Entanglement Entropy Caveats

For a pure bipartite state, the two marginal entropies agree even when their common value is infinite. That extended value requires care: it does not define an infinite-dimensional maximally mixed state, justify subtracting divergent entropies, or make entropy continuous under arbitrary norm approximations. The canonical Entanglement Entropy article explains the finite-dimensional interpretation; Entropy and Trace develops the scalar divergence and continuity counterexamples.

Helpful background. Schmidt Decomposition in Hilbert Space gives the common nonzero marginal spectrum. Quantum Entropy supplies the information-theoretic quantities and inequalities.

Equal marginal entropies can both be infinite

Use natural logarithms, so entropies are in nats. For a normalized pure state with squared Schmidt coefficients $p_n$,

$$E(\Psi)=S(\rho_A)=S(\rho_B) =\sum_n-p_n\log p_n\in[0,\infty].$$

The equality is an equality of nonnegative series with the same terms. It remains valid when the series diverges; it does not arise by subtracting two infinite quantities.

Any probability distribution with infinite Shannon entropy gives an infinite-entanglement-entropy vector $\sum_n\sqrt{p_n}|n,n\rangle$. Its Hilbert norm is still one. The explicit slow-tail distribution supplies such an example without changing the normal-state framework.

An infinite value is not a uniquely maximally entangled state. There is no normalized density operator proportional to the identity on an infinite-dimensional subsystem. Different infinite-entropy distributions can have different largest coefficients, different energy behavior, and different operational conversion properties.

Infinite rank, entropy, and negativity are different tests

Consider squared Schmidt coefficients

$$p_n=\frac{n^{-2}}{\zeta(2)},\qquad n\geq1.$$

Their entropy is finite because $\sum_n(\log n)/n^2$ converges. More explicitly,

$$E=\log\zeta(2)-\frac{2\zeta'(2)}{\zeta(2)}.$$

The state nevertheless has infinite Schmidt rank. Its Schmidt coefficients are proportional to $1/n$, so its partial-transpose trace norm is infinite by the partial-transpose summability calculation.

This one example separates three notions: an infinite number of occupied modes, finite entropy of entanglement, and infinite negativity. A statement that entanglement is “large” is incomplete until the quantity and constraints are identified. The operational comparison of different measures belongs to Entanglement Measures.

Conditional entropy cannot always be a difference

For finite-dimensional systems one writes $S(A|B)=S(AB)-S(B)$ and $I(A:B)=S(A)+S(B)-S(AB)$. With infinite marginal entropies, these displays may contain $\infty-\infty$.

The robust starting point for mutual information is the extended relative entropy

$$I(A:B)_\rho =D(\rho_{AB}\|\rho_A\otimes\rho_B).$$

It is defined as a nonnegative extended quantity without separately subtracting divergent entropy traces. If $S(\rho_A)<\infty$, the general bound $I(A:B)_\rho\leq2S(\rho_A)$ makes

$$S(A|B)_\rho=S(\rho_A)-I(A:B)_\rho$$

a finite, well-defined conditional entropy, with $-S(\rho_A)\leq S(A|B)_\rho\leq S(\rho_A)$. It agrees with the entropy-difference formula whenever that formula is legitimately finite. The extension and its properties are established by Kuznetsova.

For example, let $\rho_{AB}=(I_2/2)\otimes\tau_B$, where $S(\tau_B)=\infty$. Both $S(AB)$ and $S(B)$ are infinite, but the mutual information is zero because the state is product. The conditional entropy is therefore $\log2$, an ordinary finite answer inaccessible to naive subtraction.

For a pure bipartite state of finite entanglement entropy, $I(A:B)=2E$ and $S(A|B)=-E$. If $S(A)=\infty$, the finite-marginal definition just given no longer applies; an expression of the form $\infty-\infty$ cannot be assigned a value by analogy.

Local energy bounds control pure-state entanglement

Let $H_A\geq0$ have $Z_A(\beta)=\operatorname{Tr}e^{-\beta H_A}<\infty$. The scalar Gibbs entropy bound, proved in Entropy and Trace, immediately gives

$$E(\Psi)\leq\beta\operatorname{Tr}(\rho_AH_A) +\log Z_A(\beta).$$

One local energy constraint is enough to bound pure-state entanglement. No assumption about the other factor’s dimension is required.

For a useful uniform continuity statement, require $Z_A(\beta)<\infty$ for every $\beta>0$ and restrict the local mean energy by a fixed finite bound. Entropy is then uniformly continuous on that constrained marginal state set. Since partial trace contracts trace distance, pure bipartite approximations whose marginals satisfy the same energy bound inherit this continuity. Winter provides quantitative bounds.

Finiteness of a partition function at one temperature ensures an entropy bound at fixed mean energy; it is not by itself the stronger uniform-continuity hypothesis. Arbitrary finite-rank approximations must also be checked against the declared energy constraint.

An oscillator gives a physical bound

For $H_A=\hbar\omega N_A$, write the mean occupation bound as $\langle N_A\rangle\leq\bar n$. Optimizing the Gibbs bound gives

$$E(\Psi)\leq g(\bar n), \qquad g(x)=(x+1)\log(x+1)-x\log x.$$

If $B$ contains a countably infinite orthonormal family, the bound is attained by a pure bipartite vector whose Schmidt probabilities are geometric with $q=\bar n/(1+\bar n)$. Its marginal is the oscillator thermal state. For finite $d_B=\dim\mathcal H_B$, the Schmidt rank instead gives the additional bound $E\leq\log d_B$; a pure joint state cannot have this infinite-rank thermal marginal when $\bar n>0$. The oscillator bound still holds, but its attainment depends on the available Schmidt partners.

Exercises

1. Can a pure state with infinite entanglement entropy satisfy a finite oscillator mean-occupation bound?

Solution

No. The oscillator bound gives $E\leq g(\bar n)<\infty$ for every finite $\bar n$. Infinite entropy forces failure of that finite local-energy condition for this Hamiltonian.

2. If a numerical pure-state approximation has very small trace-distance error, is its entanglement entropy automatically accurate?

Solution

No. Trace distance controls bounded measurements, whereas entropy can receive a large contribution from a small probability spread across many modes. A finite-dimensional bound with a fixed dimension, or an appropriate uniform energy/tail bound, is needed to convert that state error into an entropy estimate.

References

• Eisert, Jens, Christoph Simon, and Martin B. Plenio, “On the Quantification of Entanglement in Infinite-Dimensional Quantum Systems,” Journal of Physics A 35, 3911 (2002) — entanglement entropy and energy constraints. Paper.
• Kuznetsova, Anna A., “Conditional Entropy for Infinite-Dimensional Quantum Systems,” Theory of Probability and Its Applications 55, 709–717 (2011) — conditional entropy with a finite marginal entropy. Paper; preprint.
• Winter, Andreas, “Tight Uniform Continuity Bounds for Quantum Entropies: Conditional Entropy, Relative Entropy Distance and Energy Constraints,” Communications in Mathematical Physics 347, 291–313 (2016) — quantitative continuity with explicit Hamiltonian assumptions. Paper; preprint.