Sidney Coleman's opinion on interpretation in his Dirac lecture

[martinbn]

I meant constructed in a sense I have constructed my operator, as a sum/integral over all possible single outcomes.

Why does it have to be like that?

Let us look at the 4-dimensional Hilbert space for Demystifier's simplified scenario. On L is defined on the "corresponding" basis via
L |neutral_L, neutral_R⟩ = 0
L |ionized_L, neutral_R⟩ = |ionized_L, neutral_R⟩
L |neutral_L, ionized_R⟩ = |neutral_L, ionized_R⟩
L |ionized_L, ionized_R⟩ = 0
Now the Bohmian (or Zurek) wonders why we never observe a state
c2 |ionized_L, neutral_R⟩ + c3 |neutral_L, ionized_R⟩
L cannot answer that question. It can only tell us that we won't observe any of the states
c0 |neutral_L, neutral_R⟩ + c3 |ionized_L, ionized_R⟩
(for arbitrary complex coefficients c0 and c3).

That is not what Coleman says. Look at the box on page 10.

 

[gentzen]

Why does it have to be like that?

That is not what Coleman says. Look at the box on page 10.

Let us first agree on Demystifier's example. He thinks it generalizes to more complicated scenarios. And what Coleman says on page 10 is certainly more complicated.

 

[Demystifier]

Why does it have to be like that?

So that it can be compared with my argument, so if my argument is wrong I can see why it is wrong.

 

[martinbn]

Let us first agree on Demystifier's example. He thinks it generalizes to more complicated scenarios. And what Coleman says on page 10 is certainly more complicated.

So that it can be compared with my argument, so if my argument is wrong I can see why it is wrong.

Which example is that? The one in post #48?

 

[gentzen]

Which example is that? The one in post #48?

Yes. And if Demystifier agrees that his Hilbert space should have been (at least) 4 dimensional instead of 2 dimensional, then his example in some "suitably fixed" form (like the one I gave in #60).

 

[martinbn]

Yes. And if Demystifier agrees that his Hilbert space should have been (at least) 4 dimensional instead of 2 dimensional, then his example in some "suitably fixed" form (like the one I gave in #60).

Well, no, it cannot be this Hilbert space. You need to include the observer or the aparatus.

 

[Demystifier]

Yes. And if Demystifier agrees that his Hilbert space should have been (at least) 4 dimensional instead of 2 dimensional,

I think I disagree. If there is exactly one scattered particle, then there is no state $|{\rm neutral}_L,{\rm neutral}_R\rangle$, because such a state would correspond to zero scattered particles. Likewise, there is no $|{\rm ionized}_L,{\rm ionized}_R\rangle$, because it would correspond to two scattered particles.

 

[gentzen]

I think I disagree. If there is exactly one scattered particle, then there is no state $|{\rm neutral}_L,{\rm neutral}_R\rangle$, because such a state would correspond to zero scattered particles. Likewise, there is no $|{\rm ionized}_L,{\rm ionized}_R\rangle$, because it would correspond to two scattered particles.

Note that Coleman's operator L operates on the state of the cloud chamber |C⟩, not on the initial or scattered particle. L looks at the ionized atoms (or molecules), and has Eigenvalue +1 if there is at least one ionized atom and all ionized atoms lie approximately on a straight line.

So for your simplified example, you need at least one atom on the left side, and one atom on the right side. Otherwise you don't get anything analogous to the Mott-Coleman argument.

 

[Demystifier]

Note that Coleman's operator L operates on the state of the cloud chamber |C⟩, not on the initial or scattered particle. L looks at the ionized atoms (or molecules), and has Eigenvalue +1 if there is at least one ionized atom and all ionized atoms lie approximately on a straight line.

So for your simplified example, you need at least one atom on the left side, and one atom on the right side. Otherwise you don't get anything analogous to the Mott-Coleman argument.

Fine, suppose that there are two atoms, one on the left and one on the right. And suppose that you observe that no atom is ionized. Would you conclude that you have found the system in some strange non-classical superposition? I wouldn't, instead I would conclude that there is no scattered particle in the chamber at all.

 

[PeterDonis]

Can you constructively discribe the Hilbert space of a human? Probably not, but that never stops you from using it in a general argument.

Maybe it should.

Hilbert space is a mathematical tool. It's not reality. Claiming that the entire universe can be modeled with a Hilbert space as QM does is equivalent to claiming that QM, as it stands, is a complete theory of everything. But nobody actually believes that.

 

[gentzen]

And suppose that you observe that no atom is ionized.

In the given setup, L is the measurement operator. So you don't observe indiviual atoms at all. You just observe L. And in the simple scenarios, L is build up from configurations that seem to allow a classical interpretation.

Of course, you can also include the (classical) state where no atoms are ionized among the states with Eigenvalue +1 (if you want). Coleman decided to not include it, but this detail is not really important.

 

[Demystifier]

@gentzen you have not understood my point. Let me explain it by a different, but mathematically equivalent example. Instead of the ion chamber, consider the Stern-Gerlach apparatus, oriented in the vertical z-direction. Suppose that exactly one particle approaches the apparatus. Instead of my previous states $|l\rangle$ and $|r\rangle$, now the relevant states are $|up\rangle$ (corresponding to the macroscopic dot seen up) and $|down\rangle$ (corresponding to the macroscopic dot seen down). Even though the dots are macroscopic, the relevant Hilbert space is 2-dimensional. There are no states $|none\rangle$ (corresponding to no macroscopic dots seen at all) and $|both\rangle$ (corresponding to one macroscopic dots seen down and another macroscopic dot seen up). Sure, one can consider a superposition such as $|up\rangle+|down\rangle$, but this is not the state $|both\rangle$.

 

[PeterDonis]

L is build up from configurations that seem to allow a classical interpretation.

The individual ones do. But the actual quantum state that's prepared is not one of them. It's a linear superposition of all of them. And, by the argument that Coleman gives in the lecture, that linear superposition is also an eigenstate of L with eigenvalue +1. And that means that, if we make a measurement with L as the measurement operator, we should not see an individual straight line--we should see the whole superposition, because it's an eigenstate of L!

In other words, I don't think this argument does what it's claimed to do: it doesn't explain why we see a single straight line track in a cloud chamber. In fact it implies that we shouldn't!

 

[gentzen]

Even though the dots are macroscopic, the relevant Hilbert space is 2-dimensional.

Not for Coleman. For him, the relevant Hilbert space is that of the dots.

There are no states $|none\rangle$ (corresponding to no macroscopic dots seen at all) and $|both\rangle$(corresponding to one macroscopic dots seen down and another macroscopic dot seen up).

For Coleman, there are states $|none\rangle$ and $|both\rangle$. And Coleman does not mean a superposition such as $|up\rangle+|down\rangle$ by this.

If you want to convince me or martinbn that Coleman's argument is wrong (which is quite possible), you first have to understand how I (or martinbn) understood his argument. And if you think I misunderstood Coleman, then you first have to convince me that I did indeed misunderstand him.

 

[PeterDonis]

I don't think this argument does what it's claimed to do: it doesn't explain why we see a single straight line track in a cloud chamber. In fact it implies that we shouldn't!

Let me expand on this, because there's a seemingly obvious response that Coleman gives a little later in the lecture, using David Albert's "definiteness" operator, showing that, just as the linear superposition of all the straight-line tracks is an eigenstate of the "linearity" operator L, the linear superposition of all the states of the observer, in each one of which they observe a straight line track (or whatever other set of possible definite outcomes we are looking at in a particular experiment) is an eigenstate of the "definiteness" operator D. (I first saw this argument in one of David Albert's books many years ago.)

The problem is that, when we look at the outcome of an experiment, we don't just get a feeling that there is a definite outcome--we observe which outcome it is. And Coleman (and Albert in his book, from what I remember) never analyze that at all. The "definiteness" operator D is the wrong one to look at; the right operator is a "which outcome occurred" operator, W, which does not have just two eigenvalues, 1 and 0, as D (and L) do; it has a whole spectrum of eigenvalues, each one corresponding to a possible outcome that occurred. In the case of the cloud chamber, the spectrum of W is continuous, since there is a continuous range of possible straight-line tracks that could be observed.

So W is the operator that needs to be analyzed to explain why we observe definite outcomes--particular definite outcomes in each case. But Coleman never even talks about it.

Earlier in the lecture, Coleman says this:

"[T]here’s an implicit promise in here that, when you put the whole theory together and start calculating things, that the words “observes” and “observable” will correspond to entities that act in the same way as those entities do in the language of everyday speech under the circumstances in which the language of everyday speech is applicable."

But he never actually makes good on that promise; the version of "observe" that he actually analyzes does not work the same way we use the word "observe" in everyday speech.

 

[gentzen]

In other words, I don't think this argument does what it's claimed to do:

Yeah, after my reply to Demystfier #48, I also became unsure.

it doesn't explain why we see a single straight line track in a cloud chamber. In fact it implies that we shouldn't!

If his argument should be too weak to explain why we see a single straight line track, then I feel it should also be too weak to imply the opposite.

 

[PeterDonis]

W is the operator that needs to be analyzed to explain why we observe definite outcomes. But Coleman never even talks about it.

And to take this further: if we assume that states of the observer in which they observe a single straight-line track in a cloud chamber are eigenstates of W, it's easy to show that a linear superposition of all those states is not an eigenstate of W. Why? Because the states all have different eigenvalues.

 

[PeterDonis]

If his argument should be too weak to explain why we see a single straight line track

I didn't say it was too weak to explain why we see a single straight line track. I said it (quite strongly) leads to the opposite prediction, that we should not see a single straight line track. And I've now amplified that further with a few follow-up posts.

 

[Demystifier]

The problem is that, when we look at the outcome of an experiment, we don't just get a feeling that there is a definite outcome--we observe which outcome it is.

Exactly!

 

[gentzen]

I didn't say it was too weak to explain why we see a single straight line track. I said it (quite strongly) leads to the opposite prediction, that we should not see a single straight line track. And I've now amplified that further with a few follow-up posts.

Good, but then I simply disagree with you on that point.

I mean, as a proponent of the consistent histories interpretation, I wonder whether those strange measurement operators L and D will not be inconsistent with more mundane measurement operators just observing whether some specific classical configuration occurred. But even if they should be inconsistent, this still doesn't suggest that we should observe something strange. (Maybe it raises the valid question how Coleman actually intents to measure his strange operators L and D.)

 

[PeterDonis]

I simply disagree with you on that point.

I mean, as a proponent of the consistent histories interpretation

In other words, you're using a different interpretation from Coleman. That means we should be discussing your viewpoint in a separate thread. In this one we're using the interpretation Coleman is using, and we should be addressing his arguments using that interpretation. The fact that a different interpretation might make different arguments is off topic for this thread. Indeed, if you are using a different interpretation, you should be just fine with saying that Coleman's arguments, using Coleman's interpretation, are incorrect--after all, you don't agree with his interpretation anyway!

 

[Demystifier]

Not for Coleman. For him, the relevant Hilbert space is that of the dots.

Fine, but I claim that this space, in the given setup, is effectively 2-dimensional as well.

For Coleman, there are states $|none\rangle$ and $|both\rangle$. And Coleman does not mean a superposition such as $|up\rangle+|down\rangle$ by this.

That's the crucial thing. Why do you think that for Coleman there are states $|none\rangle$ and $|both\rangle$? In addition, should we also include states like $|3\; dots\rangle$?

 

[gentzen]

In other words, you're using a different interpretation from Coleman.

No, I tried to guess what you might have in mind when you claimed: "I said it (quite strongly) leads to the opposite prediction, that we should not see a single straight line track."

I didn't use any interpretation of quantum mechanics when I tried to follow Coleman's argument. I just looked at what he calculated, what he defined, and how he drew conclusions.

Coleman's argument certainly does not convince me that I "should not see a single straight line track"!

 

[gentzen]

That's the crucial thing. Why do you think that for Coleman there are states $|none\rangle$ and $|both\rangle$?

Because he cares about the state of the cloud chamber, i.e. the state of the measurement apparatus. And if the possible dots constitue the measurement apparatus, then both "no dots" and "two dots" are possible results, and both have corresponding states.

 

[PeterDonis]

I didn't use any interpretation of quantum mechanics when I tried to follow Coleman's argument.

Your post #80 sure looked to me like you were using the consistent histories interpretation to analyze it.

 

[Demystifier]

Because he cares about the state of the cloud chamber, i.e. the state of the measurement apparatus. And if the possible dots constitue the measurement apparatus, then both "no dots" and "two dots" are possible results, and both have corresponding states.

But these states have zero probability of appearing, given that exactly one particle interacts with the apparatus. The Hilbert space spanned by such states is orthogonal to anything of physical relevance in the given setup. Otherwise, we should also include states like $|3\; dots\rangle$, or even $|one\; star\rangle$ corresponding to the result of measurement in which a star-shaped outcome appears, rather than a dot-shaped one.

 

[PeterDonis]

Coleman's argument certainly does not convince me that I "should not see a single straight line track"!

If I need to spell it out, sure. It's simple:

If a quantum system is in an eigenstate of an observable, then observing it does not change the state.

The linear superposition of all the straight line tracks, which is the state that Coleman argues gets produced when a particle is emitted in an initial s wave in a cloud chamber, is an eigenstate of L.

Therefore, observing L on that linear superposition leaves that state the same--i.e., as a linear superposition of all the straight line tracks, not a single straight line track.

So this predicts that we should not observe a single straight line track in a cloud chamber; we should observe a linear superposition of all of them, since that's the state after the measurement, and the state after the measurement is supposed to represent what we observed.

 

[gentzen]

we should observe a linear superposition of all of them, since that's the state after the measurement,

The observation is just the eigenvalue and its associated eigenspace. Even so there is a state after measurement, I don't always learn that state during measurement. If there were only one eigenvector corresponding to the measured eigenvalue, then I would learn the state after measurement. But this is not the case here.

and the state after the measurement is supposed to represent what we observed.

That is the part where I don't agree with you. For me, only the eigenspace represents what I observed. The concrete state after measurement within that eigenspace remains unknown to me.

 

[PeterDonis]

The observation is just the eigenvalue and its associated eigenspace.

Well, the associated eigenspace of L with eigenvalue +1 is the entire space of possible linear superpositions of straight-line tracks. It certainly is not just one single straight-line track.

For me, only the eigenspace represents what I observed. The concrete state after measurement within that eigenspace remains unknown to me.

Even if you know that the initial prepared state was a particular state within that eigenspace? AFAIK standard QM says that if you know the prepared state is one particular eigenstate within an eigenspace, measuring that operator leaves the state in that initial prepared state.

That said, if we accept that all you know after measurement is the eigenspace, regardless of what state was initially prepared, then observing Coleman's L tells you nothing useful whatever. It certainly does not tell you that the concrete state after measurement is a single straight-line track. But Coleman is claiming that it does.

 

[gentzen]

I mean, as a proponent of the consistent histories interpretation, I wonder whether those strange measurement operators L and D will not be inconsistent with more mundane measurement operators just observing whether some specific classical configuration occurred.

Turns out L should be consistent with those more mundane measurement operators. They should be a simple refinement of L. (I have no opinion about D, because it feels complicated and hard to define to me.)