# B Robust against Decoherence

1. Oct 10, 2016

### bluecap

A question about page 6 of https://arxiv.org/pdf/quant-ph/0101077v1.pdf

"The second unanswered question in the Everett picture was more subtle but equally important: what physical mechanism picks out the classical states — face up and face down for the card — as special? The problem was that from a mathematical point of view, quantum states like "face up plus face down" (let’s call this "state alpha") or "face up minus face down" ("state beta", say) are just as valid as the classical states "face up" or "face down".

So just as our fallen card in state alpha can collapse into the face up or face down states, a card that is definitely face up — which equals (alpha + beta)/2 — should be able to collapse back into the alpha or beta states, or any of an infinity of other states into which "face up" can be decomposed. Why don’t we see this happen?

Decoherence answered this question as well. The calculations showed that classical states could be defined and identified as simply those states that were most robust against decoherence. In other words, decoherence does more than just make off-diagonal matrix elements go away. If fact, if the alpha and beta states of our card were taken as the fundamental basis, the density matrix for our fallen card would be diagonal to start with, of the simple form

density matrix = [1 0]
--------------------[0 0]

since the card is definitely in state alpha. However, decoherence would almost instantaneously change the state to

density matrix = [1/2 0]
--------------------[0 1/2]

so if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. In contrast, if we put the card in the state "face up", it would stay "face up" in spite of decoherence. Decoherence therefore provides what Zurek has termed a "predictability sieve", selecting out those states that display some permanence and in terms of which physics has predictive power."

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What I'd like clarification is the following. It says above that if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. Whereas if it is face-up.. it will stay face up in spite of decoherence.. Does it mean that the alpha and beta state are really still there just we don't see it because it says if you measure it you could still get random outcome?

2. Oct 16, 2016

### Greg Bernhardt

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Oct 16, 2016

### David Lewis

Because the front and back faces are pre-existing values intrinsic solely to the card. They don't depend on interaction with a detector to bring them into existence.

4. Oct 16, 2016

### bluecap

No, the face up and face down is just an example of quantum choices like spin up and spin down but the author use face up or face down which is clasical and may just confuse beginners. In fact I think the whole robust against decoherence thing escapes even the experts that's why the science advisers may not understand it and didn't reply to my original thread.

So let me share (after days of analyzing it). It means even though we don't see cat that is both alive and dead at the same time, the superposition of both alive and dead still exist but only for very few micro split second, then it reverts back to either dead or alive. We see dead or alive more often because it is more stable. For superposition of dead and alive. it still can exist as when Tegmark explained: "so if we could measure whether the card was in the alpha or beta-states, we would get a random outcome. In contrast, if we put the card in the state "face up", it would stay "face up" in spite of decoherence."

the "random outcome" thing means it still exists for a very very tiny fraction of second. If any science advisors understand me.. then please second the statement for our newbies here.

5. Oct 16, 2016

### David Lewis

Please correct me but my understanding is the alive and dead states (as they apply to the cat) cannot be in superposition. That is, the cat must be either dead or alive at all points in time.

6. Oct 17, 2016

### Staff: Mentor

That you find yourself in some world, and in some world its one or the other is a postulate. All theories have them. Why pick on that one rather than say why no prior geometry in GR? Its just the way nature is

Thanks
Bill

7. Oct 17, 2016

### secur

First, suppose you measure a particle and find it's spin up. Then it will stay that way unless disturbed, like the face-up card. But if you then measure the alpha / beta states - which can be done, in this case - you'll get one of them definitely. It will no longer be spin up, but a superposition of up and down (added, in the alpha case, else subtracted). And it will stay that way until disturbed. The disturbance could come from decoherence, which might force it into definite spin up, or down, state. Or, it might not.

Now, suppose you can actually have a superposition of dead and alive cat. If it were possible, decoherence undoubtedly would immediately force it into one or the other state, alive or dead. Nothing could put it back into the "alpha" state, since we have no way to measure such, and decoherence would never cause that. But since it's unknown whether superposition of states would apply to a cat, as Tegmark and MWI assume, it's all speculation.

8. Oct 17, 2016

### jartsa

Let's imagine a cat in a box, let's say we know it's dead or alive with equal probability. Or should I say it's dead and alive with equal amplitude.

Let's say we have received one photon from the cat, let's say that photon could equally well have been emitted by a dead cat or a living cat.

When we absorbed the photon, we decohered, but we did not decohere so that in the future we would be more likely to decohere to a state where we know that the cat is dead, or the other alternative.

We did not become robust against decoherence which would lead to state "cat is dead". And we did not become robust against decoherence which would lead to the other alternative.

We did not actually measure anything about the cat in the above scenario, right?

A real measurement must change the probabilities of the results of further measurements.

9. Oct 17, 2016

### PeroK

Suppose you have a card balanced on its end. How does that fit with the quantum nature of a card that demands it must either be face down or face up? If that's physically too difficult, you could use a domino instead of a card.

Or, if you pick the card up, you can orient it any way you like.

10. Oct 17, 2016

### Staff: Mentor

That's impossible but requires another thread. It is dead or alive and can never be in superposition ie dead and alive.

Thanks
Bill

11. Oct 17, 2016

### Staff: Mentor

No, it means that you've written down a two-by-two matrix with both diagonal elements equal to $\frac{1}{2}$. It also means that if alpha and beta were eigenvectors of some observable, then a measurement of that observable would leave the system in the state alpha half the time and beta half the time. It doesn't say anything about alpha or beta if no such measurement is made.

12. Oct 17, 2016

### bluecap

This is why we are supposed to see the particle moving back and forth from spin up to spin down and spin up and so on randomly. Because even after we measured it as spin up.. you said the particle will recohere into alpha-/beta states then decohere back in spin up and down.. since it's random.. then the particle should shift back and forth between spin up and spin down.. why doesn't it happen?

13. Oct 17, 2016

### PeroK

That's not what happens. You've misunderstood. Let's take a spin 1/2 particle:

1) You measure its spin about the z-axis and get a result of spin-up. Note: it's not accurate to say that you found the particle in the spin-up state, implying that it was in that state before you measured it. The act of measuring spin about the z-axis essentially forced the particle to take on a definite value of spin about that axis.

2) You measure its spin about the z-axis again. You get the same result. Measurements will remain spin-up until something disturbs the particle.

3) You measure its spin about the x-axis. This does two things.

a) The particle now has a definite value of spin about the x-axis. Before this measurement, it had no definite value of spin about the x-axis and, in fact, because you knew it was spin-up in the z-direction, it was 50-50 whether you would get a measurement of spin-up or spin-down in the x-direction. Note also that, while it was spin-up in the z-direction:

It had a state (spin-up in the z-direction), but this state is a linear combination of x-spin-up and x-spin-down states: this state, therefore, gives a definite value for a measurement of spin in the z-direction, but a random 50-50 measurement of spin in the x-direction.

b) The previous value of spin in the z-direction is effectively destroyed by the measurement of spin in the x-direction. You can see this by:

4) You measure spin in the z-direction and get spin-up or spin-down with equal probability. The previous definite value of spin-up in the z-direction was lost by the x-spin measurement.

Finally, suppose you have got a measurement of spin-up in the z-direction:

5) You measure the spin about an axis that is close to the z-axis (perhaps only a few degress off vertical). You still get spin-up or spin-down about this axis at random, but it is much more likely to be spin-up. It might be 95% spin-up and only 5% spin down. This is because z-spin-up is a linear combination of spin-up and spin-down about this axis, but is predominantly spin-up (in terms of probability amplitude).

Whereas, z-spin-up is an equal linear combination of x-spin-up and x-spin-down, so you get these measurements with equal probability for a particle in the z-spin-up state..

Finally, IMHO, playing cards simply do not behave like this, so why anyone would use playing cards to illustrate quantum behaviour is beyond me!

14. Oct 17, 2016

### bluecap

It's no less written by the legendary Max Tegmark.. and it's shared in the equally legendary arxiv https://arxiv.org/pdf/quant-ph/0101077v1.pdf

Let's replace his face up and face down with spin up and spin down and reread his statements (if it makes sense at all) :

"The second unanswered question in the Everett picture was more subtle but equally important: what physical mechanism picks out the classical states — spin up and spin down for the particle — as special? The problem was that from a mathematical point of view, quantum states like "spin up plus spin down" (let’s call this "state alpha") or "spin up minus spin down" ("state beta", say) are just as valid as the classical states "spin up" or "spin down".

So just as our particle in state alpha can collapse into the spin up or spin down states, a spin that is definitely spin up — which equals (alpha + beta)/2 — should be able to collapse back into the alpha or beta states, or any of an infinity of other states into which "spin up" can be decomposed. Why don’t we see this happen?

Decoherence answered this question as well. The calculations showed that classical states could be defined and identified as simply those states that were most robust against decoherence. In other words, decoherence does more than just make off-diagonal matrix elements go away. If fact, if the alpha and beta states of our spin were taken as the fundamental basis, the density matrix for our particle would be diagonal to start with, of the simple form

density matrix = [1 0]
--------------------[0 0]

since the particle is definitely in state alpha. However, decoherence would almost instantaneously change the state to

density matrix = [1/2 0]
--------------------[0 1/2]

so if we could measure whether the particle was in the alpha or beta-states, we would get a random outcome. In contrast, if we put the particle in the state "spin up", it would stay "spin up" in spite of decoherence. Decoherence therefore provides what Zurek has termed a "predictability sieve", selecting out those states that display some permanence and in terms of which physics has predictive power."

In Max Tegmark example. Does he only use one axis? If it doesn't make sense in one axis only (as you are emphasizing). What other observables and quantum setup can his statements make perfect sense?

15. Oct 17, 2016

### David Lewis

What we know is different from what is.

16. Oct 17, 2016

### Staff: Mentor

It doesn't. To get something sensible, you would replace his up or down card with a quantum system consisting of the particle and the measuring device. This system will have the measuring device reading either up or down after the measurement as surely as a tossed coin is heads or tails or Tegmark's card is face up or face down.

The state that you are calling alpha is spin left and the state that you're calling beta is spin right, so all we need to do get a collapse into one of those states is to perform a measurement along the horizontal axis. The same holds for a decomposition into any other basis. We don't see this happen spontaneously (that is, in the absence of any measurement or other interaction) because the spin operator commutes with the Hamiltonian; once a measurement leaves the particle in a particular pure state it stays there until something else disturbs it.

But this is the behavior of an isolated particle. It's a very different system than the combination of particle and measuring device that is analogous to Tegmark's cards.

Last edited: Oct 17, 2016
17. Oct 17, 2016

### jartsa

Let's see what kind of sense I can make out of that thing, although I don't know any quantum mechanics.

When the card is in state alpha, decoherence almost instantly changes the state to another state. In that other state the alpha and the beta are there in some way, because we would get randomly alpha or beta if we could measure whether the card was in the alpha or beta state.

But if we put the card in the state "face up", it will stay in the "face up" state — which equals (alpha + beta)/2. That state should be able to collapse back into the alpha or beta states, so why don’t we see this happen? Answer: Because that state is robust against decoherence.

So yes it seems to me that the excerpt is saying that the alpha and beta states are really still there just we don't see them. We don't see them, because that state refuses to collapse (decohere) to alpha or beta, because it just does not decohere.

18. Oct 17, 2016

### Staff: Mentor

I am at a loss to understand what could be meant by "the states are really there". It's like saying that 5 isn't really 5, it's really 3+2; yes, 5=3+2 is a true equation, but whether we write a number as "5" or "3+2" is just a matter of what's convenient at the moment as we work through whatever problem we're solving.

19. Oct 17, 2016

### stevendaryl

Staff Emeritus
I don't think that's true. As I said on another thread, decoherence does not pick out one state, dead or alive. It doesn't choose which one becomes the real state of the cat. What decoherence does is to destroy interference effects, so that the situation of the cat is described by a mixed state, rather than a superposition. So the logic of decoherence is this:
• Decoherence is a matter of the system (the cat, for example) becoming entangled with the environment (basically, the rest of the universe).
• After decoherence, the system (cat) is described by an (improper) mixed state.
• As far as observations of the cat alone is concerned, there is no distinction between an improper mixed state and a proper mixed state. They are described by the same density matrix.
• But a proper mixed state of alive cat/dead cat can be interpreted as "The cat is either alive, or dead, we just don't know which until we observe it".
• So after decoherence, you can pretend that the cat is in a definite state of alive or dead, and you won't get into trouble.
But what decoherence definitely does not do is to pick out which (alive or dead) is actually the case.

20. Oct 17, 2016

### Mentz114

The cat has always been decohered so what is the state of the cat before the first bullet point ?

I would agree with the conclusion - we get a statistical mixed state. But we had that as soon as the box closed, surely ?

21. Oct 17, 2016

### stevendaryl

Staff Emeritus
Yes, because a cat is a big object, unlike an electron, the process of the cat becoming entangled with the rest of the universe through decoherence is going on continuously. If there is a single atom whose decay triggers the death of the cat, then what's really going on is that atom's state is becoming entangled with that of the cat and the rest of the universe. The cat is never in a superposition of dead/alive, it's always entangled.

22. Oct 17, 2016

### Mentz114

OK, I can buy that - except for the entanglement, which seems to be irrelevant in any case.

23. Oct 17, 2016

### stevendaryl

Staff Emeritus
Why do you say entanglement is irrelevant? I would say that's the only reason that there are no macroscopic superpositions (of an alive cat and a dead cat).

24. Oct 17, 2016

### stevendaryl

Staff Emeritus
Decoherence as I understand it simply is entanglement with the "environment".

25. Oct 17, 2016

### Mentz114

Irrelevant to the 'robustness-against-decoherence' ?
I think you'd be wrong if you said that. 'Only' is a rather extreme term.