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  Derivation of Casher-Banks relation

+ 2 like - 0 dislike

Consider the fermion condensate in gauge theory:
\langle \bar{f}f\rangle = -i\int d^{4}x\text{Tr}[S(x,x)] 

S(x,y) = \langle x |D^{-1}(x,y)|y\rangle

is the fermion propagator and $D$ is the Dirac operator including the fermion mass $m$. Using the spectral representation of the Dirac operator,
D(x,y) = \sum_{\lambda}\frac{\psi(x)\psi^{\dagger}(y)}{\lambda + im},

one finds

$$\langle \bar{f}f\rangle = \sum_{\lambda}\frac{1}{\lambda + im}$$

How to obtain from this expression the Casher-Banks relation $\langle \bar{f}f\rangle = \pi \rho(\lambda = 0)$, where $\rho$ is the spectral density?

asked Jul 25, 2018 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Jul 25, 2018 by NAME_XXX

1 Answer

+ 2 like - 0 dislike

You have to use the limit result which says : 

$ \text{Lim}_{m \to 0} \frac{1}{m + i \lambda} = \pi \delta(\lambda) - i \mathcal{P} (\frac{1}{\lambda}) $, 

here $\mathcal{P}$ is the principal value 

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

answered Jul 27, 2018 by aperyconstant (40 points) [ no revision ]

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