# Assume QM is realistic but non-local: Explain superposition?

1. Jan 22, 2014

### greypilgrim

Hi,

The violation of Bell's inequality says that quantum mechanics can't be both local and realistic. Let's assume it is realistic but non-local. How does this explain the fact that a single particle can be in a superposition that collapses to a particular state when measured? Since we only consider a single particle at a single location I can't see how non-locality could be useful here.

To put it differently, I don't quite understand what the violation of Bell's inequality implies (if anything) if we don't look at entangled states.

2. Jan 22, 2014

### .Scott

Wouldn't assuming that a particle is always at some specific location - even when it is not being measured - be a presumption of locality?

But what a violation of the Bell inequality would show is that the results of the detection events were determined, in part, by non-local causes - that is causes that were too distant and too close to simultaneous for light to signal the cause condition to the resulting event.

This sounds like faster-than-light communication, but its only a statistical effect. No revealable information is actually communication.

The purpose of the Bell Inequality is not to directly demonstrate wave/particle duality. It is only to demonstrate that something that violates locality such as QM is required.

3. Jan 22, 2014

### Maui

I am not aware of superpositions being mentioned in the only truly realistic interpretation - the DeBB. They are superceded by a guiding wave and the popular opinion on this forum is that superpositions are not real(contrary to some ongoing quantum computing projects), so no problem exists for the BI. It's your call what you want to believe.

4. Jan 22, 2014

### Pio2001

Hi !

Not exactly. It says that it can't be both local and deterministic.
Realism is a rather philosophical concept.

It depends. If you are talking about a superposition of two different positions in space, in a double slit, for example, non-locality explains that when it is clear that one slit is open and the other closed, the particle fully meterializes in one slit and fully disappears from the other.

If you are talking about two possible states located in the same place, then non-locality is not required, but remains compatible with the collapse.

Basically the same as Newton's laws of gravity when you don't look at falling objects, I suppose.

Many-worlds and transactionnal interpretations are also compatible with "true realism".

5. Jan 22, 2014

### Maui

You need to give it some more thought - true realism requires causal, macro determinism which is not compatible with the MWI and the TI(MWI is sort of quantum realistic but classical realism and causality are emergent). Of course, the ONLY truly macro and micro realistic(and deterministic) interpretation that exists today is the MWI.

You've been reading the wrong kind of realism. Realism in physics has a different meaning.

Last edited: Jan 22, 2014
6. Jan 22, 2014

### DrChinese

There is that "realism", true, but as Maui says it has a more specific meaning with regards to QM interpretations.

It really goes back to EPR's definition of "elements of reality" for a quantum particle. Realism is then the idea that non-commuting elements of reality exist simultaneously and have well defined values (not in a superposition).

7. Jan 22, 2014

### greypilgrim

Yes, and my question is if we stick to realism (and thereby assume non-locality), how can we even write a state like
$$\left|\psi\right\rangle=\alpha\left| 0\right\rangle+\beta\left| 1\right\rangle$$
and use it to calculate expectation values which we can effectively measure in experiments?

Or do realistic theories (like Bohmian mechanics) use different maths? I was under the impression that all interpretations of QM use the same Hilbert-space-based mathematical formalism that allows superposition, but I never really looked into Bohmian mechanics.

8. Jan 22, 2014

### DrChinese

The BMers have a bit of sleight of hand to get around this. They have a "quantum equivalence hypothesis/principle" that guarantees you can use all QM math. So your formula is good as is.

9. Jan 23, 2014

### greypilgrim

Ok, but you said in your earlier post that we have no superposition in a realistic theory. This state is obviously a superposition in the $\left| 0\right\rangle,\left| 1\right\rangle$ basis. So why can we do that?