# Can a mixed state be conformal?

A vacuum in conformal theory can be defined as a state where correlation functions of primary operators take a particularly simple form. It is also known that entanglement entropy of an interval in conformal vacuum has a universal logarithmic form. Can a mixed state with similar properties exist? In particular, can one find an example of a density matrix $$\rho$$ for which some basis of local fields $$V_\Delta$$ has the usual form of correlators? For example for two-point functions this should be something like $$Tr\left(\rho V_{\Delta}(z_1)V_{\Delta}(z_2)\right)\propto \frac1{(z_1-z_2)^{2\Delta}}$$ and similarly for 3-point functions etc. Potentially the definitions of the primary operators and/or their dimensions might be different from the original theory. If such a state exists, can it be diagnosed by entanglement entropy of an interval?

Any hints and pointers are welcome!

• 1. An obvious candidate would be to take a (uniform) mixture of two or more pure vacuum states. Have you tried looking at that? 2. What does the phrase "entanglement entropy of an interval" mean? An interval of what? Commented Jul 28, 2021 at 22:55
• @ACuriousMind 1. I do not know an example where the conformal vacuum is not unique. I would be interested to know if one exists, but I had a more generic situation (non-degenerate vacuum) in mind, 2. Entanglement entropy of an interval is a standard setting in 1+1 QFT. One chooses an interval on a spatial axis and traces out degrees of freedom outside. The resulting density matrix has von Neumnamm entropy proportional to the logarithm of the interval length. Commented Jul 28, 2021 at 23:15
• @WeatherReport I haven't thought about this carefully, but it seems that if you want your density matrix to be trace-class (so that it makes sense to say that it is properly normalized), then this might be tricky. This is because trace-class operators are in particular Hilbert-Schmidt, and so they live in the tensor product $H^*\otimes H$, where $H$ is the Hilbert space. Now, you know that $H=\oplus_i D_i$ where $D_i$ are the parabolic Verma modules of local operators. If I am not mistaken, there are no singlets in $D_i^*\otimes D_j$, even for $i=j$, although I can't give you a reference. Commented Sep 13, 2021 at 14:36
• (This is because $D_i$ are infinite-dimensional; for finite-dimensional irreps there's always a singlet in $D_i^*\otimes D_i$. The logic is that the only singlet in $D_i^*\otimes D_i$ would be the inner product matrix, but it has infinite trace, because in an orthonormal basis it is just the identity matrix.) Commented Sep 13, 2021 at 14:37
• @PeterKravchuk Wouldn't you say that thermal state is actually an example, since it is conformally equivalent to a vacuum on a cylinder physics.stackexchange.com/a/234083/8910 ? Commented Sep 21, 2021 at 6:17