagger} \frac{\partial \mathcal{L}}{\partial \mathring{\phi^\dagger}} - \mathcal{L} \end{equation*} which gives: \begin{equation*} \begin{array}{ll} \mathcal{H} & = \mathring{\phi^\dagger} \mathring{\phi} + \mathring{\phi^\dagger} \mathring{\phi} - ( \mathring{\phi^\dagger} \mathring{\phi} + \partial_i \phi^\dagger \, \partial^i\phi - V(\phi) ) \\ & = \mathring{\phi^\dagger} \mathring{\phi} - \partial_i \phi^\dagger \, \partial^i\phi + V(\phi)\\ & = \mathring{\phi^\dagger} \mathring{\phi} + \vec{\nabla} \phi^\dagger . \vec{\nabla} \phi + V(\phi) \\ & = |\mathring{\phi}|^2 + |\vec{\nabla} \phi|^2 + V(\phi) \end{array} \end{equation*} We notice that the first 2 terms of the Hamiltonian are positively defined and vanish when $\phi$ is a constant field (not depending on space-time). Therefore, the minimum of $\mathcal{H}$ is reached for a constant field $\phi_0$ that minimizes the last 2 terms i.e. the potential $V(\phi_0)$.

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Hints:

  1. Then potential term $\frac{1}{2}(\nabla\phi)^2$ is semipositive definite and is only zero for a $x$-independent configuration $\phi$.

  2. If one completes the square of the potential $$V(\phi)~=~\frac{\lambda}{4}\phi^4-\frac{\mu^2}{2}\phi^2~=~ \frac{\lambda}{4} \left(\phi^2-\frac{\mu^2}{\lambda}\right)^2-\frac{\mu^4}{4\lambda},$$ then it becomes clear that the $\phi$-minimum configurations for the two potentials $V(\phi)$ and $$ U(\phi)~=~\frac{1}{2}(\nabla\phi)^2+V(\phi) $$ are the same.

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