Hi Chengshu,

This is kind of a tricky question to briefly answer on a message board and I'd really need to refer you to an article such as Schollwock's review on DMRG and MPS to give you the full answer. But the basic answer is that if you properly gauge an MPS, then the MPS tensors themselves already give you the eigenstates of the reduced density matrix of a block of sites starting from either the left or right edge of the system, up to the site that is the gauge "center". (You can see why this is hard to explain in writing - it's really something that's easier to draw with diagrams.)

On the other hand if the reduced density matrix you are asking about (you didn't specify) involves tracing out a block of sites that doesn't start from the left or right edge of the system (as defined by the MPS path) then it's highly non-trivial to get the reduced density matrix eigenstates. An example of such a hard case would be a 1d system where the r.d.m. you want is the one obtained by tracing out every even numbered site. In that case you can technically get information about the reduced density matrix by using a sampling procedure but it would be very hard to get the eigenstates of such an r.d.m.

Hope that helps you to understand the issue/question better.

Miles

Naïvely I expect the same thing happens for the reduced density matrix. Here after we "gauge" the MPS to, say, site i, then we automatically have something like \rho = U^{\dagger} D U, where D is diagonal and "right orthogonality" is used. So I guess all we need to do then is to extract the "column vector" from the U. Is that the correct way of doing this? Thanks!