# Kochen-Specker Theorem Assumptions

1. May 7, 2015

### msumm21

I read over a high-lever description of this theorem (http://en.wikipedia.org/wiki/Kochen–Specker_theorem) and was wondering how reasonable the assumptions are. Specifically, is it reasonable to assume that the value of a sum of observables is the same as the sum of the values of the individual observables? I just noticed it's only required for commuting observables, so do you know of any simple tests of this assumption? If no simple ones, then a complicated one?

2. May 7, 2015

### fzero

In standard QM (Copenhagen), it is true whether the operators are commuting or not because the observables are linear operators.

3. May 7, 2015

### Staff: Mentor

Kochen Sprecker is in fact a simple corollary to Gleason's theorem:
http://kiko.fysik.su.se/en/thesis/helena-master.pdf

For the modern version of that theorem and its proof, as well as exactly what assumptions are used, see the following (post 137):

The main assumption is non-contextuality - but there is also the implicit assumption of the strong superposition principle (its that any hermitian operator, in principle, is an observable) - although many QM proofs assume it without explicitly stating it as an assumption.

The key part of the proof is proving expectations are linear ie given any two observables A and B E(a*A+b*B) = a*E(A) + b*E(B).

It has an interesting history in that the great polymath, John Von Neumann, in his very influential - Mathematical Foundations Of Quantum Mechanics - came up with a proof that hidden variable theories were not possible. It assumed that without realising such an assumption didn't apply to hidden variable theories. A female mathematician, Grete Herman, spotted it, but was ignored:
http://en.wikipedia.org/wiki/Grete_Hermann

Not sciences finest hour.

Thanks
Bill

Last edited: May 8, 2015
4. May 7, 2015

### msumm21

fzero, I think that's a bit different, right? The observables being linear operators acting on states doesn't imply the measurement result is linear on the observables does it?

5. May 8, 2015

### Staff: Mentor

It's expectations being linear - not the operators - which are linear by definition.

Its easy to prove Born's rule (from which Kochen-Sprecker follows) from linearity.

Note for any operator O we have <bi|O|bj> = Trace (O |bj><bi|). O = ∑ <bi|O|bj> |bi><bj| = ∑ Trace (O |bj><bi|) |bi><bj|

Applying the linearity assumption:

E(O) = ∑ Trace (O |bj><bi|) E(|bi><bj|) = Trace (O ∑ E(|bi><bj|)|bj><bi|)

Define P as ∑ E|bi><bj|)|bj><bi| and we have E(O) = Trace (OP) which is the Born rule.

Thanks
Bill

Last edited: May 8, 2015
6. May 8, 2015

### aleazk

Last edited: May 8, 2015
7. May 12, 2015

### msumm21

In the Wikipedia article I mentioned it said the "values of the observables" which I took to mean the value in one run of the an experiment (instead of the average value over many runs). If we measure 4 compatible observables O1, O2, O3, O4 and the associated operators satisfy I=ΣOi then the measured values vi must satisfy Σvi=1. Is this not the assumption? It at first seems like this must be the assumption for the subsequent proof to work. On the other hand, I didn't think QM ever said Σvi=1 in one experiment, only that this must be true over many runs of the experiment (i.e. if we run the experiment 1000 times, 250 times measuring each Oi, then we should get v=1 in about 25% of the time).

Last edited: May 12, 2015
8. May 12, 2015

### Staff: Mentor

That only follows if the observables have the same eigenvectors.

The assumptions of Kochen-Sprecker is exactly the same as Gleason, since KS is a simple corollary to Gleason, but Gleason is a much more powerful result.

First you need to know what a Positive Operator Value Measure (POVM) is. A POVM is a set of positive operators Ei ∑ Ei =1 from, for the purposes of QM, an assumed complex vector space.

The key axiom used in the proof of the modern version of Gleason is:

An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

Only by Ei means regardless of what POVM the Ei belongs to the probability is the same. This is the assumption of non contextuality, and is the well known rock bottom essence of Born's rule via Gleason, and hence KS. The other assumption, not explicitly stated, but used, is the strong law of superposition ie in principle any POVM corresponds to an observation/measurement.

Thanks
Bill

9. May 13, 2015

### msumm21

Thanks Bill. I did see your earlier posts about KS following from Gleason's theorem, but for the short term I was just trying to understand the KS argument on Wikipedia for efficiency sake (when I have more time later I'd like to try to understand Gleason's theorem). Maybe the argument on Wikipedia is oversimplified to the extent that it requires shaky assumptions (or I'm still missing something there)?

Is this always true? The operators used in the Wikipedia article are the projection operators onto orthogonal vectors (S1, S2, S3, and S4). So in this case the Si are eigenvector of all the operators Oi, but there's always a >= 3-fold degeneracy when a measurement gives 0. If O1 gives v1=0 then the quantum state could be any linear combination of S2, S3, and S4 right? So I still don't understand how they can assume Σvi=1.

It seems like the best we can say is that ∑vi<=2 for the projection operators (one run of the experiment). Does QM say more than this?

Last edited: May 13, 2015
10. May 13, 2015

### Staff: Mentor

The assumption the Wikipedia article made was:

'In order to do so it is sufficient to realize that, if u1, u2, u3 and u4 are the four orthogonal vectors of an orthogonal basis in the four-dimensional Hilbert space, then the projection operators P1, P2, P3, P4 on these vectors are all mutually commuting (and, hence, correspond to compatible observables, allowing a simultaneous attribution of values 0 or 1).'

They were commuting.

I will let you investigate what commuting means for eigenvectors.

I went through KS ages ago, but since its really just a special case of Gleason don't worry about it any more.

Thanks
Bill

11. May 13, 2015

### msumm21

Yes, I'm too rusty on the basics, I think I realize the problem now. One manifestation of what I was thinking was this
1. measure O1 and get 0
2. now we know the state is a linear combination of S2,S3,S4
3. measure O2 and get 0
4. now we know the state is a linear combination of S1,S3,S4
5. measure O3 and get 0
6. now we know the state is a linear combination of S1,S2,S4
7. measure O4 and get 0
8. now we know the state is a linear combination of S1,S2,S3

Looks like that's just not precise enough in 4,6 -- the state after measuring a degenerate eigenvalue is not just any linear combination of the eigenvectors, but the projection of the previous state onto this eigenspace (I forgot the latter). So even if 1,3, and 5 occurred this projection rule would then force the state to be S4 so that 7 could NOT happen.

Similarly, if v1=1 then immediate subsequent measurements of O2,O3,O4 must all give 0 by this projection rule (I was previously thinking only the NEXT measurement had to give 0, then the state would be any linear combination of the other three states and hence anything goes in the 3rd measurement O3).

Thanks!

12. May 13, 2015

### msumm21

Regarding this one, anyone know good references about why this would or would not be a good assumption?

13. May 13, 2015

### Staff: Mentor

Its a good assumption often used but not explicitly stated. But its an assumption.

Thanks
Bill