# I Are superposition states observable?

1. Mar 29, 2017

### mike1000

The way I am coming to understand it, the allowed states that an observable can be "observed/measured" in are defined by the eigenvectors (and associated eigenvalues) of the observable's operator. Since those eigenvectors form a basis and span the space of vectors defined by the operator, a linear combination of two or more eigenstates is also an allowed state of the observable i.e. superposition.

Does this mean that an observable can be observed/measured in a state which is a superposition of eigenstates?

Here is a quote from Dirac's book "The Principles of Quantum Mechanics". I do not have this book and I have not read this book, yet!) Someone in another thread mentioned it and that started me on a quest. In that quest I found the following quotation from Dirac's book. Here is that quote...
It seems to me that Dirac is saying, "No, we cannot observe/measure the particle in a superposition of states"

Or maybe he is saying that if we want to observe the photon in its superposition state we need a different way to measure it!

What if we did not want to observe whether the photon was polarized in one of only two states? Did we force it into one of those two states by the method we used to measure it?

Last edited by a moderator: May 8, 2017
2. Mar 29, 2017

### DrChinese

Usually, you expect a photon polarization observation to result in one of two values - on the selected basis. There are essentially an infinite number of those bases, for starters (rotating around 360 degrees). Also, if you have full knowledge of state on one basis, there is at least one basis which is now completely unknown. Further, there is no requirement that you have complete knowledge on any basis. You could theoretically have partial knowledge on one basis, and partial knowledge on another basis - that would not violate the HUP.

3. Mar 29, 2017

### Jilang

So does the measurement force it to be in one of the available states for that basis?

4. Mar 29, 2017

### Staff: Mentor

It depends on whether you adopt a collapse or a no collapse interpretation of QM. On a collapse interpretation, yes, the measurement collapses the state onto one of the eigenstates of the measurement operator. On a no collapse interpretation, no, it doesn't; all of the branches of the superposition remain (but each branch gets entangled with the corresponding state of the measuring device, observers, etc.).

5. Mar 29, 2017

### Staff: Mentor

As PeterDonis says, that depends on your interpretation. You can avoid the interpretational swamp by saying that your subsequent measurements will behave AS IF the measurement had forced the system to be in one of the particular states, namely the one that you measured.

6. Mar 29, 2017

### mikeyork

The latter. Any physical state is represented by an eigenvector in some observable and frame of reference -- but a superposition in others. In particular, any observable state will have been prepared (either by our apparatus or by nature) in an eigenstate of some basis even if a superposition in another. So that preparation basis is the basis in which a superposition in another basis could be "observed" as such.

As a fairly simple example, prepare an electron state with spin +1/2 along the z-axis. Measuring the spin projection on the z-axis is effectively the same as measuring a superposition in any other direction (except the opposite direction where it is an eigenstate with spin -1/2). Repeating such an experiment of measuring in another direction over and over will then give the appropriate statistical frequency that converges on that corresponding to the superposition.

Having said that, "observing a superposition state" is a very convoluted and strange way to describe this situation.

7. Mar 29, 2017

### mike1000

Let me try saying it this way. I am really asking a very simple, and I think, very direct question, that to me, should only have one answer and the answer should be "Yes".

Reading a little of Dirac and beginning to learn the basics of Matrix Mechanics, it is the eigenstates of an particular operator that become the states in which the observable can be found (measured). Also, the eigenstates form an orthonormal basis for the observable. It has been pointed out many times that a state formed by a linear combination of eigenstates(superposition) is also a pure state for the observable. Doesn't this imply that we should be able to observe a particle in a superposed state?

But I gather from all of the responses, that the answer is no, we cannot observe a particle in a superposed state. This implies that a quantum particle in a superposed state is NOT an observable, doesn't it? And if a particle in a superposition of eigenstates is not an observable, then shouldn't the superposed state not be considered a "pure" state?

Last edited: Mar 29, 2017
8. Mar 29, 2017

### mikeyork

Then the answer depends on what you mean by the question. If by "observing a superposition" you mean what I just described, then the answer is yes. If you mean something else like, for instance "measuring several different states at the same time" then the answer is no because it is obviously self-contradictory nonsense.

No. It is an observable that might be but has not yet been observed.

It seems to me you are getting tied up in the limitations of English. Study the math and be content with that.

9. Mar 29, 2017

### mike1000

No, I do not think I am getting tied up in the limitations of English. I find your response convoluted. It is a very simple question, can we observe a particle in a state which is a superposition of eigenstates?

The reason I said before that the answer should be "Yes" is because of the definition of observable and the foundation of Matrix Mechanics ( as I understand it as derived for observable quantities only) All pure states should be observable. That is why I said before the answer should be "yes".

10. Mar 29, 2017

### mikeyork

And I gave a simple answer. If you mean can you measure the relative frequencies of different eigenstates over multiple repetitions of the same experiment, or if you mean can a superposition be an eigenstate in another observable, then the answer is yes. If you mean something else then you had better be more specific about what you mean.

If you repeat the same question again, I shan't bother to repeat my answer.

11. Mar 29, 2017

### mike1000

It appears to me that you are changing the definition. I am not talking about the outcomes of many trials. I am talking about the outcome of a single trial. I think that is what is implied by the definition of eigenstates and superposition for an operator. Please look at the Dirac quote above, I will repeat it here

12. Mar 29, 2017

### mikeyork

Then as I said 3 posts ago, and repeated in my last post, the answer is yes if you can find the right observable for which the state is an eigenstate*. Otherwise the answer is no.

*Or, if we understand that as far as the logic of QM is concerned, observation can mean either detection or preparation, then the appropriate observation of the superposition was already made when the state was prepared.

13. Mar 29, 2017

### mike1000

[
I am sorry but I do not understand that answer. It should be either yes or no, there should be no "if" in it.

I think what you are saying is that for every superposition in one basis there must be a different basis, in which, the superposition in the first basis becomes an eigenvector in the new basis.

14. Mar 29, 2017

### mikeyork

That is correct.

15. Mar 29, 2017

### mike1000

Well, it is not quite correct, because I forgot to specify that there must be an observable that has the new eigenvector as one of its states.

Lets take the spin direction for an electron. It can be up or down. What you are saying implies that there must be an observable for which the electron can be measured both up and down simultaneously.

16. Mar 29, 2017

### mikeyork

No. The -z direction is not the same as the +z direction. But detecting one is equivalent to detecting the other.

17. Mar 29, 2017

### mike1000

That is not the same thing. You can try to use it in this case because its a binary system, ie there are only two , mutually exclusive, states. For any observable where there are more than two eigenstates your answer would not be correct.

Last edited: Mar 29, 2017
18. Mar 29, 2017

### mikeyork

Ok. If you meant two different eigenstates, then each has a different superposition in any other direction (obtained by a rotation) so a unique superposition in the original basis uniquely selects an orientation in which s = +1/2 and not -1/2. I do not understand why you think the number of eigenstates makes any difference. Even a position eigenstate generates a unique superposition in a momentum basis.

19. Mar 29, 2017

### Staff: Mentor

No, it isn't a simple question, because the question is ambiguous. See below.

You don't have to specify it. It's already a theorem in linear algebra.

No, it doesn't. It just implies that, if "up" and "down" are orthogonal eigenstates of some observable, then there will be some other observable that has the linear combination "up plus down" (with appropriate normalization) as an eigenstate. In the case of a spin-1/2 particle, that observable is spin-x (assuming that "up" and "down" refer to eigenstates of the spin-z observable). But spin-x "+" (or "left", or whatever you want to call it) is not the same as "spin-z up and spin-z down simultaneously".

So the question I quoted from you at the top of this post, as I said there, is ambiguous; it can be interpreted two ways, one of which leads to the answer "yes" and the other of which leads to the answer "no":

(1) Given an observable O, can we make a measurement of that observable which gives a result that is a superposition of eigenstates of that observable? The answer to that is "no".

(2) Given an observable O and a state which is a superposition of eigenstates of that observable, can we make a measurement of some other observable which will give that state as a possible result (i.e., the state is an eigenstate of the other observable)? The answer to that is "yes".

20. Mar 29, 2017

### mike1000

I am not talking about changing the basis for a given observable and I do not think that is what Dirac is talking about in this quote.

I think the answer is "no". States which are superpositions of eigenstates are not observable, meaning that the observable in question cannot be observed in that superpositioned state.

The best analogy I have is, again, flipping a coin. When the coin is flipping you can say that it is in the superposition of the head and tail state, because that is the basis, or the only two states the coin can actually be in. But when it is flipping it is in an underdermined state which is represented as 50% heads and 50% tails. When we measure it, by stopping the flipping, we find one of the eigenstates. The superposition allows us to calculate the probability but it is not an observable state that the coin can be found in.

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