Sidney Coleman's opinion on interpretation in his Dirac lecture

[PeterDonis]

Turns out L should be consistent with those more mundane measurement operators. They should be a simple refinement of L.

Please show this explicitly, or give a reference that does so. It does not seem at all obvious to me, but I'm not very familiar with the concepts you're using.

 

[Sambuco]

In case it helps, I'll share my interpretation of what Coleman is trying to say. He begins by defining the operator $L$ as having straight line for the ionization track in the cloud chamber as an eigenstate with eigenvalue +1. Then, by linearity, this means that a superposition of straight line tracks is also an eigenstate of the operator with eigenvalue +1. In principle, this seems to suggest that we should see a superposition of straight line tracks, not just one. However, Coleman argues that this is incorrect because the observer is also a quantum system, and that by incorporating it into the state vector, time evolution will result in an eigenstate consisting of a superposition of states, each of which has an observer who has seen a straight line track. In this way, he demonstrates that QM predicts the "experience" of seeing a straight line track. At this point, I don't know what else can be said about it, because Coleman isn't clear on the matter. That is, if he assumes that the quantum state represents physical reality, then this is nothing more than the many-worlds interpretation.

Lucas.

 

[gentzen]

Please show this explicitly, or give a reference that does so. It does not seem at all obvious to me, but I'm not very familiar with the concepts you're using.

For "my fixed version of" Demystifier's simplified scenario, L was defined as

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

The "more mundane measurement operators" would just measure a single classical state. For example, an operator M_{iL,nR} could be defined via
M_{iL,nR} |neutral_L, neutral_R⟩ = 0
M_{iL,nR} |ionized_L, neutral_R⟩ = |ionized_L, neutral_R⟩
M_{iL,nR} |neutral_L, ionized_R⟩ = 0
M_{iL,nR} |ionized_L, ionized_R⟩ = 0

M_{iL,nR} measures, whether the state of the chamber is such that the left atom is ionized and the right atom is still neutral.
M_{iL,nR} commutes with L, i.e. L M_{iL,nR} = M_{iL,nR} L, because there exists a basis of shared eigenstates. And if I would define similar operators M_{nL,nR}, M_{iL,iR}, and M_{nL,iR}, similar statements would apply to them.

In conclusion, measuring L doesn't put up any restriction for talk about what would have happened if M_{iL,nR} or any of the other mundane measurements operators would have been measured. (Well, consistent histories doesn't say "would have been measured", but is just concerned with talk about the state as if it had the measurable properties.)

 

[Sambuco]

As a addition to my post #92, Coleman's argument is not equivalent to what Mott presents in his seminal 1929 work. Mott demonstrates that if an $\alpha$ particle is emitted in a cloud chamber such that its initial state is spherically symmetric, and we assume that it ionizes atoms in its path, the probability of two or more ionization events is negligible unless these events, along with the point of emission, lie on a straight line. Mott uses the probabilistic interpretation of the wave function, he never presupposes that the wave function represents the physical reality or anything of the sort.

Lucas.

 

[PeterDonis]

measuring L doesn't put up any restriction for talk about what would have happened if M_{iL,nR} or any of the other mundane measurements operators would have been measured.

That's all very well (assuming it's correct--I'll have to cogitate some more about the states and operators you give), but it doesn't seem relevant to Coleman's argument. Coleman's argument appears to be that measuring L, all by itself, is sufficient to ground a claim that we observe single outcomes. Coleman doesn't even introduce any of these other measurement operators.

I understand that the consistent histories interpretation might take a different viewpoint, but again, this thread is about Coleman's interpretation as it appears in the lecture. Discussion of how consistent histories explains, for example, straight tracks in a cloud chamber belongs in a separate thread.

 

[PeterDonis]

M_{iL,nR} commutes with L, i.e. L M_{iL,nR} = M_{iL,nR} L, because there exists a basis of shared eigenstates.

I'm not sure this actually works with zero eigenvalues. For example, if the state is |n_L, i_R>, applying ML to it is straightforward enough, but how do I apply LM? M annihilates the state, so there's nothing to apply L to.

 

[martinbn]

I'm not sure this actually works with zero eigenvalues. For example, if the state is |n_L, i_R>, applying ML to it is straightforward enough, but how do I apply LM? M annihilates the state, so there's nothing to apply L to.

If M annihilstes it, it means that it takes it to the zero vector. So L is applied to the zero vector and returns the zero vector.

 

[PeterDonis]

If M annihilstes it, it means that it takes it to the zero vector. So L is applied to the zero vector and returns the zero vector.

Zero times some other vector isn't the zero vector, is it? It's zero, the number. That's the weird thing about a zero eigenvalue.

Granted, I'm normally ok with a physicist's level of mathematical rigor (i.e., not much), so this probably shouldn't bother me. I'll need to cogitate some more.

 

[martinbn]

Zero times some other vector isn't the zero vector, is it? It's zero, the number. That's the weird thing about a zero eigenvalue.

Zero, the number, times any vector gives the zero vector. Think of any specific vector space, say $\mathbb R^3$, then $0\cdot(a, b, c) = (0, 0, 0)$.

Granted, I'm normally ok with a physicist's level of mathematical rigor (i.e., not much), so this probably shouldn't bother me. I'll need to cogitate some more.

 

[PeterDonis]

Zero, the number, times any vector gives the zero vector.

Ah, got it. Thanks for the reminder of basic vector space theory.

 

[Demystifier]

Here is a more general version of my argument that the projector $L$ with the desired properties does not exist. Since $L$ is a projector (in some space including the states of the apparatus and the observer), it can be written in the form
$$L=\sum_{i\in V} |i\rangle\langle i|$$
where $V$ is some index set that labels a set of mutually orthogonal unit vectors, $\langle i|j\rangle = \delta_{ij}$. Then for any superposition of the form
$$|s\rangle = \sum_{j\in V} c_j|j\rangle$$
we have
$$L|s\rangle =|s\rangle \neq 0$$
Hence there is a plenty of possible superposition states $|s\rangle$ that cannot be excluded as possible measurement results of the measurement of $L$, contrary to the intention of the Mott-Coleman argument. Q.E.D.

Note: For the argument it is not important what the set $V$ is. All that matters is that it has at least 2 elements, but it can be arbitrarily large. In fact, larger $V$ means more possibilities for superpositions $|s\rangle$ with the property above.

@gentzen and @martinbn , did I convince you?

 

[Demystifier]

One additional note. The error in the Mott-Coleman argument is conceptually similar to the error of failing to distinguish proper and improper mixture, the error often appearing in the erroneous claims that decoherence alone explains the wavefuncion collapse. For that purpose, consider the mixed state
$$\rho = \sum_{i\in V} p_i |i\rangle\langle i|$$
which has a form similar to the $L$ operator in my previous post. Such a mixed state can be viewed in two ways. In the proper mixture view, the system is in some pure state $\rho_i= |i\rangle\langle i|$, but we don't know which state it is, so $p_i$ are our Bayesian probabilities assigned to different possibilities. In the improper mixture view, the system is in the state $\rho$. The improper mixtures appear frequently as results of straightforward interpretation-independent computations. By contrast, a proper mixture view requires some additional interpretation-dependent mechanism explaining why only one of the states $|i\rangle\langle i|$ is real. The conceptual error is committed when one obtains an improper mixture as a result of interpretation-independent computation, and then concludes that the mixture is proper, i.e., that the system is in some definite pure state $\rho_i$.

The error in the Mott-Coleman argument is of a similar nature. From the fact that a superposition $|s\rangle$ is an eigenstate of $L$ with eigenvalue 1, it does not follow that it is an eigenstate of some projector $|i\rangle\langle i|$ with eigenvalue 1. But it looks as if Mott and Coleman somehow think that it does, corresponding to the belief that the system is in the pure state $\rho_i=|i\rangle\langle i|$.

 

[martinbn]

@gentzen and @martinbn , did I convince you?

Not me, because what you wrote is obvious and irrelevant.

 

[Demystifier]

Not me, because what you wrote is obvious and irrelevant.

Why irrelevant?

I have a very elementary question for you, related to the Mott-Coleman argument. Let $|C_1\rangle$ be a state (of the chamber together with the apparatus and the observer) corresponding to one straight trajectory. Let $|C_2\rangle$ be a state corresponding to another straight trajectory, $\langle C_1| C_2\rangle=0$. Does the state
$$|S\rangle = \frac{|C_1\rangle + |C_2\rangle}{\sqrt{2}}$$
correspond to a straight trajectory?

 

[martinbn]

Why irrelevant?

I have a very elementary question for you, related to the Mott-Coleman argument. Let $|C_1\rangle$ be a state (of the chamber together with the apparatus and the observer) corresponding to one straight trajectory. Let $|C_2\rangle$ be a state corresponding to another straight trajectory, $\langle C_1| C_2\rangle=0$. Does the state
$$|S\rangle = \frac{|C_1\rangle + |C_2\rangle}{\sqrt{2}}$$
correspond to a straight trajectory?

If you mean that these two are the initial states that lead to the observation of a straight trajectory. Then, yes, the superposition will too.

 

[Demystifier]

If you mean that these two are the initial states that lead to the observation of a straight trajectory. Then, yes, the superposition will too.

OK, now I understand where the controversy comes from. @PeterDonis has already explained it in one of the posts, so I will just rephrase what he said.

The claim that $|C_1\rangle$ corresponds to a straight trajectory can be interpreted in two ways:

Interpretation A: The observer thinks that there is a straight trajectory, and that this trajectory is $T_1$.

Interpretation B: The observer thinks that there is a straight trajectory, but he has no idea which trajectory it is.

In interpretation A, the states $|C_1\rangle$ and $|C_2\rangle$ are distinguishable by the observer. In interpretation B, they are not distinguishable by the observer. The Mott-Coleman argument deals with the interpretation B, not with the interpretation A. In other words, the Mott-Coleman argument explains why the observer thinks that there is a straight trajectory, provided that he is not able to distinguish one straight trajectory from the other. But a realistic observers is able to distinguish one straight trajectory from the other. Hence the problem solved by the Mott-Coleman argument is not the real problem corresponding to a realistic observer.

 

[martinbn]

OK, now I understand where the controversy comes from. @PeterDonis has already explained it in one of the posts, so I will just rephrase what he said.

The claim that $|C_1\rangle$ corresponds to a straight trajectory can be interpreted in two ways:

Interpretation A: The observer thinks that there is a straight trajectory, and that this trajectory is $T_1$.

Interpretation B: The observer thinks that there is a straight trajectory, but he has no idea which trajectory it is.

In interpretation A, the states $|C_1\rangle$ and $|C_2\rangle$ are distinguishable by the observer. In interpretation B, they are not distinguishable by the observer. The Mott-Coleman argument deals with the interpretation B, not with the interpretation A. In other words, the Mott-Coleman argument explains why the observer thinks that there is a straight trajectory, provided that he is not able to distinguish one straight trajectory from the other. But a realistic observers is able to distinguish one straight trajectory from the other. Hence the problem solved by the Mott-Coleman argument is not the real problem corresponding to a realistic observer.

How does the ability to distinguish trajectories matter?

 

[Demystifier]

How does the ability to distinguish trajectories matter?

The state $|C_1\rangle$ corresponds to the trajectory $T_1$, while the state $|C_2\rangle$ corresponds to the trajectory $T_2$. If the observer distinguishes trajectories, then his observation involves the identification of a specific trajectory. But there is no specific trajectory $T$ associated with the superposition $|C_1\rangle+|C_2\rangle$, so such a superposition cannot correspond to a straight trajectory if the trajectories are distinguished. The trajectories must be indistinguishable in order for your post #105 to make sense.

 

[Sambuco]

But it looks as if Mott and Coleman somehow think that it does, corresponding to the belief that the system is in the pure state ρi=|i⟩⟨i|.

As a quick aside, as I mentioned in post #94, Mott has nothing to do with Coleman's argument. Mott's analysis is correct, and in fact it is considered one of the first works on decoherence, even in the early years of QM (his paper is from 1929).

Lucas.

 

[martinbn]

The state $|C_1\rangle$ corresponds to the trajectory $T_1$, while the state $|C_2\rangle$ corresponds to the trajectory $T_2$. If the observer distinguishes trajectories, then his observation involves the identification of a specific trajectory. But there is no specific trajectory $T$ associated with the superposition $|C_1\rangle+|C_2\rangle$, so such a superposition cannot correspond to a straight trajectory if the trajectories are distinguished. The trajectories must be indistinguishable in order for your post #105 to make sense.

Is this about whether it is a srtaight line? I dont know if it will be a straight line. I dont know if straight lines are possible.

 

[Demystifier]

Is this about whether it is a srtaight line? I dont know if it will be a straight line. I dont know if straight lines are possible.

Given that you don't know that, are you still certain about your claim in the post #105?

 

[gentzen]

@gentzen and @martinbn , did I convince you?

Thanks for trying to drive this forward to closure. Your "more general version of your argument" nicely captures the reasons why

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

However, this only convinces me that Coleman's argument is too weak, in the way he presented it. The concrete weak point is pointed out here:

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.

The problem is that Coleman's operators L and D are never actually measured, they are only "talked about". The whole thing gives me "Zurek vibes", even so it now seems that the argument was due to David Albert. Whenever I accidentally read Zurek, he was a great fan of slick computations (without clarifying their context):

Now we’re going to answer that question, and in a faster and slicker way then Neville Mott did.

And those vibes were the reason for my original question. The whole argument only makes sense to me as a slick Zurek computation for an initially valid argument with more cumbersome computations.

The error in the Mott-Coleman argument is of a similar nature. From the fact that a superposition $|s\rangle$ is an eigenstate of $L$ with eigenvalue 1, it does not follow that it is an eigenstate of some projector $|i\rangle\langle i|$ with eigenvalue 1. But it looks as if Mott and Coleman somehow think that it does, corresponding to the belief that the system is in the pure state $\rho_i=|i\rangle\langle i|$.

I don't think that Mott himself had any unjustified beliefs that caused him to commit serious errors.

What one probably has to to is to look at all measurement operators measuring a classical result where the ionized atoms are not on a straight line. For each such operator, the eigenvalue argument is (approximatively) valid. Then you only have the problem to turn those approximate results for a huge number of classical results into an approximate result for the actually observed outcome. (I guess Mott's original argument managed to do that, and Zurek's slick argument simply ignores this ugly detail. But of course, this explanation presupposes that Zurek somehow influenced Coleman's argument.)

 

[martinbn]

Given that you don't know that, are you still certain about your claim in the post #105?

Yes, the claim is that Coleman's argument shows that if two initial states result in the observer seeing a definite trajectory, so will their superposition.

 

[Demystifier]

The problem is that Coleman's operators L and D are never actually measured, they are only "talked about".

Yes, my post #106 is also written in that spirit, so we basically agree on that.

 

[Demystifier]

Yes, the claim is that Coleman's argument shows that if two initial states result in the observer seeing a definite trajectory, so will their superposition.

And my claim is that it depends on what one means by "seeing a definite trajectory". It is correct in interpretation B (in which different trajectories are not distinguished), but not in interpretation A (in which they are distinguished). But if you refuse to specify which interpretation do you have in mind, then your claim is just too vague to associate a truth value to it.

 

[martinbn]

And my claim is that it depends on what one means by "seeing a definite trajectory". It is correct in interpretation B (in which different trajectories are not distinguished), but not in interpretation A (in which they are distinguished). But if you refuse to specify which interpretation do you have in mind, then your claim is just too vague to associate a truth value to it.

Why is it wrong in case A? Why does it matter?

 

[Demystifier]

Why is it wrong in case A? Why does it matter?

You already asked me and I already answered. See #108.

 

[martinbn]

You already asked me and I already answered. See #108.

But you just claimed that in the case if superposition there is no specific trajectory. That is not an abswer, it is just repeating the claim.

 

[Demystifier]

But you just claimed that in the case if superposition there is no specific trajectory. That is not an abswer, it is just repeating the claim.

Do you think that there is a specific trajectory?

 

[martinbn]

Do you think that there is a specific trajectory?

Let's say yes. Why does it matter?