High School The Wave Function of Our World: How Does It Emerge?

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Particles possess a wave function that collapses to a specific eigenvalue upon measurement, leading to the question of whether macroscopic objects, like tables, exist in a collapsed state or remain part of a non-collapsed quantum wave function. The discussion highlights that while individual particles cannot have precise values for both position and momentum, macroscopic objects behave classically due to averaging effects over many particles, making quantum uncertainties negligible. Decoherence is presented as a key concept explaining why macroscopic objects do not exhibit superpositions of distinguishable states, as it renders such superpositions unobservable rather than causing actual collapse. The conversation also touches on the implications of decoherence for understanding the wave function of the universe and the nature of measurement in quantum mechanics. Overall, the emergence of classical behavior from quantum mechanics remains a complex and debated topic.
  • #91
Mentz114 said:
Calling the mixture a super-position is just a matter of choice.

No, it isn't. Saying that "the value of spin-x is definite" is a much stronger statement than saying "there is a 50-50 probability of measuring spin-z up or down". There are an infinite number of possible states that can give the latter probabilities. But there are only two states (the spin-x eigenstates) in which the value of spin-x is definite.

Saying that a state is a "mixture" of 50-50 spin-z up and spin-z down only tells you that you are in one of the infinite number of possible states that can give those probabilities. Saying that you are in a "superposition" of spin-z up and spin-z down implies that you know which of those infinite number of possible states is the one that was actually prepared.

In other words, a "superposition" is a pure state, whereas a "mixture" is not. At least, that's the standard terminology. If you are going to use non-standard terminology, that's your choice, but it doesn't change the physical distinction between pure states and mixed states.

Mentz114 said:
it isn't relevant to my assertion that the state ##\alpha |u\rangle + \beta |d\rangle## can only be interpreted as a prediction of the reletave frequencies of certain observations

That's a matter of interpretation. Some interpretations limit the meaning of the state to this, and some don't. But the distinction between pure states and mixed states is not interpretation dependent.
 
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  • #92
PeterDonis said:
No, it isn't. Saying that "the value of spin-x is definite" is a much stronger statement than saying "there is a 50-50 probability of measuring spin-z up or down". There are an infinite number of possible states that can give the latter probabilities. But there are only two states (the spin-x eigenstates) in which the value of spin-x is definite.

Saying that a state is a "mixture" of 50-50 spin-z up and spin-z down only tells you that you are in one of the infinite number of possible states that can give those probabilities. Saying that you are in a "superposition" of spin-z up and spin-z down implies that you know which of those infinite number of possible states is the one that was actually prepared.

In other words, a "superposition" is a pure state, whereas a "mixture" is not. At least, that's the standard terminology. If you are going to use non-standard terminology, that's your choice, but it doesn't change the physical distinction between pure states and mixed states.
Thanks for the explanations. If I understand correctly - suppose we prepare 2 beams A (spin-z up) and B(thermal) and subject them to measurements in the three orientations. For beam A we get 100% z-up, x-thermal, y-thermal, For B all three will be thermal. The first x and y measurements are on a superposition and the second two on a mixture, but they give the same results. Thus I find it hard to distinguish physically between the cases.

However, this has never been a serious issue, but the next one is.

That's a matter of interpretation. Some interpretations limit the meaning of the state to this, and some don't. But the distinction between pure states and mixed states is not interpretation dependent.
Do you know in what way (other than probabilistic) can the state ##\alpha |u\rangle + \beta | d \rangle## be interpreted, according to those who interpret it differently ?
(that is a terrible sentence - I hope you understand)
 
  • #93
Mentz114 said:
Do you know in what way (other than probabilistic) can the state ##\alpha |u\rangle + \beta | d \rangle## be interpreted

As an eigenstate of a different operator. In other words, as the real, physical state of the system, which leads to probabilistic results with the operator you originally chose, but which leads to perfectly definite, non-probabilistic results with another operator.

To put this another way: there is a theorem which says that given any state in a Hilbert space, there will be some Hermitian operator that has that state as an eigenstate. To some people, that theorem is sufficient to show that the state the actual, physical state of the system, not just something probabilistic. You can agree or disagree with that interpretation, but it is a permissible interpretation at our current state of knowledge.
 
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  • #94
Mentz114 said:
The first x and y measurements are on a superposition and the second two on a mixture, but they give the same results. Thus I find it hard to distinguish physically between the cases.

In other words, you ignore the crucial measurement--spin-z--that actually does easily distinguish physically between the cases. Of course that will make it hard to distinguish physically between the cases.

The fact that the spin-z eigenstate is not actually a superposition in the spin-z basis is a red herring. The crucial point, as I have said a couple of times now, is that the spin-z eigenstate is a pure state, whereas your "thermal" state is a mixed state.
 
  • #95
PeterDonis said:
As an eigenstate of a different operator. In other words, as the real, physical state of the system, which leads to probabilistic results with the operator you originally chose, but which leads to perfectly definite, non-probabilistic results with another operator.

To put this another way: there is a theorem which says that given any state in a Hilbert space, there will be some Hermitian operator that has that state as an eigenstate. To some people, that theorem is sufficient to show that the state the actual, physical state of the system, not just something probabilistic. You can agree or disagree with that interpretation, but it is a permissible interpretation at our current state of knowledge.
Thank you.
 
  • #96
PeterDonis said:
In other words, you ignore the crucial measurement--spin-z--that actually does easily distinguish physically between the cases. Of course that will make it hard to distinguish physically between the cases.
More important is that the state of z-spin will not affect orthogonal directions. It gives no information about anything but z. So the state is perfect ie.it does not change under a measurement - in a chosen direction only. But this is a mere quibble.
 
  • #97
Mentz114 said:
More important is that the state of z-spin will not affect orthogonal directions.

What do you mean by "affect"? If you start with a particle that is spin-up in the x-direction, and then measure the spin in the z-direction, it will no longer be spin-up in the x-direction. So it's unclear what it means to say that it's z-spin will not affect its x-spin.
 
  • #98
Mentz114 said:
state of z-spin will not affect orthogonal directions. It gives no information about anything but z.

This statement is much too strong. An eigenstate of spin-z gives no information (in the sense of giving 50-50 probabilities) about spin-x or spin-y or linear combinations of those two states. But it does give information (i.e., does not give 50-50 probabilities) about any other observable, i.e., any observable that includes any amplitude at all for spin-z.
 
  • #99
stevendaryl said:
What do you mean by "affect"? If you start with a particle that is spin-up in the x-direction, and then measure the spin in the z-direction, it will no longer be spin-up in the x-direction. So it's unclear what it means to say that it's z-spin will not affect its x-spin.
Sorry that was loosely worded, I mean 'affect predictions about'.
 
  • #100
PeterDonis said:
This statement is much too strong. An eigenstate of spin-z gives no information (in the sense of giving 50-50 probabilities) about spin-x or spin-y or linear combinations of those two states. But it does give information (i.e., does not give 50-50 probabilities) about any other observable, i.e., any observable that includes any amplitude at all for spin-z.
Sure.

The first question I asked about superposition of the measurement apparatus has been answered and I now understand more about how one may choose to interpret a state.
 
  • #101
I've recommended this book so many times that I'm beginning to think I should get a commission on the sales... But if you can get hold of a copy of David Lindley's "Where does the weirdness go?", give it a try. It's a pretty good layman-friendly treatment of how our world emerges from the quantum world.

Ditto on that. :smile:
 

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