# Spontaneously symmetry breaking from QM perspective

I am a bit confused about the idea of spontaneously symmetry breaking (SSB), from the point of view of QM. I am talking here about the energy plot looking like a mexican hat in 2D (not the general Higgs one in the complex plane). I am mostly thinking of it by making an analogy with the ammonia molecule problem. The 2 lowest energy states in the SSB case don't respect the symmetry of the interaction, hence they can't be stationary states. Which means that they are not eigenstates of the 2 level Hamiltonian of the problem. Assuming that the energy of these 2 lowest energy levels is $$E_0$$ and the off diagonal term is $$A$$ (please see the link to Feynman lectures for more details about the notation), then, the actual stationary states, obtained by diagonalizing the hamiltonian (and which explicitly manifest the symmetry of the physical interaction) will have energies of $$E_0 ± A$$ . But this implies that there is an even lower in energy level, $$E_0 − A$$ , than the 2 equal ones. Yet in the plot I mentioned above there is no lower energy than the 2 equal ones. Can someone explain to me what is wrong with my reasoning. Why don't we have that lower energy level? Thank you!

• "Yet in the plot I mentioned above there is no lower energy than the 2 equal ones." Potential energy or total energy? – Chiral Anomaly May 31 '20 at 20:38
• I don't think it's useful to think about SSB from a QM perspective. If I am not mistaken SSB is impossible in QM. – fewfew4 May 31 '20 at 21:35

Like you mentioned, the eigenstates are the ones that respect symmetry. Thus the two SSB states are a linear combination of them. This means that the energy expectation in the SSB states must naturally be greater than or equal to (in case of degeneracy) the energy eigenvalues. The lowest energy level is $$E_0-A$$.