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The Fabri-Picasso (FP) theorem states that if a symmetry is spontaneously broken the corresponding conserved charge operator $Q$ does not exist in the Hilbert space. The state $Q|0\rangle$ will have infinite norm!

Assuming situations where FP theorem holds, what conclusion should one draw from this theorem? Will it be correct to say that the charge conservation fails? But if that is true it will be in contradiction with Weinberg's statement here

The conservation of a current is usually a symptom of some symmetry of the underlying theory and holds whether or not the symmetry is spontaneously broken.

I am thoroughly confused. What is FP theorem really trying to tell (when it holds, of course)?

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  • $\begingroup$ It just expresses in yet another form that there are 1-particle Goldstone boson generated by the current acting on the vacuum, and since 1-particle states are not normalizable so it will be the would-be state obtained by Q ( which would be the zero momentum limit of one bison state $\langle p\rightarrow0 | p\rightarrow 0\rangle$). So, FP theorem is just a red herring. $\endgroup$
    – TwoBs
    Oct 23, 2018 at 15:47
  • $\begingroup$ @TwoBs So FP theorem is of no use? $\endgroup$
    – SRS
    Oct 23, 2018 at 17:12
  • $\begingroup$ I think it conveys nothing more than something one already knows, that the spontaneously broken current generate massless one particle states when acting on the vacuum, these states being non-normalizable as for any other plane wave. As I said, I think it’s a red herring $\endgroup$
    – TwoBs
    Oct 23, 2018 at 17:18
  • $\begingroup$ Think of the GB's decay constant $f$ defined by the non-vanishing broken-current matrix element between the vacuum and one-GB state, namely $\langle 0| J^\mu(x) |\pi(p)\rangle= f p^\mu e^{-ipx}$, with $f\neq 0$. It says that $J^\mu$ generates, among others, the one-particle state $|\pi(p)\rangle$. That's why any broken current starts linearly in the GB's field, $J^\mu=-f \partial^\mu\pi+\ldots$. The charge $Q$ can thus be written as the soft limit $p\rightarrow 0$ of $\int d^4x i f p^0 e^{ipx} \pi(x)+\ldots$ which acting on the vacuum generate a non-normalizable plane wave state at $p=0$ $\endgroup$
    – TwoBs
    Oct 23, 2018 at 20:22
  • $\begingroup$ adding more details from previous comment, we have that $J^0 \propto f \int d^3 x e^{i\vec{p}\vec{x}} p^0 \pi(x)$ with $p\rightarrow 0$ (and $p^0=|\vec{p}|$), and therefore the norm of tentative state that would be generated by $Q$ acting on the vacuum results $||Q|0 \rangle ||^2=\lim_{k,p\rightarrow 0} f^2 \langle k| p\rangle$ which is IR divergent but just as any other state of definite momentum: $\langle k| p \rangle=(2\pi)^3 2p^0 \delta^{3}(k-p)$ which is volume diverging as $k\rightarrow p$. $\endgroup$
    – TwoBs
    Oct 24, 2018 at 6:13

1 Answer 1

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These two statements are not contradicting. Charge is conserved regardless of the symmetry being broken or being not broken.

The usual wisdom goes like this: assume you have a conserved current $j^\mu(x)$ with

$$ \partial_\mu j^\mu = 0 \ . $$

Now pick a constant-time ball $\Sigma$ and define

$$ Q = \int_\Sigma j^0(x) d^3 x \ . $$

This exists if the charge density decays fast enough. This is a conserved charge since:

$$ \frac{d Q}{d t} = \int_\Sigma \partial_0 j^0(x) d^3 x = - \int_\Sigma \nabla \cdot j(x) d^3 x = - \int_{\partial \Sigma} j(x) \cdot d \sigma $$

Where in the last equality we used that the currents decay at infinity.

Now this is classical reasoning. On the quantum level, the current becomes an operator, and all the statements about the fall-off of functions at infinity are statements about the states we wish to apply these operators on. In particular, it suffices to look at the vacuum, since we usually only consider states which differ from the vacuum in a finite region, so once the vacuum is fine, all excited states will be fine as well.

But in the case of spontaneous symmetry breaking, it is just not true that

$$ j^0(x)|0\rangle $$

decays at infinity, hence the integral defining the charge does not exist. Note however that while $Q$ doesn't make sense in itself, one may still give meaning to commutators of $Q$ with field operators $\phi(x)$:

$$ [ Q, \phi(x) ] $$

And in particular, these still implement the symmetry.

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  • $\begingroup$ "And in particular, these still implement the symmetry." Can you expand your last sentence a bit more? @LorenzMayer $\endgroup$
    – SRS
    Oct 23, 2018 at 17:46
  • $\begingroup$ Does this help: physics.stackexchange.com/questions/137499/… ? $\endgroup$ Oct 23, 2018 at 17:58
  • $\begingroup$ What can we say about the commutator $[H,Q]$ when the symmetry is spontaneously broken? $\endgroup$
    – SRS
    Oct 24, 2018 at 19:21
  • $\begingroup$ This commutator is zero. (Consider the definition of a spontaneously broken symmetry) $\endgroup$ Oct 25, 2018 at 9:59
  • $\begingroup$ So charge conservation still applies when the symmetry is spontaneously broken? @LorenzMayer $\endgroup$
    – SRS
    May 6, 2019 at 3:52

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