Derivation of Casher-Banks relation

Consider two-point function $\langle \bar{\psi}\psi\rangle$ in a model with massive fermions $\psi$ and gauge field: $$\langle \bar{\psi}\psi\rangle =\frac{1}{V}\sum_{n} \frac{1}{\lambda_{n} +im},$$ where $\lambda_{n}$ is the eigenvalue of the corresponding Dirac operator (with $\lambda_{0} = 0$). The expression can be rewritten in the form $$\tag 1 \langle \bar{\psi}\psi\rangle = \int d\nu \rho(\nu)\frac{1}{\nu + im},$$ where $\rho(\nu) = \frac{1}{V}\sum_{n}\delta(\nu - \lambda_{n})$.

On p. 56 of the paper is stated that by applying thermodynamic limit and second the limit $m \to 0$ to $(1)$, we'll obtain zero; this sounds reasonable since we have then finite system where spontaneous symmetry breaking corresponding to non-zero order parameter can't occur. However, some small (although general) detain is unclear for me, and I don't understand why numerically it is so.

Can anyone please explain it for me?

• The other way around: You get zero if you take $m\to 0$ first, before $V\to\infty$. – Thomas Jul 22 '18 at 15:15
• @Thomas: yes, I meaned exactly this case. But I still don't understand why it is zero. – Name YYY Jul 22 '18 at 15:17

Consider a theory with chiral symmetry breaking condensate $\Sigma$. If we study the spectrum of the euclidean Dirac operator in a finite volume $V$, then on average the first non-zero eigenvalue is located at $\lambda\sim 1/(V\Sigma)$, and a finite density of eigenvalues at $\lambda=0$ only emerges in the thermodynamic limit $V\to\infty$. This has been studied in great detail in random matrix theory, see the survey article.

Postscript: Consider a euclidean volume $V$ with $N_+$ instantons and $N_-$ anti-instantons. The topological charge is $Q=N_+-N_-$, and by the index theorem the number of exact zero modes is $Q$. The remaining $N_++N_--|Q|$ instanton zero modes get lifted and become small eigenvalues. Because of the fermion determinant the probability of encountering $Q$ eaxct zero modes scales as $m^{N_f|Q|}$, so the contribution to the spectral density is at most $\rho(\lambda)\sim m^{N_F}\delta(\lambda)$. This means that for $N_f>1$ the chiral condensate is not related to exact zero modes, but to the quasi-zero modes.

• Thank you! But I don't clearly understand two things. First, why the first non-zero mode is determined by the condensate and not only by the volume. Second, why the VEV is non-zero if the zero mode exist. – Name YYY Jul 24 '18 at 7:07
• Well, you can read the article and it will explain the details. Roughly, regarding question 1, a non-zero condensate means that there is an accumulation of small eigenvalues, and this is why the first non-zero eigenvalue scales as $1/\Sigma$. I am not sure what you are asking in the second question. – Thomas Jul 25 '18 at 3:24
• There is a simple observation you should keep in mind: Consider a discretized system (Dirac operator in latticed QCD at finite volume). Then any configuration, and indeed any ensemble, has a smallest eigenvalue. Then, if you make m smaller than that, the condensate will go to zero. This means that chiral symmetry breaking requires $V\to \infty$ so that the sum over eigenvalues becomes a continuous integral, and the density of eigenvalues at zero can in principle be non-zero. – Thomas Jul 25 '18 at 3:33
• Thank you again! However, on p. 56 of the paper mentioned in my question there is a statement that "...We have excluded zero modes since they do not contribute to the quark condensate in the limit $m\to 0$ (for $N_{f}>1$)...". I don't understand how the zero mode can simultaneously determine the condensate and do not contribute into it. – Name YYY Jul 25 '18 at 17:44
• @NameYYY ... added a postscript – Thomas Jul 25 '18 at 23:33

I am not completely sure about what you are asking, but may be, this is useful --

Using the result,

$$\displaystyle\lim_{m \to 0} \frac{1}{m + i \lambda} = \pi \delta(\lambda) - i \mathcal{P} (\frac{1}{\lambda})$$

where $\mathcal{P}$ is the principal value, we can show that,

$$\langle \overline{\psi} \psi \rangle = \int_{-\infty}^{+\infty} d\lambda ~ \rho(\lambda) \frac{1}{m + i \lambda} = \pi \rho(0)$$

• And what will be in the case when the limits are finite and $m$ is set to zero? – Name YYY Jul 27 '18 at 14:08
• If the finite limits are symmetric around zero (including it) , then I think this still holds. – R.G.J Jul 27 '18 at 15:01