# B How do we know superposition exists?

1. Jun 10, 2017

### Lunct

When measured/viewed particles in superposition collapse and take the form of one state. So how do we know they are in superposition in the first place. And do we know why they collapse into one state?

2. Jun 10, 2017

### PeroK

This question is based on a fundamental misunderstanding of superposition. Every state is a single state and, at the same time, a superposition of other states.

Essentially a state is a vector and a vector, as you know, is a single vector and linear combination of other vectors.

3. Jun 10, 2017

### Lunct

I'm sorry I did not understand that.

4. Jun 10, 2017

### PeroK

I would say it is one of the most common misconceptions about superposition. Also, "collapse" is a really misleading word, IMHO. In the Copenhagen interpretation, the state changes suddenly after a measurement to the relevant eigenstate of the observable.

This "collapsed" state is just as much a state as any other. In fact, in terms of position, say, the eigenstates are not
physically realisable, so after a position measurement the state "collapses" into a continuous distribution of position eigenstates - which is sometimes known as a "wave-packet". THis is just a regular state, albeit localised about a particular point.

5. Jun 10, 2017

### Lunct

So essentially the superposition only "collapses" after a specific measurement to the particle. Is that how it works?

6. Jun 10, 2017

### Staff: Mentor

It's analogous to the way that I can describe the position of something as "one kilometer to the north and one kilometer to the east" or as "1.414 kilometers northeast, zero kilometers northwest". The first description is more convenient if I'm looking at a map with north up and north/south and east/west gridlines on it; the second is more convenient if I'm in a city whose streets are laid out in a grid pattern with downtown/uptown avenes running from southwest to northeast and crosstown streets at right angles to the avenues. One way, my basis vectors are north/east, the other ways they're uptown/crosstown. But it's the same point with the same physical relationship to me either way.

7. Jun 10, 2017

### PeroK

That's the Copenhagen interpretation. There are other interpretations, but to get you started on QM Copenhagen is as good as any. So, yes, a measurements "collapses" the system into an eigenstate of the observable, with the eigenstate correspondiong to the eigenvalue returned by the measurement.

8. Jun 10, 2017

### Lunct

The Copenhagen interpretation says that we can never find out why it "collapses", right?

9. Jun 10, 2017

### PeroK

Yes, there's no why in that sense.

10. Jun 10, 2017

### vanhees71

The Copenhagen Interpretation doesn't exist. There are many flavors of it, some with collapse, some without. IMHO QT is much less weird when avoiding collapse arguments. There are tons of arguments for and against this opinion in this forum. Just use the Search function.

11. Jun 10, 2017

### Lunct

Thank you
Thank you as well

12. Jun 11, 2017

### Physics Footnotes

Taken at face value (which may not be quite what you meant, as I'll explain shortly), @PeroK gives the correct answer, which I quote again because this fact can never be repeated too much:
However, here is a modification of your question which I think you may have intended (and if not, I'm pretty sure it will teach you something useful anyhow!).

Suppose we have an observable $A$ which has two eigenstates $|a\rangle$ and $|b\rangle$ corresponding to eigenvalues $a$ and $b$, respectively. We now make a series of $A$-measurements on a beam of particles and get a series of readings like this:$$aababaabbbbaababaabbaaaabbbaba$$ Your teacher says that the particles in the beam have all been prepared in a superposition state of the form $\alpha|a\rangle + \beta|b\rangle$ and that upon measurement each particle's state 'collapses' randomly to either $|a\rangle$ or $|b\rangle$ with probabilities $\alpha^{2}$ and $\beta^{2}$, respectively, resulting in the observed series of $a$ and $b$ values.

But You Ask: How do we know that the particles were originally in the assumed superposition state? Couldn't we just assume that the particles in the beam are each in either the state $|a\rangle$ or $|b\rangle$, resulting in a mixture of these states in proportions $\alpha^{2}$ and $\beta^{2}$, rather than a superposition? After all, we never get to 'directly observe' the superposition state $\alpha|a\rangle + \beta|b\rangle$ because the measurement process keeps projecting the damn thing onto a single eigenstate!

The Answer: You can't distinguish a superposition of eigenstates of $A$ from a mixture of those eigenstates using the observable $A$ itself, because it will give just the same statistics in each case! The way to distinguish the superposition $\alpha|a\rangle + \beta|b\rangle$ from the corresponding mixture (which, by the way, we represent as $\alpha^{2}|a\rangle\langle a| + \beta^{2}|b\rangle\langle b|$) is to measure instead a different observable $B$ that (to use a technical term) does not commute with $A$.

The difference between the two statistical patterns of $B$-measurements on the superposition $\alpha|a\rangle + \beta|b\rangle$ versus the mixture $\alpha^{2}|a\rangle\langle a| + \beta^{2}|b\rangle\langle b|$ is precisely what we mean by the term interference effect, motivated by our good old friend the double-slit experiment. (In that experiment $A$ corresponds to the pair of detectors placed at the slits, while $B$ corresponds to the screen placed some distance away.)
Forgetting any quibbles about how you've phrased the question, the answer is, no we don't.

Last edited: Jun 11, 2017
13. Jun 11, 2017

### Lunct

Is that where the interpretations of QT come in, like the Many Words interpretation?