# I Meaning of superposition

1. Mar 14, 2017

### mike1000

[Mentor's note: Split off from this thread]
I hope I am not hijacking this thread, but I noticed something. When a system is in a superposition of quantum states, as in the six atoms example, isn't that an expression that, from the probability viewpoint, they do not know what state it is in? What the experiment was saying is that, quantum mechanically, they did not know what state it was in and it could equally likely be in the spin up or the spin down state. Saying it is in a superposition of states is merely a mathematical way of saying we don't know what state it is in and the amplitude of the superposed states just tells us the likelihood of being in one or the other?[/URL]

Last edited by a moderator: Mar 14, 2017
2. Mar 14, 2017

### vbrasic

Basically. We won't know until we make a measurement, and then, the superposition state will just collapse into any of its constituent states. The superposition state, as you correctly posit, simply tells us what we might see (in your case up/down spins). As you will note, this is indeed a very probabilistic statement.

Well, to be super technical, it tells us the likelihood of being in one of the states IF we measure the system. Otherwise, it just remains in the superposition state. But yes, you are correct, it just gives us a probability.

3. Mar 14, 2017

### Strilanc

Superpositions aren't probability distributions. They don't represent uncertainty.

When a qubit is in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, there's no uncertainty. The von neumann entropy is zero. We know the exact state. There are measurements that you can perform on a qubit in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$ that have uncertain results, but the state itself is not uncertain. It's exactly $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$.

Rational agents can't even disagree on the pure superposition that a system is in (but they can disagree on the mixed state, since those can represent uncertainty). If Alice says the state is $\sqrt{\frac 3 5} |0\rangle + \sqrt{\frac 4 5} |1\rangle$ and Bob says the state is $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, at least one of them just unnecessarily risked losing infinite Bayes points. That's just not the kind of thing that rational agents do.

4. Mar 14, 2017

### mike1000

When you make a measurement do you ever actually measure a state which is a superposition of the possible eigenstates?

5. Mar 14, 2017

### Strilanc

Of course you may. That would be an example of a measurement that has an uncertain outcome. But it's the measurement that's uncertain, not the state. We can be sure about what the state is.

6. Mar 14, 2017

### Staff: Mentor

Only on a collapse interpretation. There are QM interpretations that don't have collapse, such as the MWI.

7. Mar 14, 2017

### mike1000

Well, is the state that you measured a superposition of eigenstates? I am no expert, but if the system can only exist in one of the allowed eigenstates, it seems to me that when you make a measurement you are going to find the system is, indeed, in one of those states. Before hand, when you do not know what state is going to be in, the wave function might be represented by a superposition of eigenstates, each with an amplitude, that is proportional, in some way, to the likelihood of the measurement being in one of the superposed states.

8. Mar 14, 2017

### Staff: Mentor

That is not the case. A system can exist in any state in its Hilbert space. The eigenstates of a particular measurement operator are only a very small subset of all the possible states of the system.

9. Mar 14, 2017

### mike1000

When you say in its Hilbert Space, you mean the Hilbert Space of the operator? And would that mean any state which is a linear combination of the eigenstates? Does this mean you can measure the spin of an electron to be both up and down at the same time?

10. Mar 14, 2017

### Staff: Mentor

No, of the system. The operator is a mapping from the Hilbert space to itself.

If the measurement operator is complete, then its eigenstates form a basis of the Hilbert space, so any state would be expressible as a linear combination of the eigenstates, yes.

All of this is laid out in QM textbooks; if you haven't studied one, I would try Ballentine.

11. Mar 14, 2017

### mike1000

Well, does that mean that you can measure the spin of an electron to be both up and down at the same time? I assume that is one of the states in the Hilbert Space of the system.

How do you get the Hilbert Space of the system from the eigenstates of the operator?

12. Mar 14, 2017

### Staff: Mentor

Not by measuring spin, no. But see below.

Yes; if we have chosen a particular axis for measuring the spin, the "up" and "down" eigenstates of that measurement operator could be written as $\vert \uparrow \rangle$ and $\vert \downarrow \rangle$. The state you describe as "up and down at the same time" (which is actually not a good description of a superposition) would be something like $\frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle + \vert \downarrow \rangle \right)$.

Now, this state is a perfectly valid state, and so there will be some Hermitian operator that has this state as one of its eigenstates. So if we were to measure using that operator (which will not be one of the usual spin operators, of course), we could measure the system to be in this state.

By forming all possible complex linear combinations of those eigenstates.

13. Mar 14, 2017

### PeroK

If the up and down states are in the z-direction, then the operator $S_x$ would have that combination as an eigenstate!

14. Mar 14, 2017

### Staff: Mentor

Yes, good point. So actually, the answer to the question is "yes" even more simply: the state "spin up about the x axis" is in fact a state which is both "up" and "down" about the z axis (if we allow ourselves to use that sloppy language).

Actually, now that I think of it, if we are just considering spin, any linear combination of eigenstates of some spin operator will be an eigenstate of some other spin operator. We would have to consider more general states (for example, states where both spin and position degrees of freedom are present) to find cases where linear combinations of eigenstates of a "simple" operator (like spin or position) are not also eigenstates of a "simple" operator.

15. Mar 14, 2017

### mike1000

My question is about measurement, what we would measure in an experiment. It seems rather simple, even to someone at my level, that one cannot measure the spin of an electron to be both up and down at the same time, it will be either up or it will be down, not both. I suppose you can manipulate the basis so that up spin or down spin do not coincide with an actual basis direction, but that, it seems to me, is exactly the kind of stuff we should try to avoid. It is that kind of discussion that leads lay people, (like myself) to misunderstand what quantum mechanics is doing, such as the misunderstanding that a system can be in two states at the same time. Some things are simple. Lets not intentionally make them more complex than they already are.

16. Mar 14, 2017

### Staff: Mentor

And if you take a particle in the state I wrote down (in the spin-z basis), and measure its spin about the z axis, you will measure either spin up or spin down. If you take a particle in that state and measure its spin about the x axis, you will always measure spin up--that state is an eigenstate of spin-x with the eigenvalue "up". But given the way that state is written in the spin-z basis, you could also interpret that spin-x measurement as telling you that the particle has z spin "both up and down at the same time". I don't think this would be a very fruitful interpretation, but mathematically how are you going to refute it? The state is what it is, and if you insist on using the spin-z basis it is a superposition, and remains a superposition after you do a spin-x measurement.

See above.

Why? The basis I use will depend on the measurement operator I want to apply, which will depend on the experiment I want to run. Why should I prevent myself from running the experiment I want to run, just because the state of particles coming into the experiment is or is not an eigenstate of the measurement operator?

17. Mar 14, 2017

### mike1000

You are obviously the expert, but I think you over complicate the situation because you want to win a debate. You even admit it is not fruitful. And you are right, it is not fruitful, it is actually harmful. Because you and even I know, that the spin can only be either up or down but not both. I would say this. If the mathematics can be interpreted in a way you describe above, that the electron spin can be both up and down at the same time, then that shows the model is not quite exactly right.

18. Mar 14, 2017

### Staff: Mentor

No, I don't know that as you state it, because as you state it, it makes it seem like the only possible states of the system are the eigenstates of one particular spin operator. Which is how this whole subthread got started in the first place.

If you really want to be clear, use math. Math is unambiguous. If you absolutely have to have an ordinary language description, you're going to have to make sure to make it as unambiguous and minimal as possible. Something like this, perhaps: "Whenever you measure the spin of a spin-1/2 system, you will get one of two results: up or down." And then stop. Don't say anything about what the state is, or whether it "can be up or down at the same time", or anything like that. Because anything you say, beyond the bare minimum, can and will be misinterpreted. That's why physicists use math when they actually want to make predictions.

The model makes correct predictions, regardless of how it is interpreted or whether the interpretation is considered fruitful. All that is shown by the possibility of bad interpretations, IMO, is the possibility of bad interpretations. The solution, IMO, is to avoid interpretations altogether. But very few people are willing to do that.

19. Mar 14, 2017

### Staff: Mentor

I would say that you are overcomplicating the situation by trying to overlay an imprecise ordinary language description on top of a precise physical model.

20. Mar 14, 2017

### mike1000

As you have shown, math is not unambiguous. It can be ambiguous too.

I like math very much and I really like to write equations in these posts using the equation editor. As I progress I will write some equations.

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21. Mar 14, 2017

### Staff: Mentor

It looks like your post got cut off here. In any case, I'm sorry you're frustrated, but I don't think the things I'm pointing out are "little things". You appear to agree that misinterpretation of QM is frequent. So it seems like trying to reduce the chance of misinterpretation would be a good thing. I'm just telling you what I think that requires. Yes, it's pretty stringent--but then again, if it weren't, I wouldn't expect it to fix a problem that is frequent.

Where did I show that?

22. Mar 14, 2017

### mike1000

Ignore the first part that was cut off. I was going to write something and decided not to so you can ignore that part.

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23. Mar 14, 2017

### Staff: Mentor

Exactly.

These things are much better expressed in math.

When Feynman met his hero Dirac - you know what Dirac said - 'I have an equation do you have one to'. Its amusing but contains much truth.

I suggest the following relatively cheap texts:
https://www.amazon.com/Quick-Calculus-Self-Teaching-Guide-2nd/dp/0471827223
https://www.amazon.com/Theoretical-Minimum-Start-Doing-Physics/dp/0465075681
https://www.amazon.com/Quantum-Mechanics-Theoretical-Leonard-Susskind/dp/0465062903

Thanks
Bill

Last edited by a moderator: May 8, 2017
24. Mar 14, 2017

### mike1000

Yes things are better expressed in math but that is no excuse for not expressing things clearly and simply in english. Here is a paper by Dirac that I found today. What impressed me was the simplicity with which he expressed his thoughts in the first few pages. You have to do both.

http://www.imotiro.org/repositorio/...vention of the second quantization method.pdf

Last edited by a moderator: May 8, 2017
25. Mar 14, 2017

### Staff: Mentor

Well I think you have been given clear answers - its just your background makes them not as easy to understand as they should be.

If you read the books I listed you will understand things a lot better.

Please take the time and effort to do it.

Thanks
Bill