3

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)?

SRS
  • 26,333
  • 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. – TwoBs Oct 23 '18 at 15:47
  • @TwoBs So FP theorem is of no use? – SRS Oct 23 '18 at 17:12
  • 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 – TwoBs Oct 23 '18 at 17:18
  • 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 $p=0$ – TwoBs Oct 23 '18 at 20:22
  • 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 $k\rightarrow p$. – TwoBs Oct 24 '18 at 06:13
  • So charge conservation continues to hold when a symmetry is spontaneously broken? @TwoBs – SRS May 06 '19 at 03:55
  • not quite, the charge is simply not well defined quantity that one can associate to the states of the system. However, one can still talk of the current conservation which is unaffected by SSB. Another way of phrasing this is noticing that the commutator of the charge operators remains instead well defined, and so one can assign a charge to fields, in contract to the ill-defined charge of a state. – TwoBs May 06 '19 at 10:01
  • @TwoBs It seems that there is a difference between your last comment and Lorenz Mayer's answer which claims " Charge is conserved regardless of the symmetry being broken or being not broken." Now I'm thoroughly flummoxed! – SRS May 06 '19 at 14:24
  • sorry @SRS but I don't see any contradiction, actually it seems to me we agree on everything substantial. Current conservation still hold, the commutator of the charge operator is well-defined so that one can assign charges to operators (and hence have charge preserving dynamics, with a slight abuse of language, just by building hamiltonian that it's neutral out of charged fields), but one can't assign a charge to a state (because Q formally creates infinitely many soft quanta of Goldstone bosons). – TwoBs May 06 '19 at 14:44

1 Answers1

3

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.

Lorenz Mayer
  • 1,459
  • "And in particular, these still implement the symmetry." Can you expand your last sentence a bit more? @LorenzMayer – SRS Oct 23 '18 at 17:46
  • Does this help: https://physics.stackexchange.com/questions/137499/connection-between-conserved-charge-and-the-generator-of-a-symmetry ? – Lorenz Mayer Oct 23 '18 at 17:58
  • What can we say about the commutator $[H,Q]$ when the symmetry is spontaneously broken? – SRS Oct 24 '18 at 19:21
  • This commutator is zero. (Consider the definition of a spontaneously broken symmetry) – Lorenz Mayer Oct 25 '18 at 09:59
  • So charge conservation still applies when the symmetry is spontaneously broken? @LorenzMayer – SRS May 06 '19 at 03:52
  • The second sentence of my answer is: "Charge is conserved regardless of the symmetry being broken or being not broken". It is also the content of the quote by Weinberg in your question. The statement of the FP theorem is that the charge contained in an infinite volume is infinite. This does not contradict the conservation of charge. – Lorenz Mayer May 06 '19 at 06:51
  • Weinberg's statement is about current conservation, not charge! Please, also see the last comment by TwoBs @LorenzMayer It seems that the issue is unclear in the community.(?) – SRS May 06 '19 at 14:26
  • Charge Conservation = Current Conservation. The latter is simply a covariant and local way of expressing the fact that there are conserved quantities in the system. The comment of TwoBs is not in contradiction to that, as he just says that there is no Q operator defined as in, for example, my answer. The problem is that the charge of all states is infinite, not that it is not conserved by the dynamics. – Lorenz Mayer May 06 '19 at 14:32