Wondering about QM whether things are actually all deterministic?

[Pio2001]

When you say that something "is done in A" you probably mean that there is an object C (probably a physicist) that interacts with the subsystem A.

Not necessarily. For example the charges in A can emit an electromagnetic wave by themselves if they are in an initial configuration that has enough potential energy.

But those laws require that the motion of any point charge/mass is determined by the resultant field of both subsystems, A and B so that they are not and they will never become independent.

That's true in electrostatic. In electrodynamics, those laws require that electromagnetic waves travel at a speed equal to c.

 

[jeebs]

good grief... it's going to be years before I have any real appreciation of half of this thread. the phrase "can of worms" comes to mind.

 

[Pio2001]

For the interactions involved, the final relative state of the joint (Alice, Alice's particle, Bob, Bob's particle) system is completely determined by its initial state, is it not?

The mechanism that transforms $$|O\rangle_i$$, the observer and its measurement device in its initial orientation, into $$|O\phi\rangle_i$$, the observer and its measurement device in its final orientation $$\phi$$, is not explicit, but yes, we can assume that this is the result of a quantum evolution (with myriads of other sub-universes created meanwhile).

And in regards to the local structure of space-time, the final state of your experiment is completely determined from information contained in the past light-cone of the events over which the experiment takes place.

Yes, but since by definition of an EPR setup the space-time region into which the "events take place" is basically a space-like slice of space-time (the reunion of the spatially separated A and B regions), looking at its past light cone is not relevant. We can demonstrate that if these events are described by a state vector, then the sub-regions A and B of this space-time region evolve in violation of locality :

After the emission of the particles and before they reach Alice and Bob, our quantum state is

$$|O\rangle_1 \otimes |O\rangle_2 \otimes (|+-\rangle - |-+\rangle )$$

Then, Alice and Bob choose new orientations for their detectors. Our state splits because of all the quantum events that turning the devices involve, but one of the resulting copies is

$$|O\alpha\rangle_1 \otimes |O\beta\rangle_2 \otimes (|+-\rangle - |-+\rangle )$$

The evolution is local.

Then, after the particles has entered the detectors (and have been destroyed), but before the future light-cones of each measurement meet, our new state is

$$f_{++}(\alpha, \beta)(|O\alpha +\rangle_1 \otimes |O\beta +\rangle_2)$$
$$+f_{+-}(\alpha, \beta)(|O\alpha +\rangle_1 \otimes |O\beta -\rangle_2)$$
$$+f_{-+}(\alpha, \beta)(|O\alpha -\rangle_1 \otimes |O\beta +\rangle_2)$$
$$+f_{--}(\alpha, \beta)(|O\alpha -\rangle_1 \otimes |O\beta -\rangle_2)$$

The description of the observer 1 (Alice) has become a function of $$\beta$$ (the angle chosen by Bob), that is itself the result of the evolution of $$|O\rangle_2$$ into $$|O\beta\rangle_2$$, that occurred outside its past light-cone.

 

[Pio2001]

Said more properly, the non-separable object that Alice and Bob represents evolves in a time t that is inferior to x/c, with x being its spatial extention, which is a violation of special relativity.

 

[Hurkyl]

The description of the observer 1 (Alice) has become a function of $$\beta$$ (the angle chosen by Bob),

What you wrote is^* a description of the joint state of both Alice and Bob.

To get a description of Alice only, you have to take a partial trace to eliminate the second component. If I've computed correctly, this is a statistical mixture of (the states named by)

• $$f_{++}(\alpha, \beta) |O\alpha+ \rangle + f_{-+}(\alpha, \beta) |O\alpha- \rangle$$
• $$f_{+-}(\alpha, \beta) |O\alpha+ \rangle + f_{--}(\alpha, \beta) |O\alpha- \rangle$$

weighted with probability 50% on each state.

This could depend on beta. However, doesn't f_{\cdot \cdot}(\alpha, \beta) factor into g_{\cdot}(\alpha) g_{\cdot}(\beta)? In this case, the above state would be independent of \beta. (Because multiplying a ket by a scalar doesn't change what state it represents)


*: I'll take your word for it that the state is pure -- I hate computing partial traces

 

[Pio2001]

You're right, the confusion of Alice's state with the joint state of Alice and Bob should be avoided.

That's why I prefer the second formulation : the joint state completely evolves in a time t, that is strictly inferior to x/c, where x is the distance between Alice and Bob : this is non local.

 

[Hurkyl]

the joint state completely evolves in a time t, that is strictly inferior to x/c, where x is the distance between Alice and Bob : this is non local.

The evolution of their joint state over this region of space-time depends only on what's going on in the union of the two laboratories, correct? So the evolution of this state which is spread out across the two laboratories depends on nothing that is outside the past light-cone of this region. This is consistent with locality, not a proof of non-locality!

It is true that there is one part of this region (Alice's laboratory) whose past light-cone fails to contain other parts of this region (e.g. Bob's laboratory), and the fact that the joint state depends on the entirety of the region. But you only look at the global state; you can't have a reasonable claim that states evolve non-locally if you don't look at localized portions of the state.

Now, relative states are not a MWI-specific thing; they are basic tools of the mathematics of quantum mechanics. We have them because they are useful. They make a host of issues obvious, such as why Bob cannot use their entangled particles to send a FTL signal to Alice. (Any measurement Alice can perform depends only on the relative state of her half of the entangled pair) You're free to ignore them if you wish, but I think you're hurting yourself.

 

[Pio2001]

The evolution of their joint state over this region of space-time depends only on what's going on in the union of the two laboratories, correct? So the evolution of this state which is spread out across the two laboratories depends on nothing that is outside the past light-cone of this region.

Yes, I agree.

This is consistent with locality, not a proof of non-locality!

We can't tell anything about the locality or not of this situation : imagine that Alice sends a supraluminic message to Bob. She obviously violates locality.

However, all your constatations still hold : The evolution of their joint state over this region of space-time depends only on what's going on in the union of the two laboratories, and on nothing that is outside the past light-cone of this region.
...Except that in this particular case, there is no way we can tell that "this is consistent with locality".

You're free to ignore them if you wish, but I think you're hurting yourself.

Sorry, that's not stubbornness from me, just ignorance

 

[Hurkyl]

We can't tell anything about the locality or not of this situation : imagine that Alice sends a supraluminic message to Bob. She obviously violates locality.

The result is consistent with locality, but of course does not prove it. (That's why I was careful to say "consistent with locality", not "proves locality")

But in your example, if we looked at Bob's relative state, we would see a time evolution that is not determined by the conditions in Bob's laboratory, which would disprove locality.

We only see that your example is non-local by looking at the localized bits.

 

[ueit]

Not necessarily. For example the charges in A can emit an electromagnetic wave by themselves if they are in an initial configuration that has enough potential energy.

That's true in electrostatic. In electrodynamics, those laws require that electromagnetic waves travel at a speed equal to c.

For moving charges you must integrate over their past positions taking into account the finite speed of light. This means A will be dependent on the past state of B and B on the past state of A. However, given the fact that the evolution of both A and B is deterministic it follows that A and B cannot be independent.

 

[Pio2001]

In the hypothesis "If A and B are two space-time regions separated by a space-like interval, then nothing that is done in A can have an effect in B and conversely", A and B are not place, nor objects. They are space-time regions. The "past of A" is a space-time region that is completely outside of A.

What you are saying is that A depends partly on some parts of the past light-cone of B. Right, but A itself is completely independant of B.
Once a signal emitted in B reaches "an object that took part in the A space-time event", A is over.

you can't have a reasonable claim that states evolve non-locally if you don't look at localized portions of the state.

Can't we say that writing the state with kets describing eigenstates of Alice or Bob, knowing that they were in A and B respectively, and knowing that they didn't move faster than light, is itself a way of looking at localized portions of the state ?

 

[ueit]

In the hypothesis "If A and B are two space-time regions separated by a space-like interval, then nothing that is done in A can have an effect in B and conversely", A and B are not place, nor objects. They are space-time regions. The "past of A" is a space-time region that is completely outside of A.

What you are saying is that A depends partly on some parts of the past light-cone of B. Right, but A itself is completely independant of B.
Once a signal emitted in B reaches "an object that took part in the A space-time event", A is over.

Perhaps it would be better if you could propose a way in which your assumptions could be fulfilled in practice for two systems of point charges.

 

[Pio2001]

Exactly the way you stated above : For moving charges you must integrate over their past positions taking into account the finite speed of light.

It seems to me that this is the meaning of the locality hypothesis used in Bell's theorem : the movement of a charge in A $$(x_A, y_A, z_A, t_A)$$ can't affect the movement of a charge in B $$(x_B, y_B, z_B, t_B)$$.
It can however affect the movement of a charge at $$(x_B, y_B, z_B)$$, but only at time $$t > t_B$$.

 

[ueit]

Exactly the way you stated above : For moving charges you must integrate over their past positions taking into account the finite speed of light.

It seems to me that this is the meaning of the locality hypothesis used in Bell's theorem : the movement of a charge in A $$(x_A, y_A, z_A, t_A)$$ can't affect the movement of a charge in B $$(x_B, y_B, z_B, t_B)$$.
It can however affect the movement of a charge at $$(x_B, y_B, z_B)$$, but only at time $$t > t_B$$.

I agree on that. The issue is not the locality assumption but the independence assumption. If each system depends on the past state of the other system they cannot be said to be independent. The motion of Earth at t does not depend on the position of the Sun at t but at t-8 min. However, we do not conclude that the motions of the two bodies are independent. The reason we cannot go from locality to independence is the deterministic structure of the theory. Saying that A depends on the past of B is no different than saying that A depends on B.