Does Busch's Theorem Offer a Simplified Proof of Gleason's Theorem?

[Fredrik]

An effect, by definition, is a positive operator that belongs to a POVM. If E3 is an effect then another effect E (it may be zero) must exist such that E3 + E = I by this definition. Then of course by the fact they are mapped to probability ie since E3 + E is a two element POVM it has two outcomes ie its a two element event space, then f(E3) + f(E) = 1, again from the Kolmogerov axioms, or, if you want to use measure theory language, its a measure space of total measure 1 - which of course is exactly what the Kolmogorov axioms are. Similarly f(E1) + f(E2) + f(E) =1. Equating the two you end up with f(E1) + f(E2) = f(E3) by cancelling f(E).

I don't see why the E in f(E3) + f(E) = 1 should be the same as the E in f(E1) + f(E2) + f(E) =1. Also, to get that second equality, you must have used f(E1+E2)=f(E1)+f(E2), which is fine if E1 and E2 are part of the same POVM and therefore correspond to mutually exclusive outcomes, but what if they're not?

Regarding LaTeX - LaTeX is truth, but I find it far to time consuming and cumbersome so try to avoid it.

It's up to you, but it's really very easy. It takes a little bit of time the first and second time, but then it doesn't slow you down, except when you're using it to say things that you otherwise wouldn't. I think it would take you less time to learn these LaTeX codes than it did to write that last post above.

x^y
x^{yz}
E_1
E_{12}
\sin\theta
\cos{3x}
\sum_{k=1}^\infty x_k
\sqrt{1-v^2}
\frac{u+v}{1+uv}
\int_a^b f(x) dx
\mathbb R
\mathbb C

$$\begin{align}
&x^y\\
&x^{y+z}\\
&E_1\\
&E_{12}\\
&\sin\theta\\
&\cos{3x}\\
&\sum_{k=1}^\infty x_k\\
&\sqrt{1-v^2}\\
&\frac{u+v}{1+uv}\\
&\int_a^b f(x) dx\\
&\mathbb R\\
&\mathbb C\\
&\end{align}$$ We have a FAQ post on LaTeX if you decide to give it a try. You you can type something into a reply and just preview it if you want practice. https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[bhobba]

I don't see why the E in f(E3) + f(E) = 1 should be the same as the E in f(E1) + f(E2) + f(E) =1.

We assume E1 + E2 = E3 and all three are effects. Since E3 is an effect by definition it must be part of a POVM ie some ∑Ui = 1, Ui positive operators. WOLG (without loss of generality) we can assume E3 = U1 so E3 + ∑Ui = 1 where the Ui are summed from 2. We let E = ∑Ui so E3 + E = 1. E is obviously also an effect since E3 + E is a POVM. Now since E3 = E1 + E2 we have E1 + E2 + E = 1.

Also, to get that second equality, you must have used f(E1+E2)=f(E1)+f(E2), which is fine if E1 and E2 are part of the same POVM and therefore correspond to mutually exclusive outcomes, but what if they're not?

In my basic assumption I have assumed the probability of outcome i depends only on the Ei its mapped to, in particular I have assumed it does not depend on what POVM it is part of ie this is the assumption of non-contextuality which is the real key to Gleason in either variant. This means f(E), which I have defined as the probability of E, is the same whether E is part of the POVM E1 + E2 + E = 1 or E is part of the E3 + E = 1 POVM.

Now since the E1 + E2 + E POVM has three outcomes and one of those outcomes must occur, f(E1) + f(E2) + f(E3) = 1 and similarly for the POVM E3 + E we have f(E3) + f(E) = 1. Equating and cancelling f(E) we have f(E1) + f(E2) = f(E3).

It's up to you, but it's really very easy.

Must get around to it when I get a bit of time.

Have few things to do today so will leave it to a bit later to look at your other issues in the main thread.

Thanks
Bill

 

[naima]

To Fredrik,

Remember that Busch's theorem needs no choice of topology on Effects to be true. Only sigma additivity.

 

[Fredrik]

Remember that Busch's theorem needs no choice of topology on Effects to be true. Only sigma additivity.

When we're dealing with classical probability measures, the σ-addititivity condition looks like this: $\mu\big(\bigcup_{i=1}^\infty E_i\big)=\sum_{i=1}^\infty\mu(E_i)$. Here $(E_i)_{i=1}^\infty$ is a pairwise disjoint sequence of sets. When we're dealing with probability measures on the set of effects, the σ-addititivity condition looks like this: $\mu\big(\sum_{i=1}^\infty E_i\big)=\sum_{i=1}^\infty\mu(E_i)$. The sum on the left is defined as the limit of the sequence of partial sums, and the limit is defined using a topology.

I don't think this is a big issue, since $\mathcal E(\mathcal H)$ is a subset of $\mathcal B(\mathcal H)$, which has several useful topologies. The word "several" in that sentence is perhaps something to be concerned about. I haven't really thought that through.

 

[naima]

The core of the proof is that the dual of compact operators is the set of trace class operators

http://en.wikipedia.org/wiki/Trace_class#Trace_class_as_the_dual_of_compact_operators