# I Some said 1000's of experiments support superposition but

1. Apr 11, 2017

### mike1000

In another thread, that is now closed, someone said that 1000's of experiments support the proposition that a particle can be in two, mutually exclusive states at the same time.

The results of experiments show just the opposite. The measured result is not a superposition state. They never measure a result that is a superposition of eigenstates.

For an observable with a finite (=n) , discrete spectrum, there can be at most n possible outcomes of a quantum experiment, regardless of the basis chosen to represent those outcomes. If you consider the superposition states to be allowed outcomes then there would be an infinite number of allowed outcomes for all operator (which I think implies a continuous spectrum as well)

Last edited: Apr 11, 2017
2. Apr 11, 2017

### Greg Bernhardt

3. Apr 11, 2017

### Oudeis Eimi

That's not what superposition means. Other posters have expressed doubts that you understand what a vector space is, and that understanding happens to be critical to this issue.

ANY state vector (that's the simplest kind of quantum state) is in an infinite number of different superpositions when expressed in different bases. If a state vector is an eigenstate of operator A with eigenvalue a, it will be in a superposition in the eigenbasis of non-commuting operator B.

The status of being "in superposition" is NOT a characteristic of the state itself (which is ALWAYS unique, up to a phase factor, for vector ("pure") states), but of the arbitrary basis chosen to represent it.

Example: you measure z-component of a 1/2 spin particle, you get the "up" value, if we adopt a collapse point of view, immediately after measurement the particle is in the "spin-up along z" state, but is ALSO in a superposition in ANY other basis.

You will continue to have trouble with this until you study abstract linear algebra (the maths of vector spaces).

Last edited: Apr 11, 2017
4. Apr 11, 2017

### mike1000

I understand there are an infinite number of bases. I understand that an eigenvector of one basis may resolve into a superposition of basis vectors in a different basis. What I do not see, is how that somehow justifies the statement that a particle can be measured to be simultaneously in two (or more) mutually exclusive states.

Maybe what you are saying is that given the result of any quantum measurement, that is a single eigenstate of an operator, you can always find a different basis in which that single eigenstate is expressed as a linear combination of other basis vectors. That does not mean that the particle was actually in those two(or more) mutually exclusive states.

5. Apr 11, 2017

### Oudeis Eimi

It does not justify the statement, because that statement is not correct. The state is unique up to a phase factor, a particle is never in two different states (from two different rays in the Hilbert space anyway).

Using the example of a spin 1/2 system, a state | z up > can be expressed as a linear combination of | x up > and | x down >, but both the "original" and the "combination" state are one and the same. It's the same vector in two different bases. That's all there is to it.

The "z basis" will be the more convinient if further measurements will be done along the z axis, the "x basis" would be the one to use if they are to be performed along the x axis. The superposition in that case would mean that both x up and x down results are possible outcomes.

6. Apr 11, 2017

### mike1000

The way I see it, if you are using a basis to express your result and in that basis the result is expressed as a linear combination of mutually exclusive states, then that should tell you to look for a different basis.

7. Apr 12, 2017

### Staff: Mentor

Try this then. You prepare a spin in the state $| +x \rangle$, i.e. spin up along x. You then want to measure the spin along z. In what state is the spin just before measurement, and right after measurement?

8. Apr 12, 2017

### vanhees71

That's not true! At least I (and I think also several others) have told you that being in a superposition doesn't mean that the particle is in two states at the same time. Choosing another basis you could even say it's in infinitely many states at the same time. That's wrong, and repeating wrong nonsense doesn't help you nor others reading this forum!

As we have also explained several times, it is of utmost importance to dinstinguish clearly between observables and states. You don't measure states but observables. To completely determine a state, is much more complicated than just measuring a single observable. You rather have to make many different measurements on an ensemble and make a statistical evaluation of the outcomes. For a very detailed discussion of all these fundamental things like state determination, measurement, etc. read Ballentine, Quantum Mechanics - A modern development, Addison-Wesley.

9. Apr 12, 2017

### vanhees71

Of course this is correct, but a little subtlety is also important (also on a much more advanced level than we discuss at the moment to eliminate misconceptions of the OP he seems not to be willing to overcome). A state is uniquely determined not only up to phase. More precisely: states are uniquely represented by statistical operators (i.e., self-adjoint positive semidefinite operators with trace 1). A statistical operator represents by definition a pure state if and only if it's a projection operator, i.e., if $\hat{\rho}^2=\hat{\rho}$. This determines (now indeed up to phase factor!) a normalized vector in Hilbert space, $|\psi \rangle$, such that $\hat{\rho}=|\psi \rangle \langle \psi|$. This makes it evident that any phase factor drops out when forming the projector that represents the statistical operator in this case.

On the other hand, you can as well give an arbitrary "unit ray" in Hilbert space, i.e., an equivalence class of normalized Hilbert-space vectors $[|\psi \rangle]$ where all vectors that deviate from $|\psi \rangle$ only by a phase factor are identified, and in this way uniquely determine the state, represented by the above give projector. So it's the rays (or equivalently the projection operators as statistical operators) in Hilbert space that actually represent the pure states.

That's very important, because, e.g., it enables phenomena like half-integer angular momenta (spins), and all matter surrounding us is made of spin-1/2 particles on a fundamental level (quarks and leptons). It also enables Galilei invariant non-relativistic quantum theory that makes physical sense, because for that you need unitary ray representations of the Galilei group. Proper unitary transformations of the Galilei group do not lead to any useful quantum dynamics to work with! The true Galilei group is a non-trivial central extension of the covering group of the classical Galilei group with mass as a non-trivial central charge! See Ballentine, QM for details.

10. Apr 12, 2017

### Oudeis Eimi

Ballentine is indeed an invaluable reference. Yes, I was restricting the discussion to pure states, so as not to further confuse the OP, but the distinction is always worth making indeed.

11. Apr 12, 2017

### mike1000

Exactly what did I say that made you think I do not understand what is superposition? When I said the following in my original post
I was playing devils advocate. I was trying to make my point that superposition states do not represent possible outcomes, by which I mean, a particle cannot be found it two opposing states at the same time, even when you are using the spin z basis. That is a side-effect of the basis you used to measure the spin. By measuring in that basis and staying in that basis you promote the idea of "Schrodingers cat", ie the mis conception that a quantum particle can physically be in two mutually exclusive states at the same time.

Last edited: Apr 12, 2017
12. Apr 12, 2017

### Oudeis Eimi

Hello OP, the above makes IMHO a good example of your present confusion on this subject. I'll try to clarify some things here:

Outcomes, for the simplest, ideal case of a "projective measurement", are eigenvalues of observable operators (I'm restricting discussion to discrete spectra here for simplicity). They are in no way represented by or dependant on the state, except in the sense that the state contains information about the probabilities of different outcomes being realised. We could indeed say that consists a "representation", but that'd be sloppy language (and would cause confusion with other meanings of the term "representation").

The claim you've made that makes me think you don't understand superposition is your insisting that it implies the system being in more than one state at the same time. It doesn't, in the same way that someone having 7 grapes doesn't mean he or she has simultaneously 21 and minus 14.

Your example of the Schroedinger's cat is not a good one, as a cat cannot be in a superposition of living and dead due to its being a macroscopic object whose degree of interaction and entanglement with its environment forces a selection rule preventing any state other than "alive" or "dead". But, assuming we had a system with a two dimensional Hilbert state space, and calling "alive" and "dead" the two eigenvalues of some "L" observable, with eigenvectors | alive > and | dead >, a general vector state in that space would NOT represent a simultaneous state of being "alive" and "dead", but something other for which we have no name. More to the point, there is an observable, let's call it "N", for which the state vector, whatever it happens to be, is an eigenvector with eigenvalue "n". If we measured said observable "N" instead of "L", we would obtain "n", with probability 1, and perform a non-collapse measurement as the initial state was already an eigenvector of "N".

13. Apr 12, 2017

### mike1000

I am not insisting that the superposition be interpreted as the system being in more than one state at the same time. I am arguing completely against that. I sense that others, in this forum, have argued that. I am trying to counter that argument.

1)The vector space in which the state vector lives is a probability amplitude vector space.
2)The dimension of the vector space is determined by the number of possible outcomes.(Lets call it n)
4)The n possible outcomes form a natural orthogonal basis for the vector space. (Because all outcomes are mutually exclusive)
3)A arbitrary point in that space would be defined by a vector that has n components.
4)Each component of that vector represents the probability amplitude that the associated outcome will be observed.
5)Each vector in that space defines a probability amplitude distribution, This is superposition, ie. each vector defines a particular way that the probability amplitudes are partitioned among the n possible outcomes.
6)There are an infinite number of probability distributions.
7)There are only n possible outcomes.

You can change bases til your hearts content but you will not change the number of possible outcomes or what those possible outcomes are. Regardless of basis,you can only observe one of the n possible outcomes.

(While writing this the thought came to me, are there any cases where the set of possible outcomes are not mutually exclusive?)

Last edited: Apr 12, 2017
14. Apr 12, 2017

### MichPod

Ok. Let me have a try. Say that we measure a position of the particle. The result of the measurement is a position. We can think that our measurement measures the state of the particle being at a particular position - simple as that. That is what classical mechanics says.

Now QM says that there actually exist many more other, superposed states of the particle, i.e. states which "combine" a state of the particle being at one position and some other(s). That combination even includes some complex coefficients so that we may have "very many" possible combinations of the particle being at coordinate X1 and coordinate X2. Now, each such a "combination of states" is considered in QM as a state by itself. It is not "mix", not really a "combination", just a pure state, as pure as the state of being at one particular point. It is called "superposition".

Now, what happens when we measure a coordinate of such a superposition state? The answer is - we get back just one coordinate - either X1 or X2.

That is basically all. The only remaining question is why do we believe at all that the superposition states exist if we always get just one coordinage in our measurement? But the reasons to believe that the superposition exists is a separate topic.

15. Apr 12, 2017

### Staff: Mentor

No. There are n possible outcomes for each possible measurement operator. The outcomes for different measurement operators are different outcomes; the outcome "spin z +1/2" is not the same outcome as "spin x +1/2".

You have (despite this being pointed out multiple times in previous threads) left out the measurement operators entirely from your discussion. You need to put them back in.

16. Apr 12, 2017

### mike1000

I have always only been speaking of one measurement operator. Of course there are different outcomes for different measurement operators. That was not the topic of this thread. This thread is about the definition of superposition.

17. Apr 12, 2017

### zonde

I completely agree that superpositions are not measured. As I see superpostions are inferred from observations of interference effects i.e. one has so called superposition state and then different modes of superposition are mixed in some experimental arrangement. At different outputs of that setup we observe interference effects and consider that as confirmation of superposition.
So I rather think about superposition as a tool to track relative phase between different modes of complete state.

18. Apr 12, 2017

### mike1000

I would like to get very specific.

In the double slit experiment interference effects are obviously observed. Do we observe interference because the electrons have a certain probability of going through either slit and because of that, many different electrons end up taking different paths, some through one slit others through the the other slit and because of that they have different path lengths and different phase? (or something like this?)

Or, do we observe interference effects because the electron is actually in a superposition state of some sort?

19. Apr 12, 2017

### Mentz114

The barred patterns that are observed can be produced without interference. If the experiment is sending 1 particle at a time throught the slits, there can be no single-shot interference without superposition.

20. Apr 12, 2017

### mike1000

Your statement is very ambiguous. When you say there can be no interference without superposition what do you mean? Are you saying that a single electron goes through both slits at the same time?

21. Apr 12, 2017

### Mentz114

I meant to say "there can be no single-shot interference without superposition".

I do not believe the single-shot experimental results are caused by interference. I have no overwhelming evidence so it is a minority view.

22. Apr 12, 2017

### mike1000

By single shot you mean just firing one electron at the screen? Are you saying that when they fire one electron at a double slit they observe interference with just that one electron?

23. Apr 12, 2017

### Mentz114

Yes. For instance, if each electron leaves a dot on a screen, then a barred pattern emerges. ( You should get out more )

24. Apr 12, 2017

### mike1000

But if they only fire one electron they cannot tell anything. You mean they fire many electrons one at at time.

25. Apr 12, 2017

### Mentz114

Normally 'each electron' implies many electrons. Yes, obviously.