Reading "From Linear SUSY to Constrained Superfields" by Komargodski and Seiberg, I got a bit confused regarding the existence of the conserved charges in a theory with spontaneous symmetry breaking (SSB) of a global symmetry:

More precisely, in the second-to-last paragraph on page 1 we have

"When a global symmetry is spontaneously broken, the corresponding conserved charge does not exist because its correlation functions are IR divergent. However the conserved current and even the commutators with the conserved charge do exist."

I know that in the case of global SSB we have $Q|0\rangle\neq0$ for the conserved charge $Q$. However, I don't have any insight about the correlation functions. Could somehow $Q|0\rangle\neq0$ imply something like $||Q|0\rangle||=\infty$ or $\langle Q\rangle\rightarrow\infty$? And how could one see that?


1 Answer 1


This is called the Fabri-Picasso theorem. Their argument requires both the vacuum and the charge $Q$ to be translationally invariant: $P |0\rangle = 0$, and $[P, Q] = 0$.

The argument goes as follows: Since the charge originates from a symmetry, then according to Noether's theorem:

$$Q = \int d^3x j_0(x)$$

Consider the correlation function of the charge with itself:

\begin{align} \langle 0| QQ |0\rangle& = \int d^3x \langle0|j_0(x) Q|0\rangle \\ & =\int d^3x \langle0|e^{iPx} j_0(0) e^{-iPx} Q |0\rangle \\ & =\int d^3x\langle0| e^{iPx} j_0(0) e^{-iPx} Q e^{iPx} e^{-iPx}|0\rangle\\ & =\int d^3x \langle0| j_0(0) Q |0\rangle \end{align} The integrand in the r.h.s. does not depend on the position, therefore its value is proportional to the total space volume, thus $$||Q|0\rangle||^2 = \infty$$ Thus the operator $Q$ does not exist in the Hilbert space unless $Q|0\rangle = 0$.

However, the commutators of $Q$ with the fields for example exist because by Noether's theorem, they generate the symmetry, in other words the right hand sides of:

$$[Q, \phi] = \delta \phi$$

exist, since they are the symmetry transformed fields.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.