To obtain finite temperature correlators in a theory that is conformally invariant it is common to map the Euclidean time into the circle via $\tau\to f(\tau)=\frac{\pi}{\beta}\tan\left(\frac{\pi\tau}{\beta}\right)$. For example, in 1D this maps

$$G(\tau)=\frac{b}{|\tau|^{2\Delta}} \to \mathcal{G}_{\beta}(\tau)=b\left[\frac{\pi}{\sin\left(\frac{\pi\tau}{\beta}\right)}\right]^{2\Delta}$$

On the other hand, in many-body theory it is common to obtain finite temperature propagators from zero temperature euclidean ones by summing over charges,


Where for concreteness here we considered the Fermionic case. Supposing these two procedures agree for a conformal theory, we would have the identity

$$\left[\frac{\pi}{\sin(\pi\theta)}\right]^{2\Delta} = \sum\limits_{n\in\mathbb{Z}}\frac{(-1)^n}{\left|n+\theta\right|^{2\Delta}}$$

Where for convenience we introduced $\theta=\tau/\beta\in[0,1]$. I know that in particular this identity is true for $\Delta=1$, but cannot show for general $\Delta$. In particular I am interested in the cases $0<\Delta<1/2$ and $\Delta>1$.

Could someone either point me towards a reference where this is discussed or help establishing/dismissing this identity?

  • $\begingroup$ This identity looks wrong. On the other hand, by looking at poles and residues, you can show $\frac{\pi}{\sin\pi\theta} = \sum_{n\in\mathbb{Z}}\frac{(-1)^n}{n+\theta}$. $\endgroup$ – Sylvain Ribault Nov 5 '16 at 20:04
  • $\begingroup$ Sylvain, I agree with you. Indeed one can check in Mathematica by plotting one against the other that although they behave similarly they do not agree for $0<\Delta<1/2$. The error decreases as you approach $\Delta=1/2$ when they coincide as you mentioned. However I would like to understand conceptually why this is the case. $\endgroup$ – kurtachovo Nov 7 '16 at 11:33

Consider an (Euclidean) fermionic correlator satisfying the property above:

\begin{equation} \tag{1} \label{prop} \mathcal{G}_{\beta}(\tau)=\sum_{n\in\mathbb{Z}}(-1)^n G(\tau+n\beta). \end{equation}

Taking the Fourier transform on both sides, \begin{align*} \mathcal{G}_{\beta}(i\omega_n)&=\int_{0}^{\beta}d\tau ~e^{i\omega_n\tau}\mathcal{G}_\beta(\tau) = \int_{0}^{\beta}d\tau ~e^{i\omega_n\tau}\left[\sum_{n\in\mathbb{Z}}(-1)^n G(\tau+n\beta)\right]\\ &=\sum_{n\in\mathbb{Z}}(-1)^n \int_{0}^{\beta}d\tau ~e^{i\omega_n\tau}G(\tau+n\beta)=\sum_{m\in\mathbb{Z}}(-1)^m \int_{m\beta}^{(m+1)\beta}du~e^{i\omega_n(u-m\beta)}G(u)\\ &=\sum_{m\in\mathbb{Z}}\int_{m\beta}^{(m+1)\beta}du~e^{i\omega_nu}G(u)=\int_{-\infty}^{\infty}du~e^{i\omega_nu}G(u)=G(\omega_n). \end{align*} Where we have changed variables $u=\tau+n\beta$ and used that for fermionic Matsubara frequencies $e^{i\omega_n} = -1$. Therefore we conclude that a correlator satisfying Eq. \ref{prop} implies that the finite temperature solution in frequency space is simply the zero temperature solution evaluated at the Matsubara frequency $\omega=\omega_n$. Since we have the spectral decomposition of the analytically continued correlator $$ G(z) = \int\frac{\rho(\omega)}{\omega-z} $$ This in particular implies that the spectral function $\rho(\omega)$ of correlators satisfying Eq. \ref{prop} must be temperature independent.

This solves indirectly solves the problem since one can check that the finite temperature conformal correlator

\begin{align*} \mathcal{G}_{\beta}(\tau)=b\left[\frac{\pi}{\beta \sin\left(\frac{\pi\tau}{\beta}\right)} \right]^{2\Delta}, && 0<\Delta<\frac{1}{2} \end{align*} Has spectral function \begin{align*} \rho(\omega)\propto \beta^{1-2\Delta} \cosh\left(\frac{\beta\omega}{2}\right)\Gamma\left(\Delta+i \frac{\beta\omega}{2\pi}\right)\Gamma\left(\Delta-i \frac{\beta\omega}{2\pi}\right) \end{align*} Which is explicitly temperature dependent.

The confusion was to believe that Eq. \ref{prop} holds for any correlator, since in the literature this assumption is almost never made clear. It definitively apply for free theories, but almost all interesting systems have temperature dependent spectral functions.

Would be interesting to bound the theories for which this also holds, and to understand why in the case of conformal correlators for which it does not apply it provides a 'good' approximation.


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