# Does Fabri-Picasso theorem imply non-conservation of charge?

The vacuum expectation value of the square of the Noether charge operator $$\langle 0|\hat{Q}^2|0\rangle=\int_{\rm all space} d^3\textbf{x}\langle0|\hat{j}_0(0)\hat{Q}|0\rangle$$ diverges in case of a spontaneously broken symmetry i.e., when $$\hat{Q}|0\rangle=|0^\prime\rangle\neq 0$$. This argument relies on the assumptions that the vacuum $$|0\rangle$$ is invariant under spacetime translations and the integration is performed over whole space. Since the norm of the state $$\hat{Q}|0\rangle$$ is infinite, this state doesn't belong to the same Hilbert space as $$|0\rangle$$.

Question Assuming FP theorem does hold, what do we infer from that? Should we infer that we no longer have charge conservation or should we instead infer that this theorem is not the correct way to answer whether charge conserved?

• Apr 3, 2020 at 16:18
• As written, the integrand is not a function of $\bf{x}$.
– fqq
Apr 3, 2020 at 17:15
• @fqq ?? How does that answer my questions? In particular, question 2?
– SRS
Apr 3, 2020 at 17:18
• It does not. I didn't post an answer, but a comment pointing out that there is something wrong with the expression, I guess it should be $j_0(x)Q$, and is the comma supposed to be there?
– fqq
Apr 3, 2020 at 17:22
• Thanks. The comma is a typo. Translational invariance of the vacuum is used to eliminate the x dependence of $j_0$.
– SRS
Apr 3, 2020 at 17:24

Consider a state with with a unit charge that is is spread uniformly over a volume $$V$$. The total charge is unity so, $$\int_V d x j_0(x) =1.$$ but as $$V$$ becomes large $$j_0(x)=1/V$$ tends to zero. Thus $$1=\lim_{V\to \infty} \left[\int_V j_0(x) dx\right]\ne \int_{allspace} dx \left[\lim_{V\to \infty} j_0(x)\right].$$ I think this is trivial and has nothing to do with the total charge being ill defined.
Similarly I think that the Fabri Picasso theorem, while equally true, is an inappropriate taking of limits: in a translationally invariant state (not necessarily a ground state) we can have $$Q=1$$ while $$\langle j_0 \rangle=0$$, so $$Q$$ is not given by the integral of $$\langle j_0 \rangle$$.
A better way to understand the role of $$Q$$ in the degenerate ground state in SSB is to consider a bead of mass $$M$$ sliding on a ring. The Noether "charge" generator is the angular momentum $$L$$ and the energy states $$\langle \theta|l\rangle =e^{il\theta}$$ have energies $$\propto l^2/2M$$. As $$M\to \infty$$ they become degenerate and the position eigenstates $$|\theta_0\rangle=\sum_{l=-\infty}^{\infty} e^{il \theta}|l\rangle$$ are an equally good set of degenerate ground states that are sent into each other by the action of $$L$$. It is now reasonable to ask $$L\propto M\dot \theta$$ ill defined in the $$M \to \infty$$ limit? I think this is the same question that you are asking.