Infinite-Dimensional Entanglement

Infinite-dimensional entanglement retains finite-dimensional witnesses inside suitable local subspaces, but some familiar matrix operations lose their trace-norm bounds. In particular, the partial transpose of a density operator need not be trace class. This page proves both statements and explains how they constrain numerical and analytic entanglement tests.

Helpful background. Separable and Entangled Mixed States supplies the closed-convex definition. Partial Trace Revisited supplies controlled local cutoffs. Negativity and PPT Criterion explains the finite-dimensional diagnostic.

An entangled normal state has an entangled finite compression

Let both factors be separable, with increasing finite-rank projections $P_n\to I_A$ and $Q_n\to I_B$. Define $\Pi_n=P_n\otimes Q_n$ and, whenever $p_n=\operatorname{Tr}(\rho\Pi_n)>0$,

$$\rho_n=\frac{\Pi_n\rho\Pi_n}{p_n}.$$

These states live in finite local subspaces and converge to $\rho$ in trace norm. The cutoff bound established on the reduction page makes that convergence quantitative.

If every $\rho_n$ were separable, their limit would be separable because the separable set is trace-norm closed. Therefore an entangled $\rho$ must have an entangled finite compression. Indeed all sufficiently large members of this convergent sequence are entangled, since the complement of the separable set is open.

Conversely, local projection and conditioning cannot turn a separable state into an entangled one. A product input remains an unnormalized product under $\Pi_n$; finite convex mixtures remain separable, and continuity extends this to their closure when the success probability is nonzero. Thus an entangled compression certifies entanglement of the original state.

This is an existence theorem. It gives neither a universal cutoff size nor a guaranteed success probability. A state’s detectable entanglement may occur in a very weakly occupied high-energy sector.

Finite witnesses extend to the full Hilbert space

Suppose $W_n$ is an entanglement witness on $P_n\mathcal H_A\otimes Q_n\mathcal H_B$ and detects $\rho_n$. Extend it by zero to the orthogonal complement:

$$\widetilde W_n=\Pi_n W_n\Pi_n.$$

For any full-space product vector, its projection into this subspace is again a product vector, possibly zero. Hence the extension has nonnegative expectation on all separable states. Its target expectation is

$$\operatorname{Tr}(\rho\widetilde W_n) =p_n\operatorname{Tr}(\rho_nW_n)<0.$$

Thus a finite-subspace certificate is already a bounded full-space certificate. A small $p_n$ reduces the signal, a statistical cost separate from the mathematical validity of the witness.

Partial transpose is naturally bounded on Hilbert–Schmidt space

Fix an orthonormal basis on $B$ and define partial transpose first on finite-rank matrix units:

$$\bigl(|i\rangle\langle k|\otimes|j\rangle\langle l|\bigr)^\Gamma =|i\rangle\langle k|\otimes|l\rangle\langle j|.$$

This permutes an orthonormal basis of the Hilbert–Schmidt operator space, so it extends to an isometry there. Since a density operator is Hilbert–Schmidt, $\rho^\Gamma$ is well-defined as a self-adjoint Hilbert–Schmidt, hence compact bounded, operator.

There is no corresponding dimension-independent bound on its trace norm. A maximally entangled rank-$d$ pure state has $\|\rho^\Gamma\|_1=d$ although $\|\rho\|_1=1$. Allowing arbitrarily large finite subspaces already rules out a bounded partial-transpose map on the full trace-class space when both local dimensions are infinite.

The chosen basis changes the displayed transpose, but positivity and singular values are unaffected by a local change of basis.

A pure-state spectrum exposes the divergence

Use Schmidt bases for $\Psi=\sum_ns_n|n,n\rangle$. The partial transpose of its projector acts on each two-dimensional subspace spanned by $|m,n\rangle,|n,m\rangle$, for $m\neq n$, with eigenvalues $\pm s_ms_n$. On $|n,n\rangle$ its eigenvalue is $s_n^2$.

Consequently

$$\bigl\|(|\Psi\rangle\langle\Psi|)^\Gamma\bigr\|_1 =\left(\sum_ns_n\right)^2,$$

understood as an extended value. The partial transpose is trace class exactly when the Schmidt coefficients belong to $\ell^1$, which is stronger than the normalization condition $\sum_ns_n^2=1$.

For $s_n=C/n$, the state is normalized after choosing $C$, but $\sum_ns_n$ diverges. Its partial transpose is a legitimate Hilbert–Schmidt operator with infinite trace norm. An ordinary trace of that non-trace-class operator is not justified by formally summing its diagonal entries to one.

This explains why infinite-dimensional negativity can be infinite. One may define an extended quantity from this spectral norm, but must not pretend that partial transpose is an everywhere-bounded operation on trace class.

Separable states are still PPT

Finite mixtures of product states have positive partial transpose. For a general separable $\rho$, approximate it in trace norm by such mixtures $\rho_k$. Trace-norm convergence implies Hilbert–Schmidt convergence, and partial transpose preserves the latter. Thus $\rho_k^\Gamma\to\rho^\Gamma$ in Hilbert–Schmidt norm and therefore in operator norm. Positivity is closed under this limit, proving $\rho^\Gamma\geq0$.

Its nonnegative diagonal sum is one, so this positive operator is in fact trace class with trace one. No trace-norm continuity of the partial-transpose map was assumed.

Hence nonpositive partial transpose remains an entanglement certificate. Positive partial transpose is not sufficient for separability: finite-dimensional PPT-entangled examples can be embedded in the larger space. Their construction and dimensional limitations are developed in Negativity and the PPT Criterion.

Exercises

1. For $s_n=\sqrt{1-q}\,q^{n/2}$, $n\geq0$, compute the partial-transpose trace norm and negativity.

Solution

The coefficient sum is $\sqrt{1-q}/(1-\sqrt q)$, so the squared sum is $(1+\sqrt q)/(1-\sqrt q)$. With $\mathcal N=(\|\rho^\Gamma\|_1-1)/2$, the negativity is $\sqrt q/(1-\sqrt q)$. Infinite Schmidt rank therefore does not by itself imply infinite negativity.

2. Why can a separable-looking low-energy cutoff fail to certify that the full state is separable?

Solution

The unretained subspace may contain entanglement. The theorem says that some sufficiently rich finite compression detects any fixed entangled state, not that every compression does. Conversely, a certified entangled local compression is decisive because separability survives local conditioning.

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) — infinite-dimensional entanglement, topology, and energy restrictions. Paper.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, second revised and expanded edition, De Gruyter, 2019 — normal bipartite states and infinite-dimensional separability. Publisher.
• Simon, Barry, Trace Ideals and Their Applications, second edition, Mathematical Surveys and Monographs 120, American Mathematical Society, 2005 — Hilbert–Schmidt operators, trace class, and singular-value summability. Publisher.