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3

There are (at least) three options: We can retain the field's zero-mode $\phi_0\equiv \int d^3x\ \phi(x)$ as an observable and retain the term $\Pi_0^2$ in the Hamiltonian. Then the model does not have a vacuum state at all. The spectrum of the Hamiltonian does have a finite lower bound, as required by the principles of QFT, but technically the Hilbert ...

2

@Chiral Anomaly has already posted a very nice answer which I have marked correct. I just wanted to add a few more words which connect the discrete toy model provided to the model in the original question more explicitly. Essentially, it boils down to treating an IR cutoff carefully. The solution to the functional Schrodinger equation above is given by $$\... 1 Whenever a theory has a local gauge symmetry, it automatically also has a global symmetry. After all you can just look at the constant gauge transformations. So the answer to this part is yes, but the last comment (i.e. "and it has global symmetry G_{SM}") is redundant. In theories with spontaneous symmetry breaking, we start with a Lagrangian ... 1 As you might already know, (Gutzwiller) mean-field theory becomes exact in the limit z \to \infty (z is the coordination number). So, for a model like$$\frac{1}{N}\hat{H}=J_{4}\left[\frac{1}{N}\sum_{i=1}^{N}\sigma_{i}^{z}\right]^4+J_{2}\left[\frac{1}{N}\sum_{i=1}^{N}\sigma_{i}^{z}\right]^2+h\left[\frac{1}{N}\sum_{i=1}^{N}\sigma_{i}^{x}\right] (defined ...

1

The theorem is quite robust to variations in the interaction - it only requires that it does not violate local continuity. The biggest physical limitation is that one needs A LOT of infra-red fluctuations to destabilize the ordered state, which are cut off by the size of the sample in any real setting. In 2D, the effect of the theorem is only logarithmic in ...

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