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The traditional, non Matrix Product State, formulation of the Density Matrix Re-normalization Group (DMRG) algorithm can be coded in python. Such a code can be found in the following link:

https://github.com/simple-dmrg/simple-dmrg/blob/master/simple_dmrg_02_finite_system.py

Here is the algorithm in a nutshell. Suppose I have a system with Hilbert space dimension $m\_block$. When we enlarge the system by adding the next lattice site by using the kronecker product (the Hilbert space of a single site is 2 dimensional), the resulting enlarged system has a Hilbert space of dimension $2 m\_block$ and the Hamiltonian of the enlarged system is a $2 m\_block \times 2 m\_block = m\_sys\_enl \times m\_sys\_enl$ matrix. One can think of the environment as a clone of the system. If the original Hilbert space of the environment has dimension $m\_env$, then if we enlarge the environment by one site, the resulting enlarged environment has a Hilbert space of dimension $2 m\_env$ and the Hamiltonian of the enlarged environment is a $2 m\_env \times 2 m\_env = m\_env\_enl \times m\_env\_enl$ matrix. We can combine the Hamiltonians of the enlarged system and the enlarged environment by using the kronecker product, and the Hamiltonian of this super block is a $ m\_sys\_enl\cdot m\_env\_enl \times m\_sys\_enl \cdot m\_env\_enl$ matrix. The ground state of the superblock Hamiltonian is a $ m\_sys\_enl\cdot m\_env\_enl \times 1$ vector, which can be reshaped into a $ m\_sys\_enl \times m\_env\_enl $ vector. The resulting density matrix is a $m\_sys\_enl \times m\_sys\_enl $ matrix. Choosing the $m$ largest eigenvalues of the density matrix and the corresponding eigenvectors $| w_1 \rangle ... | w_m \rangle$, we can create a $m\_sys\_enl \times m$ matrix $O$. Hence we can compress the enlarged system's (not the super block's) Hamiltonian $H_{\text{sys}}$ by $O^T H_{\text{sys}} O$, which is now an $m \times m$ matrix. Hence the enlarged system's Hilbert's space dimension has been reduced from $m\_sys\_enl$ to $m$.

Question: However, how to we know that the compressed Hilbert space contains the relevant physics? I thought that the states in the Hilbert space with high entanglement entropy (which can be computed from the Von Neumann entropy formula) contained the relevant physics. Nothing in the algorithm seems to keep highly entangled states and throw away the less entangled states.

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I thought that the states in the Hilbert space with high entanglement entropy contained the relevant physics ... Nothing in the algorithm seems to keep highly entangled states and throw away the less entangled states.

Quite the opposite. Most of the relevant ground states do not require full entanglement entropy like random states in Hilbert space. Ground states of local 1D Hamiltonians with an energy gap to the first excited state follow an entanglement "area law". Bipartite entanglement eventually reaches a constant maximum value, independent of system size. Hastings proved this in An Area Law for One Dimensional Quantum Systems.

Being able to throw away low-weight highly entangled states and work in the exponentially-small region of Hilbert space relevant for 1D ground states is exactly what makes DMRG so efficient and successful.

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  • $\begingroup$ Thank you for your answer. Are you saying that the area law allows us to choose any state of the super block Hamiltonian to compress the enlarged system Hamiltonian, or is the area law applicable to the ground state only? $\endgroup$
    – user261609
    Feb 5, 2022 at 1:04
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    $\begingroup$ Also, I am a little bit confused: How does the area law make DMRG work? $\endgroup$
    – user261609
    Feb 5, 2022 at 1:11
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    $\begingroup$ You truncate the density matrix by keeping only the $m$ largest eigenvalue states. This compression will only be accurate if these $m$ largest eigenvalues contain almost all of the weight, and the other eigenvalues are almost zero. This can only be true if the entanglement is limited, because bipartite entanglement can be calculated as $\sum_i \lambda_i \log(\lambda_i)$, where $\lambda_i$ are the eigenvalues. Random (non-area law) states in Hilbert space have no reason to have such a rapidly-decaying entanglement spectrum, so they cannot be compressed. $\endgroup$ Feb 5, 2022 at 1:17
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    $\begingroup$ Thank you for your answer and comments!! $\endgroup$
    – user261609
    Feb 5, 2022 at 1:26
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    $\begingroup$ No worries :). In other words, because entanglement = $\sum_i \lambda_i \log(\lambda_i)$, entanglement is large when all eigenvalues $\lambda_i$ are equal, and smaller when they rapidly decay. So entanglement-limited (area law) states allow for compression by throwing away the small-eigenvalue states. $\endgroup$ Feb 5, 2022 at 1:29

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