About observability

  1. Hi everyone,
    I want to know why Schrodinger's wave equation is dependent to observation.If we look at the equation, I dont see any relation to dependant variable such as observer x. So if we think like looking down the wave function, are we just capturing some snapshots from the wave function of the system ? Isn't that what observability means ? If you can give some guidance, I'll be really appreciated.
  2. jcsd
  3. Demystifier

    Demystifier 5,593
    Science Advisor

    Observer or a measurement apparatus consists of many particles, so a single-particle Schrodinger equation cannot include the observer. To include the observer, you must study the Schrodinger equation of a very large number of particles. Of course, in practice such systems cannot be studied exactly, but there are well developed concepts and techniques that allow to study such systems approximately. Of course, I cannot here present the details of this theory, but let me just give you a hint: The crucial concept related to the the quantum theory of measurement is - DECOHERENCE.

    If you want to learn more, here are some good introductory reviews on decoherence, measurement and related stuff:
  4. The decoherence argument is only one possible view on the "observer effect" of quantum mechanics. In my opinion, decoherence is quite contrived and unnatural in quantum physics, and was only argued to deny the existence of wavefunction collapse. Wavefunction collapse is not only something you learn on the first pages of any quantum textbook, it also displays how observations change a system much more clearly.

    Anyway, in quantum physics, the state of any system (whether it be one particle or many) is described by a ket |ψ> in Hilbert Space. Every measurable quantity is represented by an hermitian operator Q which acts on kets in Hilbert space. Any measurement of the quantity represented by Q must result in one of Q's eigenvalues, {Q1, Q2, Q3, ...}, and corresponding to these eigenvalues, Q has a set of eigenkets {|ψQ1>, |ψQ2>, |ψQ3>, ...} which form an orthonormal basis for the Hilbert space. These eigenkets are referred to as "pure states" of the observable Q. If you measure the Q on the state |ψQm>, the measurement will result in Qm, with absolute certainty.

    Here's where the observer effect comes in: in general, a ket is NOT in a pure state |ψQm>, but is rather in a superposition of eigenkets. Here's an example: |ψ> = 2(-1/2)(|ψQ1>+|ψQ3>). Because this is made in equal parts of the first eigenket and the third eigenket, a measurement on this state will result in either Q1 or Q3 with equal probability. [This is akin to an electron passing through two slits simultaneously or a cat being dead and alive at once.]

    The thing is, the instant you measure 2(-1/2)(|ψQ1>+|ψQ3>), you get either Q1 or Q3. If you were to get Q3, the state is instantly changed from 2(-1/2)(|ψQ1>+|ψQ3>) into |ψQ3>. This is a discontinuous evolution of the state ket and is NOT consistent with the Schrodinger equation.

    Thus the act of measurement causes a general quantum state to collapse into an eigenstate of the measurement operator--this discontinuous evolution of the state is not described by the Schrodinger equation and represents a wholly different quantum evolution than Schrodinger evolution.

    A good example of this weirdness is called the Quantum Zeno Effect, or the Watched Pot Effect. As the rate of measurements on a quantum system approaches infinity, the quantum system's evolution stops. This is because the wavefunction is continually being collapsed back to an eigenstate. This has even been observed experimentally: if you observe whether a radioactive atom has decayed or not rapidly enough, its half life will get longer!
  5. lavinia

    lavinia 2,130
    Science Advisor

    The Shroedinger equation describes the evolution of the wave function. The wave function is not directly observable.
  6. thanks for the documents, I'll look over them
  7. Demystifier

    Demystifier 5,593
    Science Advisor

    Just the opposite, decoherence is something that can be derived from the Schrodinger equation itself, and in practice it is very difficult to avoid it whenever the system strongly interacts with the environment.

    Decoherence alone cannot replace or explain the collapse. But it tells WHEN and IN WHAT BASIS the collapse will occur. What it does not tell is WHY and IN WHICH PARTICULAR STATE the collapse will happen.

    That's true, but a correct QUANTITATIVE description of Quantum Zeno Effect cannot be made without decoherence.
    Last edited: Mar 20, 2012
  8. But can we really know that decoherence is really occurring, and not real wavefunction collapse? Here's a question I asked in another thread; maybe you can answer it for me.
    I thought decoherence tells us not only why a particular pointer basis is chosen for the appearance of collapse, but also how the particular choice of pointer state into which the collapse appears to have occured is chosen, but that in practice we cannot know what that pointer state is going to be in advance, because of the classical statistical uncertainty we have concerning the initial quantum state of the system and the exact details of the Hamiltonian of the system.
    Are you saying that it's impossible to get an exact quantitative description of the quantum zeno effect using real wavefunction collapse?
  9. kith

    kith 946
    Science Advisor

    Mathematically, decoherence is the decay of the off-diagonal elements of the system density matrix in a specific basis. In the end, you always have a mixed state. So decoherence doesn't tell us which state will be chosen in a measurement.

    I don't think this is due to statistical uncertainties or the Hamiltonian, because this basic feature is also present without dynamics and for well-defined initial states. If the initial state of system+environment is an entangled state, the system alone is in a mixed state already.
  10. But isn't any lack of knowledge concerning what quantum state the system+environment will appear to have collapsed into just statistical uncertainty about the initial quantum state of the system+environment, or ignorance about the precise details of the Hamiltonian?
  11. Demystifier

    Demystifier 5,593
    Science Advisor

    In some cases, yes.

    Unfortunately, decoherence does not tell us that. If it did, there would be no need for various interpretations of QM.

    Yes, that's what I am saying. But still, a collapse may often serve as a good approximation.
  12. kith

    kith 946
    Science Advisor

    Maybe, but this is more of an interpretational question. The "missing information answer" leads probably to Bohmian mechanics or other hidden-variables theories.
  13. Could you elaborate on this? Also, do you have any thoughts on the question I quoted in my post above, concerning measurements of an observable conjugate to that of the pointer basis?
    I'm a little confused. I thought in the absence of wave function collapse, everything is completely determined by Schrodinger evolution, so the pointer state into collapse appears to have occured can be predicted in advance, so that any doubt about what the specific pointer state will be must arise from classical ignorance concerning the initial state or the Hamiltonian.
    So are you saying that the Copenhagen interpretation can be disproven by the quantitative details of quantum Zeno effect experiments? That's a rather bold claim.
  14. I feel as though Demystifier has made tons of claims without any arguments to back them up. I gave a perfectly quantitative explanation of wavefunction collapse (which doesn't go very far beyond the basic postulates of quantum mechanics), and under the assumption that the Schrodinger equation evolves wavefunctions continuously through Hilbert space, it should be obvious that it immediately implies the Quantum Zeno effect. If you disagree that it constitutes a perfectly quantitative demonstration of the Quantum Zeno effect, then I refer you to Griffiths Intro to Quantum Mechanics Section 12.5 (second ed. p 431-433).

    So please back up your following claims, which you have merely stated without any support, quantitative or otherwise.


    I have already addressed 2): I think my wavefunction collapse argument is a correct quantitative description, check Griffiths.

    I am not so sure about 1). In practice, yes, there is always an environment that interacts with the system, and can be considered to cause "decoherence". However, in standard quantum theory, the entire universe can be considered to be described by a single wavefunction: there is no "environment" that causes decoherence. Please show me quantitatively how it can be derived from the Schrodinger equation without the assumption that there is some external "environment", since that that assumption is NOT one of the standard postulates of quantum theory.

    I'd appreciate if you could back up 3) and 4) as well.

    I think decoherence is a much more obscure and conceptually elaborate demonstration of the observer effect, whereas wavefunction collapse is one of the basic postulates of quantum mechanics. It seems as though you don't even disagree with wavefunction collapse: if you admit its existence, it's clearly a better example of the observer effect than any "decoherence" phenomena.

    Let me ask another question: if decoherence were the definitive explanation for the observer effect, wouldn't Schrodinger's Cat be rendered moot? Based on the various claims you've made, I guess your argument would be something like: "once a quantum system interacts with some macroscopic system, like a cat, it decoheres." As a less philosophical example, let's suppose we put a geiger counter inside a box with a radioactive atom, whereby the geiger counter has two states "Not Decayed" or "Decayed" which correspond to the radioactive element not decaying or decaying. Does the presence of the macroscopic geiger counter inside the box collapse the radioactive element into either the decayed/not decayed state? This is what that argument seems to imply. I argue that's just an interpretation of quantum mechanics: there is an alternate interpretation that states that until the geiger counter is "observed," it too is in a superposition of states. Granted, the "box" is an idealization--no box can shield a system from the environment totally, but the idea of a universal wavefunction ensures that there is at least one system that is shielded from any sort of "environment."
    Last edited: Mar 21, 2012
  15. Demystifier

    Demystifier 5,593
    Science Advisor

    Lugita15 and jolb, I have not intention to further elaborate my claims until you read at least one of the papers I mentioned in post #2. To make any further discussions on decoherence fruitful, we first need to have some minimal common background on it.
  16. kith

    kith 946
    Science Advisor

    Decoherence doesn't imply "absence of wave function collapse" and is a feature of quantum mechanics, not of specific interpretations. It simply arises if you look at open quantum systems instead of closed ones. It tells you on what time scale coherences decay (Demystifier's WHEN) and in what basis this decay occurs.

    Whether this is already enough to explain collapse, depends on the ontology of mixed states. So for some interpretations (statistical, MW), decoherence explains collapse, but not for the Copenhagen interpretation.
  17. Demystifier, is this good enough?

    I read this paper a while back, but I think it's a pretty good explanation.
    Last edited: Mar 22, 2012
  18. Demystifier

    Demystifier 5,593
    Science Advisor

  19. OK, then referring to that paper can you try and answer my questions?
  20. Demystifier

    Demystifier 5,593
    Science Advisor

    Not everything, only the unitary time evolution of the quantum state (wave function).

    It cannot be predicted, because collapse into a one definite state is a nonunitary event, which Schrodinger equation does not describe.

    No. Or if you disagree, try to find a statement in the paper you mentioned which confirms your view.

    A naive original version of Copenhagen interpretation can be disproven by quantum Zeno effect experiments, but it is still possible to introduce a refined version of Copenhagen interpretation which is compatible with quantum Zeno effect experiments.
    See also
  21. Demystifier, I don't follow. He says clearly "in the absence of collapse", then you keep saying "no, because, collapse....".

    Also, why is quantum Zeno a disproof of Copenhagen?
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