DMRG and Matrix Product States

A quantum state on $N$ sites with $d$ states each lives in a space of dimension $d^N$. Full configuration interaction (FCI) stores all $d^N$ amplitudes and diagonalizes in that space — exact, and exponentially doomed. At $d=2$, fifty spins is $2^{50}\approx 10^{15}$ amplitudes: a petabyte for one vector.

But you rarely need all of it. The ground state of a local Hamiltonian is one very particular vector in that enormous space, and it has far less structure than its length suggests. DMRG is the bet that you can throw away almost every amplitude and keep the answer — if you compress along the right seams.

Cut the chain in two and ask how entangled the halves are. For a random state the answer is "maximally" — the entanglement entropy grows with the volume of the smaller half. But ground states of local, gapped Hamiltonians obey an area law: in 1D the entropy across a cut is bounded by a constant, independent of how long the chain is.

That constant is the whole game. Bounded entanglement means each cut can be described by a fixed number $\chi$ of effective states, not the $d^{N/2}$ a generic vector would demand. The demo below shows it directly: in the gapped phase the entropy is flat and tiny, and a handful of states per bond reproduces the energy to machine precision. At a critical point the area law fails gently — the entropy creeps up like $\log N$ — and DMRG has to work harder.

Take the coefficient tensor ${\psi }_{s_1{s}_2\cdots s_N}$ and reshape it into a matrix splitting site 1 from the rest. Its singular value decomposition is the Schmidt decomposition across that cut: the singular values are the entanglement spectrum, and ${\sum }_{i>\chi }{\sigma }_i^2$ is exactly the weight you lose by keeping only the top $\chi$. Eckart–Young says that truncation is the best rank-$\chi$ approximation there is.

Now peel site 2 off the remainder and SVD again. And again, down the chain. Each step factors one more site and truncates its bond to $\chi$. What is left when you reach the end is the wavefunction rewritten as a product of small matrices — a matrix product state — with every bond an SVD you understand.

Draw the wavefunction as a single box with $N$ legs, one per site. A matrix product state replaces that one fat box with a row of small boxes — one per site — joined left-to-right by internal "bond" legs. Each box $A^{s_i}$ has one physical leg (its site's state) and up to two bond legs; contracting all the bonds reconstructs the full tensor.

The bond dimension $\chi$ — the width of those internal legs — is the single dial. $\chi =1$ is a product state, mean-field, no entanglement. Crank $\chi$ up and the train can carry more correlation between the sites; at $\chi =d^{N/2}$ it is exact again. Cost grows as $N{\chi }^2$, not $d^N$ — linear in the chain length, for fixed entanglement.

Here is the algorithm DMRG actually runs. Hold every tensor in the train fixed except one. With the rest frozen, minimizing $⟨\psi |H|\psi ⟩$ over that one tensor is no longer a $d^N$-dimensional problem — it collapses to a small effective eigenvalue problem, a few ${\chi }^{2}d$ unknowns. You solve it for the lowest eigenvector with Lanczos — the same three-term recurrence you already use everywhere on this site.

Move to the next site, rebuild the local problem, Lanczos again. Sweep left to right, then right to left, and the energy comes down every pass. It is the variational principle restricted to the MPS manifold — coordinate descent where each coordinate is one tensor and each line-search is an eigensolve. An SVD after each step re-truncates the bond back to $\chi$.