Sidney Coleman's opinion on interpretation in his Dirac lecture

[Demystifier]

Ok, so for you an interpretation doesn't have a clear ontology unless the state of the systems is ontological (whether if is #\psi## or something else). Is that right?

Yes, you can put it this way.

 

[martinbn]

I mean this thread, post #25.

Ah, didn't see it after the edit.

The beable of an interpretation is the thing that physically exists even when it is not measured. For example, when you say that particle exists even when it is not measured, then particle is supposed to be the beable in your interpretation. However, the problem with your interpretation is that your beable is not represented by a mathematical object. In particular, it cannot be a wave function, because an entangled 2-particle wave function does not represent a single particle. The beable λ must be represented by a mathematical object, otherwise it is not well defined.

I don't understand what the objection is. If you have an entangled system why should the subsystems be represented mathmatically. They are not even well-defined. One can only loosly walk about the subsystems.

 

[Sambuco]

This is hard to understand. What does it mean for the physical systems not to exist at all times? If they don't exist at a given time, then don't we just have an empty space-time? Why would they start to exist later on?

I can understand what it means for the physical systems not to be fundamental. I am strugling with the non-existence.

Within non-relativistic quantum mechanics, as you say, a tension arises between these two concepts because spacetime is assumed to be classical. Ultimately, if the gravitational field were inherently quantum and we interpreted it in the same way, there would also be no instants in time when the system doesn't exist. That is, what would exist are transitions between specific configurations of all degrees of freedom, including the gravitational field/spacetime. Indeed, the relational interpretation arose from Rovelli's earlier work on loop quantum gravity.

Lucas.

 

[Demystifier]

Ah, didn't see it after the edit.

I don't understand what the objection is. If you have an entangled system why should the subsystems be represented mathmatically. They are not even well-defined. One can only loosly walk about the subsystems.

Fine, but does it mean that the subsystem is not ontological, i.e., that the subsystem does not exist in a strict sense? For example, if the system consists of a proton and an electron in an entangled state, would you say that the electron does not exist?

 

[martinbn]

Fine, but does it mean that the subsystem is not ontological, i.e., that the subsystem does not exist in a strict sense? For example, if the system consists of a proton and an electron in an entangled state, would you say that the electron does not exist?

No, i wouldn't say that they don't exist.

 

[A. Neumaier]

If you have an entangled system why should the subsystems be represented mathmatically. They are not even well-defined. One can only loosly walk about the subsystems.

Of course they are well-defined. In a long distance Bell experiment, Alice sees the effects of one of the entangled photons, not even knowing whether there is an entangled partner somewhere. For her, the only system is the subsystem, and it has all the properties a single photon has.

This is the typical situation everywhere. Most systems in the world are entangled, since the Schrödinger equation entangles them. Even should they be for one moment disentangled, this is destroyed by interactions in the next moment. But we only care about the subsystem our attention is on (which may be different for different observers or agents), and it has properties as well-defined as one can expect from a quantum system.

 

[martinbn]

Of course they are well-defined. In a long distance Bell experiment, Alice sees the effects of one of the entangled photons, not even knowing whether there is an entangled partner somewhere. For her, the only system is the subsystem, and it has all the properties a single photon has.

This is the typical situation everywhere. Most systems in the world are entangled, since the Schrödinger equation entangles them. Even should they be for one moment disentangled, this is destroyed by interactions in the next moment. But we only care about the subsystem our attention is on (which may be different for different observers or agents), and it has properties as well-defined as one can expect from a quantum system.

Then the problem @Demystifier discribes doesn't exist.

 

[Demystifier]

No, i wouldn't say that they don't exist.

So you would say that they exist, but can't be represented mathematically, right? The Bell's analysis then can't be applied to them, because the Bell theorem assumes that all ontological stuff can be represented by some mathematical object $\lambda$. That's how the Bell's conclusion, that there must be some non-local action at a distance, can be avoided: the Bell's theorem is a mathematical theorem, so if there is some physical stuff not represented by mathematics, the theorem does not apply to it. That's a legitimate position, but I find it problematic. Nevertheless, it would help me understand you if you could confirm that this, more or less, is your position.

 

[Demystifier]

Then the problem @Demystifier discribes doesn't exist.

Right, it doesn't exist in the Neumaier's interpretation, but it exists in your interpretation.

 

[martinbn]

So you would say that they exist, but can't be represented mathematically, right? The Bell's analysis then can't be applied to them, because the Bell theorem assumes that all ontological stuff can be represented by some mathematical object $\lambda$. That's how the Bell's conclusion, that there must be some non-local action at a distance, can be avoided: the Bell's theorem is a mathematical theorem, so if there is some physical stuff not represented by mathematics, the theorem does not apply to it. That's a legitimate position, but I find it problematic. Nevertheless, it would help me understand you if you could confirm that this, more or less, is your position.

I thought we were talking about the ontology of RQM. How did it turn to Bell's theorem and nonlocality!

 

[martinbn]

Right, it doesn't exist in the Neumaier's interpretation, but it exists in your interpretation.

What interpretation of mine! Aren't we talking about RQM!

 

[Demystifier]

We are talking about the Coleman's interpretation (which is not RQM), for which you said that you find convincing, so I thought that's your interpretation too. For any interpretation, one of the most important questions is how it interprets the Bell theorem.

 

[martinbn]

We are talking about the Coleman's interpretation (which is not RQM), for which you said that you find convincing, so I thought that's your interpretation too. For any interpretation, one of the most important questions is how it interprets the Bell theorem.

No, i said that i find that particular argument convincing (Mott's from 1929), which is interpretation indipendent. Coleman's interpretation is some version of MWI.

 

[gentzen]

No, i said that i find that particular argument convincing (Mott's from 1929), which is interpretation indipendent.

Coleman's version of Mott's argument just uses a self-adjoint "straight line"/"linearity" operator L and the corresponding spectral version of Born's rule.

I don't think that this is an interpretation independent argument, because it doesn't directly apply to MWI or BM.

Coleman's interpretation is some version of MWI.

Whether Coleman's opinion really tends towards MWI is part of my initial question. My impression after following the discussion here and reading the corresponding sections is rather that Coleman tends towards a mix between Mott and Zurek. And Zurek in particular has always been pretty ambiguous with respect to MWI, which Coleman seems to mirror a bit.

 

[Demystifier]

Can someone (e.g. @gentzen or @martinbn ) explain the Mott's argument to me in his own words? Because from the paper I haven't understood this argument at all.

 

[gentzen]

Can someone (e.g. @gentzen or @martinbn ) explain the Mott's argument to me in his own words? Because from the paper I haven't understood this argument at all.

The argument is at the bottom of slide 15a (page 9) together with Coleman's words on page 10:

But if I have a linear superposition of eigenstates of the particle with respect to the operator L, each of which is an eigenstate with eigenvalue +1, then the combination is also an eigenstate with eigenvalue +1.

It is basically just a trivial compatation. But it shows that the argument with the radial symmetry for the Mott problem was just a simple failed intuition.

 

[Demystifier]

I still don't understand the point. The radially symmetric wave function is not supposed to be an eigenstate of $L$, so why is he talking about a wave function which is an eigenstate of $L$? Where does such a wave function come from? If such a wave function is the answer, what is the question?

 

[Demystifier]

After some thinking, I think the Mott-Coleman argument is simply wrong, because the operator $L$ does not exist.

To explain it in the simplest possible way, let me first simplify the problem by assuming that the chamber has a shape of a tube, so that classically the scattered particle scatters either to the left or to the right. When it scatters to the left, the chamber is in the state $|l\rangle$. Likewise, when it scatters to the right, the chamber is in the state $|r\rangle$. On the other hand, by the left-right symmetry, one expects that the wave function is a left-right symmetric superposition
$$|s\rangle=\frac{|l\rangle + e^{i\varphi} |r\rangle}{\sqrt{2}}$$
The goal is to explain why we do not see the system in the superposition $|s\rangle$.

Now the Mott-Coleman argument requires the projector $L$, with the property that it gives the value 1 when it acts on states $|l\rangle$ and $|r\rangle$, and the value 0 when it acts on states orthogonal to those states. But there is no such projector!!! The projector that gives the value 1 when it acts on states $|l\rangle$ and $|r\rangle$ is
$$L=|l\rangle\langle l| + |r\rangle\langle r| .$$
But this is the unit operator $L=1$, because the Hilbert space is 2-dimensional so there are no states orthogonal to both $|l\rangle$ and $|r\rangle$. In particular, the symmetric state $|s\rangle$ obeys
$$L|s\rangle=|s\rangle \neq 0$$
Thus, there is no operator $L$ that could explain why the chamber is not seen in the symmetric superposition $|s\rangle$.

Now it's straightforward to generalize the argument to the spherical chamber. Now the projector $L$ is of the form
$$L=\int d\Omega |\Omega\rangle\langle\Omega|$$
where $|\Omega\rangle$ corresponds to a classical scattering state into the direction $\Omega$. The spherical symmetric superposition is something like
$$|s\rangle \propto \int d\Omega' |\Omega'\rangle$$
Hence, assuming $\langle\Omega|\Omega'\rangle = \delta(\Omega-\Omega')$, we have
$$L|s\rangle = |s\rangle \neq 0$$
so the operator $L$ with the desired properties does not exist.

 

[gentzen]

Your problem is that you model the state of the chamber as a two dimensional Hilbert space. But even with just one ionizable atom on each side, it is already at least 4 dimensional. So even in this simple scenario, L is not forced to be the unit operator.

However, you are right that such an operator L cannot properly distinguish all different relevant scenarios, and hence cannot answer the question Coleman wants to answer.

 

[martinbn]

After some thinking, I think the Mott-Coleman argument is simply wrong, because the operator $L$ does not exist.

To explain it in the simplest possible way, let me first simplify the problem by assuming that the chamber has a shape of a tube, so that classically the scattered particle scatters either to the left or to the right. When it scatters to the left, the chamber is in the state $|l\rangle$. Likewise, when it scatters to the right, the chamber is in the state $|r\rangle$. On the other hand, by the left-right symmetry, one expects that the wave function is a left-right symmetric superposition
$$|s\rangle=\frac{|l\rangle + e^{i\varphi} |r\rangle}{\sqrt{2}}$$
The goal is to explain why we do not see the system in the superposition $|s\rangle$.

Now the Mott-Coleman argument requires the projector $L$, with the property that it gives the value 1 when it acts on states $|l\rangle$ and $|r\rangle$, and the value 0 when it acts on states orthogonal to those states. But there is no such projector!!! The projector that gives the value 1 when it acts on states $|l\rangle$ and $|r\rangle$ is
$$L=|l\rangle\langle l| + |r\rangle\langle r| .$$
But this is the unit operator $L=1$, because the Hilbert space is 2-dimensional so there are no states orthogonal to both $|l\rangle$ and $|r\rangle$. In particular, the symmetric state $|s\rangle$ obeys
$$L|s\rangle=|s\rangle \neq 0$$
Thus, there is no operator $L$ that could explain why the chamber is not seen in the symmetric superposition $|s\rangle$.

Now it's straightforward to generalize the argument to the spherical chamber. Now the projector $L$ is of the form
$$L=\int d\Omega |\Omega\rangle\langle\Omega|$$
where $|\Omega\rangle$ corresponds to a classical scattering state into the direction $\Omega$. The spherical symmetric superposition is something like
$$|s\rangle \propto \int d\Omega' |\Omega'\rangle$$
Hence, assuming $\langle\Omega|\Omega'\rangle = \delta(\Omega-\Omega')$, we have
$$L|s\rangle = |s\rangle \neq 0$$
so the operator $L$ with the desired properties does not exist.

No, you missing a crusial point. You have to consider the full Hilbert space not just of the system, but also the aparatus and physicist. This is the only part the may depend on an interpretation. That everything is quantum mechanical. The oprator acts on that Hilbert space.

 

[Demystifier]

No, you missing a crusial point. You have to consider the full Hilbert space not just of the system, but also the aparatus and physicist. This is the only part the may depend on an interpretation. That everything is quantum mechanical. The oprator acts on that Hilbert space.

It doesn't change anything in my argument. I can define
$$|\Psi_{\Omega}\rangle = |\Omega\rangle \otimes |{\rm apparatus}\; {\rm measures}\; \Omega\rangle \otimes |{\rm physicist}\; {\rm sees}\; \Omega\rangle$$
Now
$$L=\int d\Omega |\Psi_{\Omega}\rangle\langle\Psi_{\Omega} |$$
$$|S\rangle = \int d\Omega' |\Psi_{\Omega'}\rangle$$
so again
$$L|S\rangle = |S\rangle \neq 0$$

 

[martinbn]

It doesn't change anything in my argument. I can define
$$|\Psi_{\Omega}\rangle = |\Omega\rangle \otimes |{\rm apparatus}\; {\rm measures}\; \Omega\rangle \otimes |{\rm physicist}\; {\rm sees}\; \Omega\rangle$$
Now
$$L=\int d\Omega |\Psi_{\Omega}\rangle\langle\Psi_{\Omega} |$$
$$|S\rangle = \int d\Omega' |\Psi_{\Omega'}\rangle$$
so again
$$L|S\rangle = |S\rangle \neq 0$$

He doesn't define the operator like that. He says if initial state (of the whole thing) is such that when you run the experiment and get definite single outcome then the operator is define to act as the identity on such an intial state. So these states are eigenstates with eigenvalue one. If this dosn't happen when you run the experiment, for example the particle was lost and the detector didn't click, then the operator takes it to zero. Now we know that if the system we perform a measrment on is in an eigen state of the observable we measure we get a single outcome, the rest aparatus and physicist are in the ready state. So the operator that tells us whether we had one outcome will have this state as an eigenstate with eigen value one. If you prepare the system to be measured in a superposition of eigenstates of the observable, what will Coleman's operator do? It ill have that state as an eigenstate with eigen value one, because of linearity. So it tells us that we will still get a single outcome.

 

[gentzen]

So it tells us that we will still get a single outcome.

I no longer believe that Coleman‘s operator can tell us that:

However, you are right that such an operator L cannot properly distinguish all different relevant scenarios, and hence cannot answer the question Coleman wants to answer.

 

[martinbn]

I no longer believe that Coleman‘s operator can tell us that:

Why? That is how it is defined.

 

[Demystifier]

He doesn't define the operator like that. He says if initial state (of the whole thing) is such that when you run the experiment and get definite single outcome then the operator is define to act as the identity on such an intial state. So these states are eigenstates with eigenvalue one. If this dosn't happen when you run the experiment, for example the particle was lost and the detector didn't click, then the operator takes it to zero. Now we know that if the system we perform a measrment on is in an eigen state of the observable we measure we get a single outcome, the rest aparatus and physicist are in the ready state. So the operator that tells us whether we had one outcome will have this state as an eigenstate with eigen value one. If you prepare the system to be measured in a superposition of eigenstates of the observable, what will Coleman's operator do? It ill have that state as an eigenstate with eigen value one, because of linearity. So it tells us that we will still get a single outcome.

That's a lot of talk. Can you write down such an operator explicitly, in a form similar to my operator, so that we can compare them?