The Observer Effect

[Luc Turpin]

26 minutes ago, sethoflagos said:

Yes.

However, I find the word 'chaos' more misleading than helpful. Lorentz (or deterministic) chaos has the particular sense of

...which may appeal to mathematicians but not to me.

I much prefer to think in terms of systems that spontaneously trend towards high diversity. This correlates in simple proportion to both maximal entropy and quantum entanglement. As an engineer, this helps give an inituitive feel for how, for example, the system's thermodynamic and chemical equilibria are likely to evolve.  

 

 

What fascinates me most with complexity are sensitive to initial conditions and order in disorder

[KJW]

3 hours ago, Luc Turpin said:

I think that i know what linear means

Just in case you don't, an operator [math]L()[/math] is linear if and only if it satisfies:

[math]L(\psi + \phi) = L(\psi) + L(\phi)[/math]

Linearity is essential to QM because quantum superposition demands it.

 

 

[sethoflagos]

27 minutes ago, Luc Turpin said:

What fascinates me most with complexity are sensitive to initial conditions and order in disorder

The beauty of 'diversity' is that it emcompasses elemental domains of both order and disorder within a heterogenous whole, which corresponds with everyday macroscospic experience.

[Luc Turpin]

6 minutes ago, sethoflagos said:

The beauty of 'diversity' is that it emcompasses elemental domains of both order and disorder within a heterogenous whole, which corresponds with everyday macroscospic experience.

Vey well said

[geordief]

20 minutes ago, KJW said:

Just in case you don't, an operator L() is linear if and only if it satisfies:

L(ψ+ϕ)=L(ψ)+L(ϕ)

Linearity is essential to QM because quantum superposition demands it.

 

 

Are there  other  features of QM that could be described as essential  in the way that it differs from Classical Physics?

[swansont]

5 minutes ago, geordief said:

Are there  other  features of QM that could be described as essential  in the way that it differs from Classical Physics?

The probabilistic part of it. 

[Genady]

14 minutes ago, geordief said:

Are there  other  features of QM that could be described as essential  in the way that it differs from Classical Physics?

Non-commuting observables.

[joigus]

39 minutes ago, Genady said:

Non-commuting observables.

Oh, man. If something is it. That is it. 

A logic that sometimes doesn't allow you to say this and that.

That's the essence of Bell''s theorem without a doubt.

[TheVat]

1 hour ago, Genady said:

Non-commuting observables.

 

Let me see if I'm following this.

Noncommuting, this means a measurement can change the state, right ?

If we have observations A and B, we have two observables A and B that fail to commute, it is to say there is an eigenvector (direction) of A that it is not an eigenvector of B.  

So then an interaction with B isn't just some passive uncovering of preexisting information, but is an interaction that can change the state in question.  

[KJW]

1 hour ago, joigus said:

1 hour ago, Genady said:

Non-commuting observables.

Oh, man. If something is it. That is it. 

A logic that sometimes doesn't allow you to say this and that.

That's the essence of Bell''s theorem without a doubt.

I've never seen Bell's Theorem described in terms of non-commuting observables. Could you please explain how non-commuting observables relate to Bell's Theorem.

 

[Genady]

52 minutes ago, TheVat said:

Noncommuting, this means a measurement can change the state, right ?

A measurement usually changes the state. Non-commuting implies that the measurement of one observable affects the measurement of the other. The two cannot be both measured independently on the same state.

[joigus]

1 hour ago, KJW said:

I've never seen Bell's Theorem described in terms of non-commuting observables. Could you please explain how non-commuting observables relate to Bell's Theorem.

 

Well, you're right. You must assume quantum mechanics, of course. It's not just plain non-commutativity. You must assume in particular the correspondence between observables-operators and eigenvalues-spectrum of measurement.

That's why I didn't say Bell's theorem can be described in terms of non-commuting observables. I said they are the essence, but QM must be in the back of your mind.

Bell's theorem is about whatever variables that can take on definite values at the same time.

As the theoretical structure of QM forbids non-commuting operators to take on a definite value at the same time, there's the connection. But you do need the apparatus of QM, sure.

Edited December 9, 2023 by joigus
minor correction

[geordief]

37 minutes ago, Genady said:

A measurement usually changes the state. Non-commuting implies that the measurement of one observable affects the measurement of the other. The two cannot be both measured independently on the same state.

You say "usually".

Does that mean or imply that some measurements do not change the state of a system?

[Genady]

6 minutes ago, geordief said:

Does that mean or imply that some measurements do not change the state of a system?

Yes, it does.

[geordief]

6 minutes ago, Genady said:

Yes, it does.

Could I ask you for an example?

[Genady]

26 minutes ago, geordief said:

Could I ask you for an example?

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

[geordief]

28 minutes ago, Genady said:

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

Would that apply to the position/momentum of a quantum object?

(I hope "quantum object "is a meaningful description  of  something)

[Genady]

9 minutes ago, geordief said:

Would that apply to the position/momentum of a quantum object?

(I hope "quantum object "is a meaningful description  of  something)

(We can call it, "particle".)

Yes, if a particle has a definite momentum, measuring its momentum does not change its state. The same holds for its position.

[geordief]

5 minutes ago, Genady said:

(We can call it, "particle".)

Yes, if a particle has a definite momentum, measuring its momentum does not change its state. The same holds for its position.

So a particle whose momentum  is measured does not undergo a change in momentum as a consequence of the measurement?

Or is it rather that a particle whose momentum  is measured does not undergo a change in position as a consequence of the measurement.?

I feel it should be the latter as the position is unknown both before and after the measurement.

And ,if that is right ,the same would apply to a measurement of the position of the particle.That should not change its momentum ,for the same reasons.

[Genady]

2 minutes ago, geordief said:

So a particle whose momentum  is measured does not undergo a change in momentum as a consequence of the measurement?

This cannot be answered. What can be said is, 

 

13 minutes ago, Genady said:

if a particle has a definite momentum, measuring its momentum does not change its state.

If a particle has a definite momentum, measuring its momentum does not change its momentum, its position, its spin, etc.

[swansont]

7 minutes ago, geordief said:

So a particle whose momentum  is measured does not undergo a change in momentum as a consequence of the measurement?

It would have to be in a momentum eigenstate

12 minutes ago, Genady said:

If a particle has a definite momentum, measuring its momentum does not change its momentum, its position, its spin, etc.

I don’t think you can claim this. If a particle has a definite momentum its position is completely uncertain, so you can’t say measurement doesn’t change it.

The eigenstates of the momentum operator are plane waves, which you can’t normalize. It’s not physical.

[Genady]

8 minutes ago, swansont said:

If a particle has a definite momentum its position is completely uncertain

Yes, it is completely uncertain before and after the measurement. I call this, "does not change."

[geordief]

11 minutes ago, swansont said:

It would have to be in a momentum eigenstate

What might it be if it was not in a momentum eigenstate?

Any one of the other eigenstates it could be in?

Can it be in more than one eigenstate at a time ?

[Genady]

2 minutes ago, geordief said:

What might it be if it was not in a momentum eigenstate?

Superposition of states.

[geordief]

 How do we know that a particle is in  a momentum eigenstate and not in a superposition of states?

1 minute ago, Genady said:

Superposition of states.