Find a ground state by sweeping, and decide whether to believe it

DMRG holds all tensors but two neighbours fixed, finds the best pair by solving a small eigenvalue problem, splits it back with a truncated decomposition and moves one site on. A sweep runs to the end of the chain and back. Each update can only lower the energy, so the energy settles. Whether it settles on the ground state needs separate evidence: the energy variance, a comparison with an exact limit, and a check of the symmetries the true ground state must have.

Sandbox Physics · 2026-10-29

A variational search, two sites at a time

Any normalized trial state has an energy at or above the ground-state energy. DMRG searches among matrix product states of a given bond dimension for the one with the lowest energy. Optimizing all tensors at once is a hard non-linear problem, so DMRG improves a pair of neighbouring tensors with the rest held fixed.

With the other tensors fixed and kept orthonormal, the energy of the pair is a quadratic form in its entries. The best pair is the lowest eigenvector of a small effective Hamiltonian, which a few Lanczos steps find. A truncated singular value decomposition splits the pair back into two tensors and keeps at most the allowed number of Schmidt terms. Then the window moves one site. Schollwöck describes this iterative search in Section 6.3 of his review.

H_{\rm eff}^{[j,j+1]}\,\Theta=E\,\Theta,\qquad \Theta\approx U\,\mathrm{diag}(s_1,\dots,s_\chi)\,V^{\dagger}

Three diagnostics, and what each one misses

The energy is the first diagnostic. On the critical Ising chain of 32 spins with at most sixteen states per link, the error against the exact free-fermion energy falls from 0.035 after the first sweep to 0.0012, to seven millionths, to eight billionths, and settles at about two ten-billionths from the fifth sweep on. A settled energy tells you the search has stopped, not that it found the ground state.

The energy variance is the second. An exact eigenstate has zero variance, and a small variance guarantees an eigenvalue close to the measured energy. Schollwöck recommends it in Section 6.4 as a check that does not depend on the two-site setup. The third is the discarded weight of each truncation. Both shrink as the bond dimension grows. With only four states per link the same chain keeps an energy error of one thousandth and a variance of 0.0038.

\sigma_E^2=\langle H^2\rangle-\langle H\rangle^2,\qquad \exists\,E_n:\ |E_n-\langle H\rangle|\le\sigma_E

Worked example

Check the Heisenberg chain against the Bethe ansatz

Karbach, Hu and Müller give, as their Eq. 13, Hulthén's ground-state energy of the infinite Heisenberg antiferromagnet: a quarter minus the natural logarithm of two per bond, about minus 0.443147 in units of the coupling. An open chain cannot be compared with this number directly. Its two ends cost extra energy, so the energy per bond of a 32-spin chain is minus 0.4515.

Take differences instead. Adding sixteen spins adds sixteen bulk bonds but leaves the two ends the same, so the end energy cancels. Run DMRG for 16, 32, 48 and 64 spins, with the energy variance at most about one millionth in each run. The energy per added bond comes out as minus 0.442849, then minus 0.443040, then minus 0.443092. The gap to the Bethe value falls from 0.00030 to 0.00011 to 0.00006 as the chains grow. The same DMRG code also matches exact diagonalization of the sixteen-spin chain to two billionths.

\frac{E(L_2)-E(L_1)}{L_2-L_1}\ \longrightarrow\ \Big(\tfrac14-\ln2\Big)J\approx-0.443147\,J

A tiny variance that hides the wrong state

Lower the field to 0.7, into the ordered phase. Two states now lie extremely close in energy: an even and an odd combination of the two ordered patterns. The exact ground state of the finite chain is one of them and has a definite value of the spin-flip parity. DMRG at bond dimension sixteen returns a state with variance below two billionths, yet its energy is 5.6 millionths too high and its parity is 0.002 instead of a definite value.

The variance did its job: it guaranteed an eigenvalue nearby. It cannot tell you which one, and with two nearly degenerate states, a mixture of both also has a small variance. The symmetry-broken state is also cheaper to store: as the cat state of chapter two showed, the even or odd combination carries an extra logarithm of two across every cut. Only the symmetry check gives it away. On the critical chain the same check reads a parity of 1.000.

Read the entanglement it found

After the last sweep, the Schmidt weights at every cut are available for free. On the critical Ising chain the entanglement profile follows the Calabrese–Cardy form for an open chain. A fit of the central charge to the DMRG profile of 32 spins gives 0.523992, which differs from the fit to the exact free-fermion profile by less than a ten-millionth. The value sits above one half because 32 spins is a short chain; chapter two shows the exact fit approaching one half as the chain grows.

c_{\rm DMRG}(L=32)=0.523992,\qquad c_{\rm DMRG}-c_{\rm exact}\approx-9\times10^{-8}

Predict, change, explain

Open the DMRG Lab on the critical Ising chain. Predict the sweep on which the energy error stops falling.

Switch to the ordered-phase preset and open the diagnostics.

Explain why the variance is tiny while the energy stays above the exact value, and which diagnostic exposes the problem.

The Labs simulate open spin-1/2 chains with energies in units of the coupling and time in units of the reduced Planck constant divided by the coupling. Each Lab states its model and limits in its notes and can export the data behind its readouts.

Try explaining it yourself

1. After one sweep, the variance on the critical chain is 0.05. How far from the measured energy must the nearest eigenvalue lie, at most?

Read an explained answer

Within one standard deviation, the square root of the variance: about 0.22. That is a weak statement; the guarantee becomes useful only once the variance is small.

\sigma_E=\sqrt{0.05}\approx0.22

2. Why does dividing the energy of a 32-spin open Heisenberg chain by its 31 bonds not give the Bethe-ansatz value?

Read an explained answer

The bonds near the ends are not like bulk bonds. Their extra energy is spread over all bonds when you divide, and it shrinks only as one over the length. Differences between two lengths cancel it.

3. In the ordered phase, would a larger bond dimension alone fix the parity?

Read an explained answer

Not reliably. The symmetry-broken state has less entanglement and nearly the same energy, so the sweep can stay there. Schollwöck describes the standard remedy in Section 2: build the symmetry into the tensors, so that every state the sweep can reach has a definite parity.

Sources to read alongside

U. Schollwöck, The density-matrix renormalization group in the age of matrix product states, Annals of Physics 326, 96 (2011)

The standard review. Section 3 uses the area law to explain why DMRG works in one dimension; Section 4.1 builds matrix product states from successive singular value decompositions and bounds the truncation error in Eq. 58; Section 6 sets out the ground-state search as a variational sweep; Section 7 covers time evolution and its limits.

P. Calabrese and J. Cardy, Entanglement entropy and quantum field theory, Journal of Statistical Mechanics P06002 (2004)

Derives the entanglement entropy of critical one-dimensional systems. Equation 2 gives the profile along an open chain; its coefficient is one sixth of the central charge.

M. Karbach, K. Hu and G. Müller, Introduction to the Bethe ansatz II, Computers in Physics 12, 565 (1998)

A worked introduction to the Bethe ansatz for the spin-1/2 Heisenberg antiferromagnet. Equation 13 gives Hulthén’s ground-state energy of the infinite chain: a quarter minus the natural logarithm of two per bond, in units of the coupling.