Sidney Coleman's opinion on interpretation in his Dirac lecture

  • Context: Graduate 
  • Thread starter Thread starter gentzen
  • Start date Start date
  • #91
gentzen said:
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.
 
Physics news on Phys.org
  • #92
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.
 
  • Like
Likes   Reactions: Demystifier
  • #93
PeterDonis said:
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
gentzen said:
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.)
 
  • #94
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.
 
  • Like
Likes   Reactions: PeterDonis
  • #95
gentzen said:
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.
 
  • #96
gentzen said:
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.
 
  • #97
PeterDonis said:
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.
 
  • #98
martinbn said:
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.
 
  • #99
PeterDonis said:
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)##.
PeterDonis said:
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.
 
  • #100
martinbn said:
Zero, the number, times any vector gives the zero vector.
Ah, got it. Thanks for the reminder of basic vector space theory. :oops:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 376 ·
13
Replies
376
Views
24K
  • · Replies 412 ·
14
Replies
412
Views
23K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 37 ·
2
Replies
37
Views
7K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 25 ·
Replies
25
Views
6K
  • · Replies 169 ·
6
Replies
169
Views
11K
  • · Replies 76 ·
3
Replies
76
Views
9K