Topological susceptibility

Question

In QCD we have strong CP violation (and hence a $\theta$-dependence of the theory) only if the topological susceptibility of the vacuum is nonzero:

$$\langle F\tilde{F},F\tilde{F}\rangle_{q \rightarrow 0} =\text{const} \neq 0,$$

where $F$ is the gluon field strength matrix, $\tilde{F}$ is its dual, and $q$ is the momentum.

My first question: What do the notions "topological" and "susceptibility" mean in this context? I know susceptibility only from the context of electromagnetism. And what does it have to do with topology?

My second question: We know that $F\tilde{F}=dC$, where $C$ is the Chern-Simons three-form of QCD, the gauge field that generates $F\tilde{F}$. Why do we have

$$\langle C,C\rangle_{q \rightarrow 0} = \frac{1}{q^2}?$$

Answer 1

Here is an answer that assumes the question did not require a great deal of care and precision in the response.

1) ``Topology'' is used because $F \tilde F$ integrated over a manifold is a topological quantity. (One of my preferred discussions of this integral is in the second volume of Weinberg's series on QFT.)

2) A susceptibility $\chi$ tells us the response of a system to some particular perturbation, for example how the polarization $P$ changes in response to an electric $E$ field, $P = \chi E$. In this case, we are interested in how $F \tilde F$ changes in response to changing $\theta$. So we think of the $\int \theta F \tilde F dx$ term in the action just like we think of the usual source-operator coupling, where derivatives of the path integral with respect to $\theta$ generate correlation functions of $F \tilde F$. A single derivative gives the one-point function, a double derivative the two point function and so on. So, in a perhaps too pedestrian way, I can write $$ \delta F \tilde F = {\delta \langle F \tilde F \rangle\over \delta \theta} \delta \theta $$ where the ``derivative'' on the right hand side is the two-point function that is being a called a susceptibility. One can (and probably should) be much more careful and precise here.

3) I don't know how to verify that the numerator is 1, but the fact that there is $1/q^2$ scaling in the $CC$ correlator seems very natural in Fourier space where acting with $d$ is like multiplying by $q$. One just moves the $q$'s to the other side.

Answer 2

Just one comment on your second question.

Such structure of correlator, $$ \lim_{q\to 0}\int d^{4}xe^{iqx}\langle 0| C^{\mu}C^{\nu} |0\rangle = \text{sign}(\kappa (0))|\kappa (0)|\frac{g_{\mu \nu}}{q^2}, $$ requires the presence of pole which defines some state which couples directly to Chern class $C^{\mu}$. Immediately there are following consequences: since $C_{\mu}$ isn't gauge invariant, then the state is certainly unphysical, i.e., it is a ghost; since it has structure of the mass term for the gluon, it modifies the pole structure of the gluon propagator, $$ \lim_{p \to 0} D_{\mu \nu}(p) \sim -\frac{g_{\mu \nu}p^{2}}{-\text{sign}(\kappa (0))|\kappa (0)|} $$ Finally, its sign determines whether the gluon is confined. The last statement is obvious if we will look on the gluon propagator: $$ D_{\mu \nu}^{ab}(p) = -\frac{g_{\mu \nu} - (1 - \epsilon)\frac{p_{\mu}p_{\nu}}{p^{2}}}{p^2 - \frac{\text{sign}(\kappa (0))\kappa (0)}{p^{2}}} $$ You see that for the negative sign there is no pole in the propagator, i.e. the gluon can't be observed.