Contemporary Mathematics

An Unentangled Gleason’s Theorem

Nolan R. Wallach

Abstract. The purpose of this note is to give a generalization of Gleason’s theorem inspired by recent work in quantum information theory. For multipartite quantum systems, each of dimension three or greater, the only nonnegative frame functions over the set of unentangled states are those given by the standard Born probability rule. However, if one system is of dimension 2 this is not necessarily the case.

1. Introduction.

Let H be a Hilbert space with unit sphere S(H). Following Gleason ([Gleason]) we will call a function f : S(H) → ℂ a frame function of weight w if for every orthonormal basis {v_{i}} of S(H)

(1)  ∑_{i} f(v_{i}) = w.

In [Gleason] the following theorem was proved

Theorem 1. If dim H ≥ 3 and f is a frame function that takes non-negative real values then there exists a self adjoint trace class operator T : H → H such that

f(v) = ⟨v|T |v⟩, v ∈ S(H).

This theorem is of importance to quantum mechanics because it allows a significant weakening of the axioms, showing that the Born probability rule [Born] provides the unique class of probability assignments for measurement outcomes so long as those probabilities are specified by frame functions [Pitowsky]. The theorem also rules out a large class of hidden-variable explanations for quantum statistics, the so-called noncontextual hidden variables, in dimension 3 or greater. The interested reader should consult [Bell] for a discussion of this point. If the Hilbert space is of dimension 2, then the statement in the theorem is easily seen to be false.

The purpose of this note is to give a generalization of Gleason’s theorem inspired by recent work in quantum information theory. In that context the issue of local measurements and operations on multipartite quantum systems (as opposed to the full set of operations) is of the utmost importance [BDFMRSSW]. For instance,

\begin{matrix}c \\ ◯ \end{matrix}2002 American Mathematical Society

2000 Mathematics Subject Classification. Primary 81P68.

it has been pointed out that probabilities for the outcomes of local measurements are enough to uniquely specify the quantum state from which they arise if the field of the Hilbert space is complex, though this fails for real and quaternionic Hilbert spaces [Araki,Wootters].

Chris Fuchs has asked to what extent local and semi-local measurements not only uniquely specify the quantum state, but also a Born-like rule as in Gleason’s result [Fuchs]. In this regard, the following formalization appears natural. We confine our attention to finite dimensional Hilbert spaces for the sake of simplicity. Let H_{1}, ..., H_{n} be Hilbert spaces. Set H = H_{1}⊗H_{2}⊗···⊗H_{n}. Let Σ = Σ(H_{1}, ..., H_{n}) denote the subset of S(H) consisting of those elements of the form a_{1} ⊗ ⋯ ⊗ a_{n} with a_{i} ∈ S(H_{i}) for i = 1, ..., n. In the jargon of quantum information theory such states are called unentangled or product states. The ones that are not of this form are said to be entangled. An orthonormal basis {v_{i}} of H is said to be unentangled if v_{i} ∈ Σ for all i. We say that f : Σ → ℂ is an unentangled frame function of weight w if whenever {v_{i}} is an unentangled orthonormal basis of H then f satisfies (1) above. We establish the following result.

Theorem 2. If dim H_{i} ≥ 3 for all i and if f : Σ → ℝ is a non-negative unentangled frame function then there exists T : H → H a self adjoint trace class operator such that f(v) = ⟨v|T |v⟩ for all v ∈ Σ.

This theorem is an almost direct consequence of Gleason’s original theorem. We will give a proof of it in the next section. The second result in this paper shows that the dimensional condition is necessary.

It should be noted however, that despite the absence of entangled or “nonlocal” states in Σ, in [BDFMRSSW] it is asserted that not all unentangled bases correspond to quantum measurements that can be carried out by local means alone (even with iterative procedures based on weak local measurements and unlimited amounts of classical communication between the measurers at each site). The simplest kind of purely local measurement is given by an alternative type of basis adapted to the tensor product structure. This is a product basis and is defined as to be a basis of the form {u_{i_{1}1} ⊗ u_{i_{2}2} ⊗ ⋯ ⊗ u_{i_{n}n}} where u_{1j}, ..., u_{n_{j}j} is an orthonormal basis of H_{j}. We could define a product frame function in the same way as we did for an unentangled frame function except that we only assume that there exists a weight w such that ∑_{i_{1},i_{2},...,i_{n}} f(u_{i_{1}1} ⊗ u_{i_{2}2} ⊗ ⋯ ⊗ u_{i_{n}n}) = w for every product basis. One can ask whether this is all that us necessary for the conclusion of the theorem above. The answer is no and a method of “finding” a large class of examples will be given at the end of the next section (see the proposition at the end of the section). This result amasses some evidence that the structure of local measurements alone is not enough to establish the Born rule for multipartite systems, but a full answer would require consideration of the largest class of local measurements in [BDFMRSSW].

These issues also spawn another theorem.

Theorem 3. Let dim H_{1} = 2 and let f : S(H_{1}) → ℂ be a frame function of weight w_{1} and g : Σ(H_{2}, ..., H_{n}) → ℂ be an unentangled frame function of weight w_{2}. We set h(v_{1} ⊗ u) = f(v_{1})g(u) for u ∈ Σ(H_{2}, ..., H_{n}). Then h is an unentangled frame function of weight w_{1}w_{2}.

This result is a bit harder and the proof involves a method (see Theorem 5) that describes a combinatorial scheme for finding all unentangled orthonormal bases where all of the spaces, H_{i}, have dimension 2. This analysis in turn leads to a natural question. Given and unentangled orthonormal set can it be extended to an unentangled orthonormal basis? Or even stronger: Can it be a proper subset of an unentangled orthonormal set? This question was studied in [BDMSST]. We conclude the paper by giving a proof based on simple algebraic geometry of the following theorem which is related to the bound that occurs in [BDMSST].

Theorem 4. Let V be a subspace of H_{1} ⊗ ⋯ ⊗ H_{n} such that if v ∈ V and v ≠ 0 then v is entangled. Then dim V ≤ dim(H_{1})···dim(H_{n})−∑(dim H_{i} −1)−1. Furthermore, the upper bound is attained.

2. The unentangled Gleason theorem.

In this section we will give a proof of Theorem 2. If n = 1 the statement is just Gleason’s theorem. We consider the situation of H = H_{0} ⊗ V with V = H_{1} ⊗H_{2} ⊗···⊗H_{n} and dim H_{i} ≥ 3 for all i. We prove Theorem 1 by induction (i.e. assume the result for n). We note that if {v_{i}} is an orthonormal basis of H_{0} and if for each i, {u_{ij}} is an unentangled orthonormal basis of V then the set {v_{i} ⊗ u_{ij}} is an unentangled orthonormal basis of H. Thus if w is the weight of f then we have

∑_{j} f(v_{1} ⊗ u_{1j}) = w − ∑_{i≥2,j} f(v_{i} ⊗ u_{ij}).

Thus for each v ∈ S(H_{0}) the function f_{v}(u) = f(v ⊗ u) is an unentangled frame function. The inductive hypothesis implies that for each v ∈ S(H_{0}) there exists a self adjoint (due to the reality of f) linear operator T (v) such that f(v ⊗ u) = ⟨u|T (v)|u⟩ for u ∈ Σ(H_{1}, ..., H_{n}). Similarly, if {u_{i}} is an unentangled orthonormal basis of V and for each i, {v_{ij}} is an orthonormal basis of H_{0} then {v_{ij} ⊗ u_{i}} is an unentangled orthonormal basis of H. We therefore conclude as above that if u ∈ Σ(H_{1}, ..., H_{n}) then there exists S(u) a self adjoint linear operator on H_{0} so that f(v ⊗ u) = ⟨v|S(u)|v⟩ for all v ∈ H_{0}.

Let {u_{i}} be an unentangled orthonormal basis of V and let {v_{j}} be an orthonormal basis of H_{0}. Set

a_{ij}(v) = ⟨u_{i}|T (v)|u_{j}⟩

and

b_{ij}(u) = ⟨v_{i}|S(u)|v_{j}⟩, u ∈ Σ(H_{1}, ..., H_{n}).

We now observe that if v = ∑_{i} x_{i}v_{i} and if u = ∑_{j} y_{j}u_{j} then we have

∑_{p,q} a_{p,q}(v)ȳ_{p}y_{q} = ∑_{r,s} b_{r,s}(u)x̄_{r}x_{s}.

If we substitute v = v_{r} then we have

b_{rr}(u) = ∑_{p,q} a_{p,q}(v_{r})ȳ_{p}y_{q}.

Now assuming that r ≠ s and taking v = \frac{1}{\sqrt{2}}(v_{r} + v_{s}) we have

Re b_{rs}(u) = ∑_{p,q} a_{p,q}(\frac{1}{\sqrt{2}}(v_{r} + v_{s}))ȳ_{p}y_{q}−

\frac{1}{2} (∑_{p,q} a_{p,q}(v_{r})ȳ_{p}y_{q} + ∑_{p,q} a_{p,q}(v_{s})ȳ_{p}y_{q}) .

Also if we take v = \frac{1}{\sqrt{2}}(v_{r} + iv_{s}) then we have

− Im b_{rs}(u) = ∑_{p,q} a_{p,q}(\frac{1}{\sqrt{2}}(v_{r} + iv_{s}))y_{p}\overline{y}_{q}−

\frac{1}{2} (∑_{p,q} a_{p,q}(v_{r})y_{p}\overline{y}_{q} + ∑_{p,q} a_{p,q}(v_{s})y_{p}\overline{y}_{q}) .

Thus if we set

c_{rrpq} = a_{pq}(v_{r})

and if r ≠ s then

c_{rspq} = a_{p,q} (\frac{1}{\sqrt{2}}(v_{r} + v_{s}) − \frac{1}{2}(a_{p,q}(v_{r}) + a_{p,q}(v_{s}))) +

a_{p,q} (\frac{1}{\sqrt{2}}(v_{r} + iv_{s}) − \frac{1}{2}(a_{p,q}(v_{r}) + a_{p,q}(v_{s})))

Then

f(v ⊗ u) = ∑_{rspq} c_{rspq}x̄_{r}ȳ_{p}x_{s}y_{q}.

This is the content of the theorem.

We will now give a counterexample to the analogous assertion for product bases.

Proposition 5. Let H_{1} and H_{2} be finite dimensional Hilbert spaces of dimension greater than 1. Then there exists f : Σ(H_{1}, H_{2}) → [0, ∞) such that ∑_{i,j} f(u_{i} ⊗ v_{j}) = w, with w ∈ ℝ fixed, for all choices {u_{i}} and {v_{j}} of orthonormal bases of H_{1} and H_{2} respectively but there is no linear endomorphism, T, on H_{1}⊗H_{2} such that f(u ⊗ v) = ⟨u ⊗ v|T |u ⊗ v⟩ for u ∈ S(H_{1}) and v ∈ S(H_{2}).

Proof. Let for w > 0, P_{w} denote the set of all Hermitian positive semi-definite endomorphisms, A, of H_{2} such that tr(A) = w. Fix w_{o} = \frac{w}{dim H_{1}} . Let φ : S(H_{1}) → P_{w_{o}} be a mapping (completely arbitrary). Set f(u⊗v) = ⟨v|φ(u)|v⟩, for u ∈ S(H_{1}) and v ∈ S(H_{2}) . If {u_{i}} is an orthonormal basis of H_{1} and if {v_{j}} is an orthonormal basis of H_{2} then

∑_{i,j} f(u_{i} ⊗ v_{j}) = ∑_{i} (∑_{j} ⟨v_{j}|φ(u_{i})|v_{j}⟩) = ∑_{i} tr(φ(u_{i})) = dim(H_{1})w_{o}.

Note: In this argument only one factor need be finite dimensional. Also note that f can be chosen to be continuous.

3. Unentangled Bases

In this section we will develop the material on “unentangled bases” that we will need to prove Theorem 3 (in fact as we shall see a generalization). Let V be a 2-dimensional Hilbert space and let H be an n-dimensional Hilbert space. Fix Σ ⊂ S(H) such that λΣ = Σ for all λ ∈ ℂ with |λ| = 1. We will use the notation S(V) ⊗ Σ = {v ⊗ w|v ∈ S(V), w ∈ Σ}.

If a ∈ S(V) then up to scalar multiple there is exactly one element of S(V) that is perpendicular to a. We will denote a choice of such an element by â. The main result of this section is

Theorem 6. If {u_{j}}_{j=1}^{2n} is an orthonormal basis of V ⊗ H with u_{j} ∈ S(V) ⊗ Σ for j = 1, ..., 2n then there exists a partition

n_{1} ≥ n_{2} ≥ ... ≥ n_{r} > 0

of n, an orthogonal decomposition

H = U_{1} ⊕ ... ⊕ U_{r},

elements a_{1}, ..., a_{r} ∈ S(H), and for each i = 1, ..., r orthonormal bases {b_{i1}, ..., b_{in_{i}}} and {c_{i1}, ..., c_{in_{i}}} of U_{i} such that

{u_{i}| i = 1, ..., 2n} = ⋃_{i=1}^{r} ({a_{i} ⊗ b_{ij}| j = 1, ..., n_{i}} ∪ {â_{i} ⊗ c_{ij}| j = 1, ..., n_{i}}) .

Before we prove the theorem we will make several preliminary observations. Let {u_{i}} be as in the statement of the theorem. Then each u_{i} = a_{i} ⊗ h_{i} with a_{i} ∈ S(V) and h_{i} ∈ Σ.

1. For each i there exists j such that a_{j} is a multiple of â_{i}.

If not then we would have ⟨a_{i}|a_{j}⟩ ≠ 0 for all j. Since ⟨a_{i} ⊗ h_{i}|a_{j} ⊗ h_{j}⟩ = ⟨a_{i}|a_{j}⟩⟨h_{i}|h_{j}⟩, ⟨h_{i}|h_{j}⟩ = 0 for all j ≠ i. This implies that {u_{j}}_{j≠i} ⊂ V ⊗{h_{i}^{⊥}}. This space has dimension equal to 2(n − 1). So it could not contain 2n − 1 orthonormal elements. This contradiction implies that assertion 1. is true.

2. Assume that i ≠ j. If ⟨a_{i}|a_{j}⟩ ≠ 0 then ⟨h_{i}|h_{j}⟩ = 0. If ⟨h_{i}|h_{j}⟩ ≠ 0 then ⟨a_{i}|a_{j}⟩ = 0.

This is clear (see the proof of 1.)

We will now prove the theorem by induction on n. If n = 1 the result is trivial. We assume the result for all H with dim H < n and all possible choices for Σ. We now prove it for n.

For each i let m_{i} denote the number of j such that a_{j} is a multiple of a_{i}. Let m = max{m_{i}|i = 1, ..., 2n}. If we relabel we may assume that the first m of the a_{i} are equal to a_{1} (we may have to multiply h_{i} by a scalar of norm 1). By 1. above we may assume that the next k of the a_{i} are equal to â_{1} with 1 ≤ k ≤ m and if i > m+k then a_{i} is not a multiple of either a_{1} or â_{1}. This implies by 2. above that ⟨h_{i}|h_{j}⟩ = 0 for j > m + k and i = 1, ..., m. Also {h_{1}, ..., h_{m}} is an orthonormal set. Thus u_{i} ∈ V ⊗({h_{1}, ..., h_{m}}^{⊥}) for i > m+k. This implies that V ⊗({h_{1}, ..., h_{m}}^{⊥}) contains 2n − (m + k) orthonormal elements. Since dim V ⊗ ({h_{1}, ..., h_{m}}^{⊥}) = 2(n − m) this implies that k = m. We now rewrite the first 2m elements of the basis as

a_{1} ⊗ b_{1}, ..., a_{1} ⊗ b_{m}, â_{1} ⊗ c_{1}, ..., â_{1} ⊗ c_{m}.

If we apply observation 2. again we see that the elements h_{i} for i > 2m must be orthogonal to {b_{1}, ..., b_{m}} and to {c_{1}, ..., c_{m}}. A dimension count says that they must span the orthogonal complements of both {b_{1}, ..., b_{m}} and {c_{1}, ..., c_{m}}. But then {b_{1}, ..., b_{m}} and {c_{1}, ..., c_{m}} must span the same space, U ⊂ H. We have therefore shown that {u_{i}}_{i>2m} is an orthonormal basis of V ⊗ U^{⊥}. We may thus apply the inductive hypothesis to U^{⊥} and Σ ∩ U^{⊥}. This completes the inductive step and hence the proof.

If W is a Hilbert space and if Ξ is a subset of S(W) that is invariant under multiplication by scalars of absolute value 1 then a function f : Ξ → ℂ is said to be a Ξ-frame function of weight w = w_{f} if whenever {u_{i}} is an orthonormal basis of W with u_{i} ∈ Ξ (i.e. {u_{i}} is a Ξ-frame) we have ∑_{i} f(u_{i}) = w. We note

3. Let f be a Ξ-frame function. If {u_{i}} is a Ξ-frame for W and if F is a subset of {u_{i}} then f_{|F^{⊥}∩Ξ} is a F^{⊥} ∩ Ξ-frame function of weight w_{f} − ∑_{u_{i}∈F} f(u_{i}).

This is pretty obvious. Let {v_{j}} be a Ξ ∩ F^{⊥}-frame for F^{⊥}. Then {ν_{j}} ∪ F is a Ξ-frame for W.

Proposition 7. Let V be a two dimensional Hilbert space and let H be an n-dimensional Hilbert space. Let Σ ⊂ S(H) be as in the rest of this section and let g : S(V) → ℂ and h : Σ → ℂ be respectively a frame function and a Σ-frame function. Then if f(v ⊗ w) = g(v)h(w) for v ∈ S(H) and w ∈ Σ then f is an S(V) ⊗ Σ-frame function of weight w_{g}w_{h}.

Proof. Let {u_{i}} be an S(V) ⊗ Σ-frame. Then Theorem 5 implies that we may assume that there is partition n_{1} ≥ n_{2} ≥ ... ≥ n_{r} > 0 of n and elements a_{i}, b_{ij} and c_{ij} as in the statement so that

{u_{i}} = ⋃_{i=1}^{r} ({a_{i} ⊗ b_{ij}|j = 1, ..., n_{i}} ∪ {â_{i} ⊗ c_{ij}|j = 1, ..., n_{i}}) .

Thus

∑_{i} f(u_{i}) = ∑_{i} g(a_{i}) ∑_{j=1}^{n_{i}} h(b_{ij}) + ∑_{i} g(â_{i}) ∑_{j=1}^{n_{i}} h(c_{ij}).

Observation 3. above implies that for each i we have ∑_{j=1}^{n_{i}} h(b_{ij}) = ∑_{j=1}^{n_{i}} h(c_{ij}). Now g(a_{i}) + g(â_{i}) = w_{g}. Hence since {b_{ij}} is a Σ-frame the result follows.

Theorem 3 is an immediate consequence of the above proposition.

4. Entangled subspaces.

Let H_{1}, ..., H_{n} be finite dimensional Hilbert spaces and set H = H_{1} ⊗ H_{2} ⊗ · · · ⊗ H_{n}. If V ⊂ H is a subspace than we will say then V is entangled if whenever v ∈ V and v ≠ 0 then v is entangled (i.e. v cannot be written in the form v = h_{1} ⊗ h_{2} ⊗ ⋯ ⊗ h_{n} for any choice of h_{i} ∈ H_{i}). The purpose of this section is to give a proof of Theorem 4 using basic algebraic geometry. That is, we will prove that

dim V ≤ dim(H_{1}) ⋯ dim(H_{n}) − ∑(dim H_{i} − 1) − 1

and that this estimate is best possible. The reader should consult [Hartshorne] for the algebraic geometry used in the proof of this result.

Let L = {λ ∈ H^{∗}| λ(V) = 0} (H^{∗} the complex dual space of H). Let X = {h_{1} ⊗ ⋯ ⊗ h_{n} | h_{i} ∈ H_{i}}. We consider the map Φ : H_{1} × ... × H_{n} → X given by Φ(h_{1}, ..., h_{n}) = h_{1} ⊗ ⋯ ⊗ h_{n}. Then Φ is a surjective polynomial mapping. If we denote by \overline{Φ} the corresponding mapping of projective spaces we have \overline{Φ} : P (H_{1})×...×P (H_{n}) → P (H). General theory implies that the image of \overline{Φ} is Zariski closed in P (H). Since X is clearly the cone on that image we see that X is Zariski closed and irreducible. Also the map \overline{Φ} is injective so the dimension over ℂ of its image is ∑(dim H_{i} −1). Thus the dimension over ℂ of X is d = ∑(dim H_{i} −1)+1.

Since V is entangled X ∩ V = {0}. This implies that {x ∈ X ||λ(x) = 0, λ ∈ L} = {0}. Thus dim L ≥ dim X = d. Hence dim V = dim H − dim L ≤ dim H − d. This is the asserted upper bound. The fact that this upper bound is best possible follows from the Noether normalization theorem which implies that there exist λ_{1}, ..., λ_{d} ∈ H^{∗} such that {x ∈ X ||λ_{i}(x) = 0 for all i} = {0} (i.e. a linear system of parameters).

References

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[Fuchs] C. A. Fuchs, private communication.

[BDMPSST] C. H. Bennett, D. P. DiVincenzo, T.Mor, P. W. Shor, J. A. Smolin, B. M. Terhal, Unextendible product bases and bound entanglement, Phys. Rev. Lett. 82 (1999) 5385–5388.

[Hartshorne] R. Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics, 52, Springer-Verlag, New York, 1977.

(Nolan R. Wallach) University of California, San Diego E-mail address, Nolan R. Wallach: nwallach@ucsd.edu