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Projection postulate - can it be verified?

  1. Jul 19, 2008 #1


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    Many books on QM state this so called von Naumann projection postulate i.e. that after the measurement system is in eigenstate of operator whose eigenvalue is measured.
    But in Landau Quantum Mechanics in chapter 7, author explicitly says that after the measurement system is in a state that coresponds to measured value but that these state are not eigenstate of any operator. Because of this author concludes that measurement in QM cannot be reproduced. (the only exception is measurement of coordinate)
    So this two statements (neumann and landau) seem contradictory to me, and that simple experiment would give answer which statement is wrong.

    So are QM measurements reproducible?
  2. jcsd
  3. Jul 20, 2008 #2
    I don't understand your question completely. The "measurement" situation is still not resolved to the satisfaction of many.

    When discussing measurement there are two common types of outcomes. Outcomes that are discrete and outcomes that are continuous. Generally discrete results are repeatable (in the sense that they satisfy the repeatability postulate) while continuous outcomes are not.

    Repeatability postulate: If you observe the state of a system, successive measurement will yield the same result. Valid for discrete spectra.

    Results with discrete outcomes have definite eigenstates and the wavefunction collapses into these upon measurement. Measurements with continuous spectrum are a little more complicated and while you can claim a type of reduction here it is a bit more subtle.

    What I think is happening is that the books you are reading that talk about Von Neumann collapse are discussing experiments with discrete spectra where as the Landau book is discussing experiments that allow continuous spectra.
  4. Jul 21, 2008 #3


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    I was talking about discrete spectra (but I forgot to say that), Landau is also talking about discrete spectra. It seems that Landau is wrong but it is hard to believe in that.
    Any other explanations?
  5. Jul 21, 2008 #4


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    I think that you are right that Landau is wrong and that it is hard to believe in that. :biggrin:
  6. Jul 21, 2008 #5

    I wonder what this means. Can you give an example, preferably an easy one without complications such as entanglement?
  7. Jul 21, 2008 #6
    Quantum mechanics provides a non-deterministic way to handle uncertain information. Consider a yes or no question. We can express a yes or no question as a sum of two projectors
    [tex]A = \langle\psi_1,\cdot\rangle\psi_1+
    where [tex]\psi_1,\psi_2[/tex] are orthogonal.
    Given a state
    [tex]\phi\in\mathbb{R}^2[/tex] we can write it as
    [tex]\phi = \lambda_1\psi_1+\lambda_2\psi_2[/tex], with
    [tex]\lambda_i = \langle\psi_i,\phi\rangle[/tex].
    Upon measurement the projection postulate says that the state collapses into one of the two states. The probability of which state is given by lambda_i(edit) Immediately afterwards the state has not had enough time to propagate into a new state and so an immediate measurement will yield the same result.

    The problem with doing this for continuous spectra is that continuous spectra is defined differently. Continuous spectra do not come about as the result of a rigorous eigenvalue equation but rather they come about as the limit of a Weyl sequence.


    Since these limits are not elements of the Hilbert space they can not be states of quantum mechanics in the proper sense. Therefore to even talk about taking a subsequent measurement on this "state" is nonsense.
    Last edited: Jul 21, 2008
  8. Jul 21, 2008 #7
    Often times books talk about continuous spectra using the notation of discrete spectra. They do this using the Dirac notation and by considering the delta function as proper elements of a Hilbert space.

    This isn't so much wrong as it is confusing for some. I am pretty sure that the use of the notation in this way has been justified. I don't have the Landau book in front of me but I would bet that they are using the notation in this way.
  9. Jul 21, 2008 #8


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    In this chapter authors are not using Dirac notation but they say that measuring device's spectra is generally speaking continuous (but they didn't say that it has to be). I don't see why their reasoning cannot be applied to discrete spectra.
    It's really odd thing that for continuous spectra measurement can change eigenstate to new state that is much different from original one, and that it is not true for discrete spectra.
  10. Jul 21, 2008 #9
    Can you let me know what page this is on in Landau? You are using Landau-Lifschitz right?
  11. Jul 21, 2008 #10


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    I have just realized that I should write Landau's argumentation because many of you don't have his book.
    1) QM cannot be founded without classical mechanics (which is not case for relativistic and classic mechanics), because particles don't have definite path or other dynamical characteristics (so you cannot describe everyting with QM you need classical objects and that is measuring device which assigns dynamical characteristics to quantum object)
    2) Macroscopic object cannot be in coherent state (meaning that Schrodinger cat cannot be half dead half alive)
    3) Consider system "measuring device" + "electron". State of measuring device is described by quasi-classical wave function F_n(y) that corresponds to reading f_n of device. psi(x) is initial state of electron. State of whole system is:
    psi(x)*F_0(y) - F_0 initial state of measuring device
    after interaction between them this state will unitary evolute to:
    because of 2) this state will colapse to:
    where a_n(x) is wave function of electron multiplied by probability for getting result f_n according to Born rule. (a_n(x)=p_n fi_n(x) where fi(x) is normalized) Using linearity of evolution operator one can write:
    a_n(x)=K_n |psi> (K_n is linear operator, I use this notation instead of original one so I don't have to write intagral).
    Because final state of electron after the measurement doesn't depend on initial state:
    p_n=<PSI_n|psi> - from this it follows that this functions PSI_n form complete basis.
    Operator f_i*|PSI_i><PSI_i| (summation over i) is the one usually used in QM (operator that coresponds to this measurement)
    Functions fi_n are generally speaking not eigenfunctions of any operator and do not form complete basis. And fi_n can be much different from PSI_n.

    So why cannot I apply this to discrete spectra?

    PS. I am using russian edition so page numbers are not same. It's chapter 7 (he has short chapters this whole chapter is devoted to this)
  12. Jul 22, 2008 #11
    The only difference I can see is that for discrete spectra the results are the eigenfunctions of an operator.

    To be honest I never really liked most presentations of the measurement postulate. One that I do like is given


    on page 8.
  13. Jul 22, 2008 #12


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    So how do you see that for discrete spectra fi_n(x) is equal to PSI_n(x)?
    Well I don't see the connection between this and my question.
  14. Jul 22, 2008 #13
    The nature of measurement is different for discrete and continuous data. However the classical apparatus can only handle discrete outcomes.

    If your experiment yields discrete outcomes [tex]\lambda_i[/tex] then after the measurement your state will have collapsed into the state [tex]f_i(x)[/tex]. Construct the self adjoint operator
    [tex]A = \sum\limits_{i=1}^n\lambda_i\langle f_i,\cdot\rangle f_i[/tex].

    This operator has the appropriate eigenvalues and eigenfunctions.

    I don't think of outcomes [tex]f_i[/tex] as being equal to [tex]\psi_i[/tex] but rather as being completely correlated to each other. The way I interpret this is by saying that
    [tex]\psi_i[/tex] is classical information and therefore the observer has access to this information. The state [tex]f_i[/tex] is a quantum state and so the observer does not have access to this information directly but rather through the [tex]\psi_i[/tex].

    What I am trying to say is that the equality/correlation of the quantum outcomes and the state of the apparatus isn't proven but rather postulated.
  15. Jul 22, 2008 #14
    The relevance of the link is that they state the projection postulate in a way that gives equal footing to both discrete and continuous results. The weakening is that rather than ask for an explicit result from an experiment they ask that experimental results be contained in some interval.
  16. Jul 22, 2008 #15


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    I am not quite sure that I understand what you are saying. (because of notation)
    If fi_n and PSI_n are not same that means that measurement changed state that gives result f_n with certainty. (meaning it changed eigenstate)
    So you say that it is just possible for continuous spectra, but can that new state be much different from the initial eigenstate (of continuous spectra)?

    I already saw such statements about projection postulate in many books, problem is that it seems that Landau's argumentation contradicts with it.
  17. Jul 25, 2008 #16
    I think you have it backwards. The projection postulate as you normally see it works just fine for discrete spectra but DOES NOT work just fine for continuous spectra.
  18. Jul 25, 2008 #17


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    whats the issue with continuous spectra? There's *always* some experimental error (even with discrete measurements).
  19. Jul 25, 2008 #18


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    Well my question was (among others) about possibility that measurement changes eigenvalue of continuous spectra significantly. (because from Landau's derivation and don't see why should fi_n be close to PSI_n)
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