# I Can Bell's theorem contradict PBR ?

1. Dec 3, 2016

### jk22

I just suppose the Bell's Ansatz for the result of measurement to be $$A (\theta,\lambda)$$

Now the parameter lambda could be anything :

-a physical quantity like the polarization angle of the incoming photon
-the coordinate of a 'world'
- the whole wavefunction.
...
In the case of the wavefunction Bell's result is obtained and QM violates the inequalities even with this argument.

Does this mean that the wavefunction cannot be apprehended by the measurement apparatus A ?

Hence that it cannot be a measurable element of reality ?

2. Dec 3, 2016

### secur

The wavefunction is NOT measurable, regardless of its role in this Bell hidden-variable type ansatz. A measurement tells us one eigenvalue of the wavefunction, that's all. But in general it consists of a number of eigenvalues, each with its probability amplitude, including phase info. Actually that's assuming we have a basis, which is another complication. Also it can describe a number of subsystems with entangled relationships. All of that is lost when the measurement happens, all we get is one eigenvalue, the wavefunction being left in the associated eigenstate (ignoring degeneracy). So the whole wavefunction is definitely not measurable.

However we can prepare a large number of particles (ensemble) which are assumed to be all in the same state. Then, with multiple measurements, we obtain the whole wavefunction, with some uncertainty (standard deviation), which decreases as more measurements are made. Only in this statistical sense can the wavefunction be "measured".

Bottom line, you can't measure a wavefunction directly. Does that mean it's not an "element of reality"? You'd have to ask a philosopher about that, and you know what they're worth.

The wavefunction can be allowed as the HV, without the HV model violating Bell's inequality. The only caveat: if you believe in "instantaneous collapse" then the HV representing the wavefunction can't be allowed to instantaneously, nonlocally, update to reflect that collapse. Obviously.

3. Dec 3, 2016

### Staff: Mentor

The answer to the question in the title is clearly "No".
Bell's theorem states that all members of a certain class of theories must obey Bell's inequality - no more and no less. Quantum mechanics is not a member of that class, so there can be no contradiction with PBR, which is a theorem about quantum mechanics.
Bell's result is not obtained even if you choose to take the quantum mechanical wave function as $\lambda$. To get Bell's result we also need that the result at A can be written as $A(\vec{a},\lambda)$ not $A(\vec{a},\vec{b},\lambda)$ (and vice versa for the result at B), and this requirement is not satisfied by quantum mechanics, even if we take $\lambda$ to include the wave function. (This is why QM is not a member of the class of theories that must obey the inequality).

4. Dec 4, 2016

### jk22

If we put psi for lambda in the argument then Bell theorem were about averaging over all wavefunctions which of course gives Chsh <=2.

So I wonder if in this case i do not compare two different levels of averaging ?