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I am wondering that is there any restrictions for the truncation in DMRG algorithm. Currently I am using DMRG to calculate ground state energy per site of a many-body system described by on-site potential term only($H_{i} = -\mu c^{\dagger}_{i}c_{i}$). However, I got a diverging answer. This is not physical since the ground state energy per site is an intensive quantity, which is independent of the size of the system. Therefore, I re-examine my code and I found something strange there. I print out the eigenvalues of the reduced density matrix of system: \begin{equation} \text{Eigenvalues of $\rho_{sys}$} = [1.000, 6.988\text{E-31}, 8.362\text{E-32}, 6.398\text{E-33},...,0.000] \end{equation} Despite $\text{tr}(\rho_{sys}) \simeq 1$, I think that this set of eigenvalues is strange. The first element is already equal 1 and the difference between the first and second element is $10^{-30}$. This means that only the first eigenstate is most probable and other states are not. Therefore, we can truncate other states out. When I set the number of maximum states for truncation to some number(e.g. 16), I got a diverging answer. This means that I have set a wrong condition for truncation. Therefore, based on the above discussion, what truncation condition should I use if the eigenvalues of $\rho_{sys}$ like that? Besides, is there any restriction for the truncation in DMRG which I have not discussed above?

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  • $\begingroup$ Thank you for your answer, @NorbertSchuch. Besides, I want to ask a question regarding the entanglement of ground state. Suppose I use DMRG to find the ground state of such system, how do I know whether the ground state I get is an entangled state or product state? $\endgroup$ – Ricky Pang Apr 25 at 12:15
  • $\begingroup$ If you get the "strange" ES above, it is a product, otherwise entangled. $\endgroup$ – Norbert Schuch Apr 25 at 12:23
  • $\begingroup$ @NorbertSchuch, thank you for your answer. I am grateful for your help. $\endgroup$ – Ricky Pang Apr 25 at 12:44
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You target state is a product state, so the ES looks fine. However, this is not a prime application for DMRG exactly for that reason. What goes wrong depends on the details of the implementation you are using, but usually inverses of the ES appear in DMRG (e.g. for canonical forms), which become ill-conditioned. If implemented properly (e.g. discard all eigenvalues below some threshold) this should, however, not lead to instabilities of that kind.

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  • $\begingroup$ I found an argument on why the RDM eigenvalues spectrum $[1, 0,0,...]$ implies unentangled state. This argument is from note, written by Ulrich Schollwöck. On page 16.7, he says that this eigenvalues spectrum implies unentangled state since its von-Neumann entropy is zero, meaning that the no entanglement. $\endgroup$ – Ricky Pang Apr 26 at 9:43

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