# Truncation Problem of Density Matrix Renormalisation Group (DMRG)

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?

• 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? – Ricky Pang Apr 25 at 12:15
• If you get the "strange" ES above, it is a product, otherwise entangled. – Norbert Schuch Apr 25 at 12:23
• @NorbertSchuch, thank you for your answer. I am grateful for your help. – Ricky Pang Apr 25 at 12:44

• 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. – Ricky Pang Apr 26 at 9:43