How Is MPS Bond Dimension Related to Entanglement?

The bond dimension of a matrix product state (MPS) controls how much entanglement the state can represent across a cut in a one-dimensional chain. The key point is that the bond dimension is an upper bound on the number of independent Schmidt components, not a direct measurement of the entanglement actually present.

For a pure state, a bond of dimension $D$ allows at most $D$ nonzero Schmidt coefficients across that bond. The entanglement entropy is consequently bounded by $\log D$. The logarithm’s base sets the units: base 2 gives bits, while the natural logarithm gives nats.

Bond dimension and a cut through the chain

For an open-boundary MPS on $N$ sites, one common notation is

$$|\psi\rangle = \sum_{s_1,\ldots,s_N} A^{[1]s_1} A^{[2]s_2}\cdots A^{[N]s_N}|s_1\cdots s_N\rangle.$$

The matrices and boundary tensors are joined by internal indices, called bonds. Suppose one bond splits the chain into a left block $L$ and a right block $R$, and that its dimension is $D$. Grouping all left-side indices together and all right-side indices together gives the state the form

$$|\psi\rangle = \sum_{\alpha=1}^{D} |L_\alpha\rangle |R_\alpha\rangle,$$

where the block vectors need not be normalized or mutually orthogonal. This is a sum of at most $D$ product terms across the cut. The Schmidt decomposition is the corresponding expression using orthonormal vectors and nonnegative coefficients:

$$|\psi\rangle = \sum_{i=1}^{r} \lambda_i |u_i\rangle |v_i\rangle, \qquad \sum_{i=1}^{r}\lambda_i^2=1.$$

Here $r$ is the Schmidt rank, and the $\lambda_i$ are the Schmidt coefficients. Since the MPS bond can carry only $D$ independent values, $r\le D$.

The entropy bound

For a pure state, the entanglement entropy across the cut is the von Neumann entropy of either block’s reduced density matrix:

$$S(L)=-\sum_{i=1}^{r}\lambda_i^2\log(\lambda_i^2).$$

Entropy is largest when the $r$ probabilities $\lambda_i^2$ are equal. Thus

$$S(L)\le \log r\le \log D.$$

There is also a limit from the size of the physical blocks: $r$ cannot exceed either block’s Hilbert-space dimension. In general, the tight rank limit is the smallest of $D$ and those two dimensions.

For an exact open-boundary MPS, the minimum bond dimension required at a particular cut is exactly the Schmidt rank across that cut. To see this, reshape the state’s coefficients into a matrix with one row for each left-block basis state and one column for each right-block basis state. Its matrix rank is the Schmidt rank. A factorization through the bond index must have dimension at least that rank, and a rank factorization achieves it. This statement concerns the minimum exact representation; an MPS written with a larger bond may contain redundant directions.

Worked example: two qubits

Consider the normalized state

$$|\psi\rangle=\sqrt{\frac{3}{4}}|00\rangle+\sqrt{\frac{1}{4}}|11\rangle.$$

Across the bond between the qubits, this is already a Schmidt decomposition. It has two nonzero coefficients, so the Schmidt rank is $2$. The smallest exact MPS bond dimension is therefore $D=2$. Its entropy in bits is

$$S=-\frac{3}{4}\log_2\left(\frac{3}{4}\right)-\frac{1}{4}\log_2\left(\frac{1}{4}\right)\approx 0.811.$$

The general bond-dimension bound gives $S\le\log_2 2=1$ bit, which the example satisfies without reaching. A bond dimension of $4$ would allow a bound of $2$ bits, but two qubits cannot have more than $1$ bit of entanglement across this cut. A larger bond is permission to represent more entanglement, not a guarantee that the state has it.

What the bond dimension tells you in practice

A fixed bond dimension places a ceiling on entanglement across every corresponding cut. This helps explain why MPS methods work well for many one-dimensional systems: ground states of many gapped local systems have entanglement that remains bounded as the chain grows, so they can often be approximated with manageable bond dimensions. Critical systems can have entanglement that grows with subsystem size, making accurate MPS representations more demanding.

In numerical calculations, the Schmidt coefficients also guide compression. If many coefficients are very small, discarding some of them can reduce the bond dimension while making a controlled approximation. The discarded weight provides a useful measure of the truncation, although its relationship to errors in particular observables or later calculations depends on the calculation.

Common mistakes

• Treating $D$ as the entropy. Entropy is not generally $\log D$. It is computed from the Schmidt coefficients and is only bounded above by $\log D$.
• Assuming every bond direction is used. A representation with bond dimension $D$ may have fewer than $D$ nonzero Schmidt coefficients. Redundant or zero-weight directions do not add entanglement.
• Forgetting the logarithm convention. The bound is $\log_2 D$ bits with base-2 logarithms, or $\ln D$ nats with natural logarithms.
• Confusing a cut with the whole chain. Each bond bounds entanglement between the sites to its left and right. The state may have different Schmidt ranks and entropies at different cuts.
• Applying the pure-state formula without checking the state. The direct Schmidt-coefficient formula above is for pure states. Mixed-state entanglement requires other measures; the MPS bond rank alone does not specify it.