As asked in the title, is Hamiltonian containing enough information to judge the existence of spontaneously symmetry breaking?

Any examples?

The Hamiltonian of a theory describes its dynamics. Symmetry is "spontaneously broken" when a certain state of the quantum theory doesn't have the same symmetry as the Hamiltonian (dynamics). The standard example is a theory with a quartic potential. In field theory, say we have a potential $V (\phi) \sim \lambda{(\phi^2 - a^2)}^2$ in the lagrangian/hamiltonian, for some real valued scalar field $\phi$. This potential has minima at $\phi_{\pm}=\pm a$.

The theory (Hamiltonian) has a $\phi \rightarrow - \phi$ symmetry, but note that each of those vacua don't have the same symmetry i.e. vacuum states are not invariant under the symmetry. The solutions $\phi_{\pm}$ transform into each other under the transformation. So this is an example where the symmetry of the theory is broken by the vacuum state, "spontaneously" (roughly by itself, as the theory settles into the vacuum state).

So, to find whether a symmetry will be spontaneously broken (given a Hamiltonian), you have to check the symmetry of the states (typically vacua) of the theory and compare them to the symmetry of the Hamiltonian.

I'm not sure what OP has in mind, but consider e.g. the inverted harmonic oscillator $$H~=~\frac{p^2}{2m}-\frac{1}{2}kq^2, $$ which has spontaneously broken $\mathbb{Z}_2$ symmetry $(q,p) \to (-q,-p) $. The stable positions are $q=\pm \infty.$

In general, you should look to solve the potential for a function that gives a minimum. Does this minimum of the potential respect the symmetry? If not, you are very likely to find an anomaly. The classic example of this is the minimum of the higgs potential not respecting the gauge invariance of the SU(2) x U(1) fields coupled to it, but there are simpler examples, like the one @Qmechanic points out.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.