# Kochen-Specker As Simple Corollary Of Gleasons Theroem

1. Jan 26, 2012

### Staff: Mentor

I have been mucking around with Gleasons therom of late and noticed it was mentioned in some literature it has the Kochen-Specker theorem as a corollary and thought what the heck lets see if I can prove it. It was trivial which started me thinking why it is not presented that way in textbooks. All I can think is that maybe it is thought the proof of Gleasons is a bit hard. However in recent times a quite simple proof based on POVM's has been found. So maybe I am missing something. Anyone got any other ideas or is it simply the newer proof of Gleasons theorem haven't filtered down to the textbooks yet?

Thanks
Bill

2. Jan 26, 2012

### Fredrik

Staff Emeritus
The proof of Gleason's theorem doesn't appear in any QM textbooks I know of, except "Geometry of quantum theory" by V.S. Varadarajan, which is one of the hardest books I've ever seen. Most people probably wouldn't even consider it a QM book, because of all the projective geometry stuff in it.

3. Jan 26, 2012

### Staff: Mentor

Yea the proof of Gleasons Theorem as originally done by Gleason is a ripper - I have been though it - groan. But a new much simpler version has been found - see the second chapter of:
http://kof.physto.se/theses/helena-master.pdf [Broken]

My suspicion from what you wrote is its the toughness of the older proof that is the reason its not done that way.

Thanks
Bill

Last edited by a moderator: May 5, 2017
4. Jan 27, 2012

### Fredrik

Staff Emeritus
Yes, I think your suspicion is right. But it's not just that the proof is hard. If that had been the only problem, the books could at least have stated the theorem without proof, and then proceeded to derive Kochen-Specker as a corollary. But to understand the statement of the theorem, you need to understand terms like "probability measure" and "lattice", and understand why the set of closed subspaces forms a lattice. Physics students don't know these things.

That thesis looks interesting. I might read it later. I have only studied the statement of the theorem, not its proof. I was a bit inspired by your post last night and decided to take a quick look at the POVM-based proof. http://arxiv.org/abs/quant-ph/9909073. (Another member at the forum mentioned it to me a few weeks ago). It's really short, but I didn't try to understand it right away, because I didn't understand what he meant by "effects". I got the impression that the book he coauthored, "Operational quantum physics", would be a good place to read about those things, so I may have to read a chapter or so of that book first. (I know almost nothing about POVMs. I recently started a thread where I asked for suggestions about what I should read to understand them, and got a lot of suggestions, but I haven't studied any of them yet).

Have you studied this proof? Since I have only had a quick look at it, I obviously don't understand it, but one thing bothers me about it. Gleason's theorem finds all the probability measures on the lattice of closed subspaces. The first thing Busch does is to make the domain larger, so apparently we're now trying to find all probability measures on a larger lattice (it's still a lattice, right?), the lattice of all "effects". So do we really end up with Gleason's theorem, or is this just a similar theorem?

5. Jan 27, 2012

### Staff: Mentor

I have studied both proofs and interestingly you don't really need to know concepts like probability measure, lattice etc - they are not that hard really - but you don't need to know it.

If you imagine a finite dimensional (for simplicity) complex vector space partitioned into a number of sub-spaces by a set of projection operators Ei (ie as resolution of the identity or PVM - Projection Value Measure) then you know for sure that the outcome of some observation must be a vector that lies in one of those sub-spaces so you can associate a probability with each of the projection operators. A probability measure is simply an assignment of probabilities ie a number between zero and one such that if what you define it over is disjoint then the probabilities are additive. Since the Ei are disjoint the probability associated with them is additive. This probability measure in Gleason parlance is known as a frame function denoted by f. What Gleasons theorem says is there is only one way to define a probability on such a vector space - the standard trace definition of QM which is a generalization of Born's rule

Now any Hermition operator A can be uniquely expressed as A = sum Yi Ei where each Ei is a disjoint projection operator and allows us to extend f to Hermition operators by defining it as f(A) = sum Yi f(Ei). From Hermtian operators you can extend it to all operators since any operator can be expressed as the sum of a Hermtian operator and an imaginary one (ie a Hermtian operator multiplied by i). This would solve the problem except for one thing - you need to prove f is linear. That is the real hard part and Gleasons proof of it is a real doosey - and justifiably famous. He does it starting with POV's and its a long hard slog.

However what some bright spark noticed is if you start from what is called POVM's (Postive Value Measures) instead of PVM's then the proof is a lot easier. A POVM is a generalization of a PVM. A PVM is a resolution of the identity ie disjoint projection operators Ei such that Sigma Ei = 1. We have each of the Ei assigned a probability. A POVM is a set of positive operators Fi such that sigma Fi =1 - however the Fi (called Effects) do not have to be disjoint - and it is easy to show that their eigenvalues are less than or equal to one. From the extension of the frame function you can see they all have frame functions less than one and if additive should add up to one. But how do you know its additive? Here is the trick. Suppose you are in a higher dimensional vector space and you have a PVM then if you project this onto a lower dimensional sub-space it forms a POVM in that sub-space and 'induces' a probability onto the POVM so it adds up to one - ie you know they form a disjoint probability measure in the higher dimensional space - careful here - the projection can't be zero or it won't be as easy to see. Neat trick - but can this be done for all POVM's - yep - a neat theorem called Neumark's theorem says you can always do it. Basically you have now simplified the problem by physical insight. For further details check out:
http://www.theory.caltech.edu/people/preskill/ph229/notes/chap3.pdf

Yes the set of POVM's is larger than the set of PVM's but the trick is by using Neumark's Theorem you can find a higher dimensional space such it is the projection of a PVM and each effect can be assigned the frame function of the corresponding projection in the PVM.

To really get what is going on you need to go through the theorem in Chapter two of the link on Gleasons Theorem or the link you gave.

Thanks
Bill

Last edited: Jan 27, 2012