In Kochen-Specker's theorem, as presented e.g. in (Rajan 2017) or (Breuer 2002), we consider maps $\nu:\mathrm{Herm}(\mathcal H)\to\mathbb R$ that are algebra homomorphisms on compatible observables, that is, such that $\nu(AB)=\nu(A)\nu(B)$ and $\nu(A+B)=\nu(A)+\nu(B)$ whenever $[A,B]=0$. Among other things, this implies that $\nu(P)\in\{0,1\}$ for any projection operator $P$, that $\nu(I)=1$, and more generally that $\nu(A)\in\sigma(A)$ for any $A\in\mathrm{Herm}(\mathcal H)$, where $\sigma(A)$ denotes the spectrum of $A$.

In the context of Gleason's theorem, as presented e.g. in (Busch 1999), one considers frame functions, which are functions $f:\mathrm{Eff}(\mathcal H)\to[0,1]$ such that $\sum_k f(E_k)=1$ whenever $\{E_k\}_k\subset\mathrm{Eff}(\mathcal H)$ satisfies $\sum_k E_k=I$. Here, $\mathrm{Eff}(\mathcal H)\subset\mathrm{Pos}(\mathcal H)$ is the set of effects, that is, of positive operators $E$ such that $0\le E\le I$. The gist of Gleason's theorem is that frame functions extend uniquely to linear functionals $\mathrm{Lin}(\mathcal H)\to\mathbb C$, which are in bijection with quantum states via the standard isomorphism $\mathrm{Lin}(\mathcal H)^*\simeq \mathrm{Lin}(\mathcal H)$.

For the scope of this post, let me refer to the functions $\nu$ in Kochen-Specker's theorem as noncontextual assignments.

It seems clear that there are strong relations between noncontextual assignments and frame functions. Indeed, one can also easily see that any frame function satisfies $\sum_k f(P_k)=1$ for any complete set of orthogonal projections $\{P_k\}$, which is a property shared with noncontextual assignments.

One major difference between the two is that noncontextual assignments are also required to preserve multiplication on compatible observables. This implies that we have, for any projection $P$, $\nu(P)\in\{0,1\}$, whereas frame functions need only satisfy $f(P)\in[0,1]$.

Is this the only difference between noncontextual assignments and frame functions, or am I missing something? Does this mean that one can think of noncontextual assignments as a subset of frame functions? But if this is the case, why doesn't Kochen-Specker's theorem follow directly from Gleason's one?

Gleason's theorem characterises the set of frame functions. It tells us that any frame function $f$ has the form $f(A)=\mathrm{Tr}(\rho_f A)$ for some $\rho_f\in\mathrm{Herm}(\mathcal H)$. But such an $f$ clearly isn't "sharp" in the sense of Kochen-Specker: if $f(P)=1$ for some projection $P$, then $\rho_f=P$, and thus $f(Q)\notin\{0,1\}$ for some other projection $Q$, hence $f$ not being a noncontextual assignment.

Is this argument valid, or am I missing something?

  • $\begingroup$ I do not find appropriate the term "contextual assignment". If they are contextual, then they are also function of the observables you measure simultaneously. The point is that maps satisfying your axioms (which therefore should named non-contextual assigments) does not exist. $\endgroup$ – Valter Moretti Feb 18 at 8:58
  • $\begingroup$ @ValterMoretti yes you are right, I got confused between "contextual" and "non-contextual". These functions should be named non-contextual, I'll edit $\endgroup$ – glS Feb 18 at 9:55

There are many versions of the KS theorem, the one with linear and multiplicative functionals and the other with orthogonal projectors (there is no measure on the lattice of projectors taking only sharp values) are almost immediate essentially topological consequences of Gleason's theorem (the projection version is valid also in the infinite dimensional case without continuity requirements). This implication was known to Bell in particular, who provided an alternative statement of KS theorem with a proof based on a weak version of Gleason's theorem. All that simultaneously (somebody says one year before) with the appearance of KS paper.

(The proof is as follows. Suppose there is a map associating every observable $A$ of $B(H)$ to a real number $\nu(A)$ and this map preserves the sum and the multiplication values (in obvious sense) of compatible observables. Restricting $\nu$ to the lattice $L(H)$ of orthoprojectors $P$, it turns out that (a) $\nu(P) \in \{0,1\}$ and (b) either it is the zero map (if $\nu(I)=0$) or it is a probability measure (if $\nu(I)=1$). If $2<\dim(H)<\infty$, then the Gleason theorem implies that this map (if it is not trivial) has the form $\nu(P) = tr(\rho P)$ for some operator $\rho$. Hence the map is continuous on the sphere $S$ of unit vectors (defining one-dimensional orthoprojectors). As the map is continuous and the sphere is connected, its image is connected: either $\nu(S)=\{0\}$ or $\nu(S)=\{1\}$. In summary, $\nu(P)=0$ for all one-dimensional projectors or $\nu(P)=1$ for all one-dimensional projectors. In the former case $\nu$ is trivial, in the latter case $\nu(I)>1$ that is impossible for probability measures.)

However this approach is not constructive. It does not show a (possibly minimal) set of observables where the non-contextual assignment of values is impossible, nor it point out how to determine this set. It just says that this set has to exist for $\dim(H)>2$. The "elementary" combinatoric proofs as the original one by Kochen and Speker are instead explicitly constructive. There are also KS-like statements unrelated with Gleason's theorem, but constructive, as the Milman-Peres magic square.

  • $\begingroup$ this line of proof not being constructive is a good point, but isn't it trivial to make it constructive? Let $P_\psi\equiv|\psi\rangle\!\langle\psi|$ and $\sqrt2|\phi\rangle\equiv|\psi\rangle+|\psi_\perp\rangle$ for some $|\psi_\perp\rangle$ with $\langle\psi_\perp|\psi\rangle=0$. Consider the frame function $f_\psi$ corresponding to $P_\psi$. From Gleason $f_\psi(A)=\mathrm{Tr}(A P_\psi)$. Then $f_\psi(P_\phi)=|\langle \phi|\psi\rangle|^2=1/2$, hence this is a non-contextual assignment (because non-contextual assignments can only give $0$ or $1$ on projectors). What am I missing? $\endgroup$ – glS Feb 18 at 10:04
  • $\begingroup$ The idea of a contructive proof in the "elementary proofs" is different: have a look at the obstruction of the Mermin-Peres magic square. There you try to fill in a table and in view of the concrete form of the observables and the constraints on the assigment, you find an obstruction. Your (correct) counterexample uses the theoretical fact that every assigmment is a density matrix and this arises from the theoretical analysis by Gleason. $\endgroup$ – Valter Moretti Feb 18 at 10:13
  • $\begingroup$ I agree with you that the way passing through Gleason's result is more appealing. (That is my viewpoint in chapter 5 of one book of mine springer.com/it/book/9783030183455#aboutBook. ) $\endgroup$ – Valter Moretti Feb 18 at 10:13
  • $\begingroup$ thank you for the pointer. I was not aware of that reference; it is quite appreciated, especially because I found a lot of the existing literature on these topics to be quite confusing/inconsistent. If I understand, your "dispersion free probability measure" in Def 5.1 would be a frame function that is also a noncontextual assignment, correct? Still, thinking about this some more, is it true that any noncontextual assignment is also a frame function? If not, KS's thm might be more general, or at least incomparable with Gleason's. $\endgroup$ – glS Feb 18 at 12:45
  • $\begingroup$ I guess the question becomes essentially whether noncontextual assignments are also linear on POVMs. If $A+B=I$ then $[A,B]=0$, thus the answer would be positive, but for POVMs with more than two operators I'm not so sure $\endgroup$ – glS Feb 18 at 12:45

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