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.