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I Are superposition states observable?

  1. Mar 29, 2017 #1
    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
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  3. Mar 29, 2017 #2

    DrChinese

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    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.
     
  4. Mar 29, 2017 #3
    So does the measurement force it to be in one of the available states for that basis?
     
  5. Mar 29, 2017 #4

    PeterDonis

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    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.).
     
  6. Mar 29, 2017 #5

    Nugatory

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    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.
     
  7. Mar 29, 2017 #6
    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.
     
  8. Mar 29, 2017 #7
    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
  9. Mar 29, 2017 #8
    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.
     
  10. Mar 29, 2017 #9
    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".
     
  11. Mar 29, 2017 #10
    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.
     
  12. Mar 29, 2017 #11
    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

     
  13. Mar 29, 2017 #12
    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.
     
  14. Mar 29, 2017 #13
    [
    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.
     
  15. Mar 29, 2017 #14
    That is correct.
     
  16. Mar 29, 2017 #15
    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.
     
  17. Mar 29, 2017 #16
    No. The -z direction is not the same as the +z direction. But detecting one is equivalent to detecting the other.
     
  18. Mar 29, 2017 #17
    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
  19. Mar 29, 2017 #18
    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.
     
  20. Mar 29, 2017 #19

    PeterDonis

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    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".
     
  21. Mar 29, 2017 #20
    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.
     
  22. Mar 29, 2017 #21

    PeterDonis

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    See the edit I just made to post #19. I had hit "post" too early and had to revise it.

    In that case, yes, the answer is "no" since you are adopting the first of the two possible interpretations I gave in post #19. But in post #17, you said:

    The answer is "yes" if you use the second interpretation that I gave in post #19. But that requires you to change the basis, which you have now said you didn't intend to do.
     
  23. Mar 29, 2017 #22
    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.
     
  24. Mar 29, 2017 #23

    PeterDonis

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    Not for a real coin. You can say that "heads" and "tails" are the only states of the coin you are interested in, but that doesn't make those two states a basis of the coin's complete state space. There are many coin states other than "heads" and "tails" and which are not expressible as linear combinations of "heads" and "tails"; that contradicts the hypothesis that "heads" and "tails" are a basis of the coin's state space.

    If you want to consider a hypothetical "quantum coin" that acts like an actual spin-1/2 particle, you can as a heuristic analogy, but you should be aware of the limitations of such an analogy.

    Even in the limited "quantum coin" analogy, this statement is not quite correct. Assuming that the coin starts out in a superposition, if the probabilities of observing heads and tails are equal, the superposition we have been discussing up to now (but not the only possible one--see below) is ##1 / \sqrt{2}## times heads plus ##1 / \sqrt{2}## times tails, because the squares of the coefficients in the superposition are what give the probabilities and have to add up to 1 (or 100%). But the "representation" of the state is the superposition itself, with the ##1 / \sqrt{2}## coefficients.

    This matters because there is not just one state that gives 50% probabilities for heads or tails. The most general state that does so would be expressed as

    $$
    \vert \Psi \rangle = \frac{1}{\sqrt{2}} e^{i \theta} \vert \text{heads} \rangle + \frac{1}{\sqrt{2}} e^{i \phi} \vert \text{tails} \rangle
    $$

    where ##\theta## and ##\phi## are arbitrary phase angles. So there are an infinite number of possible states that can all be described as "50% heads and 50% tails". But all of those states behave differently when we perform other operations besides the heads/tails measurement. (A good recent text on quantum computing will go into this in great detail, since I am just describing the possible qubit states and operations.)

    (Also, the two basis states for a spin-1/2 particle are not "the only two states the coin can actually be in". They are just "the only two states that can result from a measurement in that basis", i.e., a measurement of the observable that has those two states as eigenstates. But there are an infinite number of possible observables and correspondingly an infinite number of possible pairs of basis states you could choose, and each choice corresponds to a distinct observable.)
     
  25. Mar 29, 2017 #24

    bhobba

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    That looks like a quote from Dirac - Principles of QM.

    A classic it is, and the best presentation of the early pioneers (with the possible exception of Von-Neumann if you want mathematical rigor), but it has errors that have been discussed here in the past. Care is required since things have moved on a lot since then. At that level Ballentine is a far better choice - then read Dirac. I know from personal experience - I did the reverse and got into all sorts of trouble. It sent me on a long sojourn into exotica like Rigged Hilbert Spaces etc, and I came out the other end with a very good grasp of certain mathematical technicalities, but it was NOT the best approach.

    Dirac states the principle of superposition very clearly - there is no ambiguity - and if you know linear algebra it simply says the states, in principle, form a vector space. He also states the space he uses is more general than a Hilbert space, but doesn't say more than that and that's exactly where certain technical mathematical issues arise that RHS's are required to resolve and what I spent a long time sorting out. Regardless they are both vector spaces which is all you need to know for the principle of superposition.

    Ballentine states the 2 axioms of QM. Get a copy and read the first 2 chapters.

    There is no collapse. The second axiom is partly dependent on the first as I explain here:
    https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

    See post 137.

    The key single axiom that implies the two of Ballentine is:
    An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

    Note nothing is mentioned about collapse. All that is mentioned is you can find this mathematical thing called a POVM and you can associate the outcomes of an observation with it. That is you take whatever the outcomes are and you can find a POVM such that each outcome can be associated with an element of the POVM. To be 100% clear I will restate it another way. Suppose you have 1, 2, 3 ....n outcomes. Then you can find a POVM with n elements E1, E2, E3, ...... En that you associate with the outcomes - E1 is associated with outcome 1, E2 outcome 2 etc. That's all there is to it formalism wise. The other implication is all you can do is predict probabilities of outcomes - but that is just a general case of determinism which is simply probabilities of 0 or 1 so whether its an actual assumption is debatable.

    Now what it means is another matter - that's where interpretations come in. In some of them you have this thing called collapse - but its not part of the formalism.

    Thanks
    Bill
     
    Last edited: Mar 29, 2017
  26. Mar 29, 2017 #25

    Nugatory

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    There are certainly worse analogies, but like all analogies this one is still quite misleading. Consider that we stop the coin from spinning by clapping it between our hands. The coin will end up with one side against our left palm and the other side against our right palm; we'll call the measurement result heads or tails according to which side is against our left palm. But we can hold our hands either vertically or horizontally when we clap them together, so we actually have two observables: heads/tails horizontally and heads/tails vertically. With the spinning coin both are random, so much so that you might wonder why I bother making the distinction between the two observables - when I'm tossing a coin to settle an argument no one cares how I hold my hands when I grab it. But here the analogy has failed to capture an essential characteristic of superposition: if the coin behaved like a superposition, I could start it spinning in such a way that it comes up vertical-heads every single time even though the horizontal measurement is 50/50 random between heads and tails.
     
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