Bond Dimension and Entanglement

The bond dimension $\chi$ of a matrix product state is not an arbitrary knob: across each cut of the chain it equals the Schmidt rank of the state at that cut — the number of non-zero singular values you found when you peeled the state apart with SVD. Bond dimension is a measure of entanglement, and this lesson makes the correspondence precise.

Schmidt rank across a cut

Split the qubits into a left block $L$ and a right block $R$ . The Schmidt decomposition writes the state as

|\psi\rangle = \sum_{\alpha=1}^{r} \lambda_\alpha\, |u_\alpha\rangle_L \otimes |v_\alpha\rangle_R, \qquad \lambda_\alpha > 0,

where $\{|u_\alpha\rangle\}$ and $\{|v_\alpha\rangle\}$ are orthonormal sets and the $\lambda_\alpha$ are the Schmidt coefficients. The integer $r$ — the number of terms — is the Schmidt rank. A product state has $r = 1$ . Any $r > 1$ signals entanglement across that cut, and a larger $r$ means richer correlations.

The bond index $\alpha_k$ that the MPS carries across the $k$ -th cut runs over exactly these $r$ Schmidt terms. Hence

\chi_k = \text{(Schmidt rank at cut } k) = r_k .

The MPS is exact precisely when every bond dimension reaches the local Schmidt rank; truncating a bond below that rank throws away the smallest Schmidt coefficients and introduces a controlled approximation.

Two extremes

Product state, e.g. $|000\rangle$ or $H^{\otimes 3}|000\rangle$ : every cut has Schmidt rank $1$ , so $\chi = 1$ . These are the cheapest states of all.
Maximally entangled across a cut: a Bell pair $\tfrac{1}{\sqrt 2}(|00\rangle + |11\rangle)$ has two equal Schmidt coefficients, so $r = 2$ and $\chi = 2$ at that cut.

The GHZ state: $\chi = 2$ everywhere

The three-qubit GHZ state

|\mathrm{GHZ}\rangle = \frac{1}{\sqrt 2}\bigl(|000\rangle + |111\rangle\bigr)

is a perfect teaching example. Cut it between qubit 0 and qubits $\{1,2\}$ :

|\mathrm{GHZ}\rangle = \frac{1}{\sqrt 2}\,|0\rangle_L|00\rangle_R + \frac{1}{\sqrt 2}\,|1\rangle_L|11\rangle_R .

Two orthogonal terms, two equal coefficients $\tfrac{1}{\sqrt 2}$ — Schmidt rank $2$ . By symmetry the cut between $\{0,1\}$ and $\{2\}$ is also rank $2$ . So an exact MPS for GHZ needs bond dimension $\chi = 2$ on every bond, regardless of how long the chain is. GHZ on $100$ qubits still has $\chi = 2$ : highly correlated but very cheap to store.

Why this governs cost

Because storage and runtime scale as powers of $\chi$ , the entanglement of the target state directly sets the price of an exact tensor-network simulation. A state whose Schmidt rank stays bounded as the system grows is efficiently representable; a state whose Schmidt rank grows exponentially with system size is not. Recognising which regime you are in — by reasoning about how entangled the state is — is the practical skill this module trains.

Try it

Build the 3-qubit GHZ state $\tfrac{1}{\sqrt 2}(|000\rangle + |111\rangle)$ . The grader checks the exact state vector. This state has Schmidt rank $2$ across every cut, so its exact MPS bond dimension is $\chi = 2$ .

Run your code to see the quantum state.

After running, the statevector panel should show amplitude $1/\sqrt 2 \approx 0.707$ on both $|000\rangle$ and $|111\rangle$ and zero everywhere else.