Entanglement Entropy from Exact States

For a pure bipartite state, entanglement entropy is obtained from the singular values of a reshaped amplitude matrix. This avoids building the full density matrix and gives a stable route to the Schmidt spectrum. The conceptual distinction between subsystem mixedness and entanglement belongs to entanglement entropy.

Helpful background. Tensor products in code fixes the index order; Schmidt decomposition explains the singular values physically.

Reshape, permute, and factor

For a bipartition $A|B$, reorder subsystem axes so that all $A$ indices precede all $B$ indices, then reshape the normalized vector into $C$ of shape $d_A\times d_B$. A singular-value decomposition gives

$$C=U\,\operatorname{diag}(s_\alpha)V^\dagger,\qquad \lambda_\alpha=s_\alpha^2,\qquad \sum_\alpha\lambda_\alpha=1.$$

The reduced state is $CC^\dagger$, so its nonzero eigenvalues are $\lambda_\alpha$. Singular values avoid small negative eigenvalues caused by explicitly diagonalizing a rounded reduced density matrix.

A noncontiguous bipartition requires an axis permutation. If the vector lives in a restricted particle-number sector, first embed through the basis map or contract occupations directly; reshaping the shorter sector vector as though it were a full tensor is wrong.

For four ordered qubits $(0,1,2,3)$ with the last index varying fastest, the partition $A=\{0,2\}$ requires reshaping to $(2,2,2,2)$, transposing axes to $(0,2,1,3)$, and then reshaping to $(4,4)$. For a product of Bell pairs on $(0,1)$ and $(2,3)$ this partition has four equal Schmidt probabilities, while $A=\{0,1\}$ has rank one. The same state therefore supplies a sharp test of the partition map.

Entropies and spectrum

Using natural logarithms,

$$S=-\sum_\alpha\lambda_\alpha\log\lambda_\alpha,\qquad S_q=\frac{\log\sum_\alpha\lambda_\alpha^q}{1-q}, \quad q>0,\ q\ne1.$$

Divide by $\log2$ for bits. Entanglement energies are $\xi_\alpha=-\log\lambda_\alpha$; zero eigenvalues correspond to infinite entanglement energies and should not be replaced by arbitrary large numbers in an unqualified plot.

For small positive $\lambda$, evaluate $\lambda\log\lambda$ only on positive entries and assign the exact limiting value zero at $\lambda=0$. A cutoff on tiny eigenvalues changes entropy; record discarded weight rather than hiding it.

Product and Bell benchmarks

For a product state $C$ has rank one, so $S=0$. For $(|00\rangle+i|11\rangle)/\sqrt2$, its two singular values are $1/\sqrt2$ and $S=\log2$. The phase does not change entanglement but catches missing conjugation in a reduced-density construction.

For $\sqrt p|00\rangle+\sqrt{1-p}|11\rangle$,

$$S=-p\log p-(1-p)\log(1-p).$$

This continuous family tests both rank-deficient endpoints and the maximum at $p=1/2$. For a pure global state, computing either side must give the same nonzero spectrum and entropy.

Error from discarded Schmidt weight

Suppose singular values below a threshold are omitted with total probability $\delta$. Renormalizing retained probabilities changes the state. Compare entropy under decreasing threshold, and keep the discarded weight as an explicit diagnostic.

There are two different trace distances. The normalized truncated global pure state has fidelity $1-\delta$ with the original and trace distance $\sqrt{\delta}$. Its reduced state retains the same Schmidt eigenvectors, with retained probabilities rescaled by $1/(1-\delta)$ and discarded probabilities set to zero. The reduced-state trace distance is exactly $\delta$; use this sharper value when bounding the entropy change.

A general continuity bound for density matrices in dimension $d$ with trace distance $T=\|\rho-\sigma\|_1/2$ is

$$|S(\rho)-S(\sigma)| \leq T\log(d-1)+h_2(T),$$

within the usual range $0\leq T\leq1-1/d$, where $h_2(T)=-T\log T-(1-T)\log(1-T)$. The dimension factor shows why tiny discarded probability can still affect entropy in a large space.

For this particular truncation one can say more. Let $S_{\rm ret}$ and $S_{\rm tail}$ be the entropies of the separately normalized retained and discarded probabilities. Grouping the original entropy sum gives the exact identity

$$S=h_2(\delta)+(1-\delta)S_{\rm ret}+\delta S_{\rm tail}.$$

If the discarded part has at most $r_{\rm tail}$ nonzero values, then $0\leq S_{\rm tail}\leq\log r_{\rm tail}$ gives an explicit interval for the error $S-S_{\rm ret}$. For probabilities $(0.7,0.2,0.1)$ and a cutoff removing $0.1$, the entropy changes from approximately $0.801819$ to $0.529706$ nats, despite the small discarded component. Evaluate the full entropy whenever those singular values remain numerically resolved; a plotting cutoff need not become a computational cutoff.

Near a degenerate energy eigenspace, different linear combinations may have different entanglement. A small eigensolver residual does not select a unique entanglement value; specify the physical state or symmetry constraints.

Mixed-state limitation

For a mixed global state, subsystem entropy contains classical and thermal uncertainty and is not an entanglement measure by itself. A maximally mixed product state has nonzero subsystem entropy but no entanglement. Use an appropriate mixed-state quantity rather than applying the pure-state interpretation mechanically.

Exercise

Two independent Bell pairs cross the same bipartition. Find the Schmidt probabilities and entropy.

Solution

The Schmidt probabilities multiply, giving four values $1/4$. Thus $S=\log4=2\log2$, or two bits. This tests tensor-product ordering and entropy additivity.

References

• Audenaert, K. M. R., “A sharp continuity estimate for the von Neumann entropy,” Journal of Physics A 40, 8127–8136 (2007), arXiv:quant-ph/0610146.
• Watrous, J., The Theory of Quantum Information, Cambridge University Press (2018), author’s text — Schmidt spectra and entropy.