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A measurement in QM.

  1. Aug 12, 2004 #1
    I'm reading Griffiths' "introduction to quantum mechanics" and there's something he has not made really clear to me. What constitutes as a measurement? I'm convinced it has got nothing to do with Wigners interpretation of the intervention of human consiousness.

    Griffiths says the general concensus among physicists is that it depends on the interaction between a macroscopic object and the quantum system. Is this true? He mentions that the combined wave function would be "monstrously complicated" and "presumably somewhere in the statistics of large numbers macroscopic linear combinations become extremely improbable"

    But what is a macroscopic object? Is it a statistical thing, the chance that a wave function collapses depending on the size of the object it interacts with?
  2. jcsd
  3. Aug 12, 2004 #2


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    It's called the measuremnt problem - i.e. the fact that though the concept of 'measurement' is a vital part of QM, but what actually consititues a measurement is not well-defined.
  4. Aug 12, 2004 #3
    also, keep in mind that a measurement inevitably changes the wavefunction of the system you want to measure. Performing a measurement is just done in QM by letting an operator work on that function. The eigenvalue squared gives you that probability that the system has for example a certain energy when you are working with the Hamiltonian.

    Like Schrodingers cat system and environment will always be entangled once we measure out a certain part of the wavefunction. Measuring breaks the superposition of the wavefunction and once the measurement is done, all other info is lost. Well, not always, look at the Stern_gerlach experiment...

  5. Aug 12, 2004 #4
    I've never read Griffiths. Nevertheless, I find this brief explanation to be rather suspicious. I mean, what does it mean to say that "linear combinations become extremely improbable"? If that is all Griffiths has to say, then he has not given the problem any of the care it requires or deserves.

    The idea which you have presented, does seem to be somewhat along the lines of what Heisenberg had to say:

    D'Espagnat responds to this argument with:


    I would say that this claim is definitely true. At the very least the measurement will depend upon the interaction between the macroscopic instrument and the quantum system. The question is, however, what is the nature of this interaction? That is, is it an interaction which follows the quantum mechanical "rules" for a compound system on a joint Hilbert space consisting of the system and apparatus (and whatever else), or is it an interaction which is not describable in those terms? It sounds like Griffiths is claiming the former. If so, he makes it seem like one of those things where, if you took the time to work out the details, then you'd see that it all works out ... but really, it doesn't.


    According to "projection-postulate" Quantum Mechanics, the measuring instrument has no description. It is just an unexplained entity which "selects" a "result" in accordance with the probabilistic rules.

    In short, according to what you have reported, Griffiths is merely "sweeping the problem under the rug". At present there is no satisfactory resolution of the problem.**

    ** I just realized that my concluding statement may have been too strong. I know of at least one formulation/interpretation where there is no measurement problem to speak of - and that's Bohm's. Bohm arranges for an understanding of |ψ|2 as a probability density in the purely classical sense where the quantum particle does have a definite position and does have definite momentum, but that we don't know what the details are on account of a "classical" sort of ignorance. So, there (and I have really not made it clear just how) there is no measurement problem. Of course, the question of whether or not Bohm's formulation/interpretation is itself considered to be "satisfactory" is a separate issue.
    Last edited: Aug 12, 2004
  6. Aug 12, 2004 #5


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    Griffiths is an introductory text. I think it is a good idea to "shortcut" the discussion on the measurement problem in introductory texts and to come back to it later, when one has learned much more about the machinery of QM.
    I suppose that what Griffiths is alluding to here (and he shouldn't), is decoherence.

  7. Aug 13, 2004 #6
    isn't this Hisenbergs Uncertany principle. The more you want to know abou position, the less you know about momentum and vice versa.
  8. Aug 14, 2004 #7


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    No the unceratinty principle and the measuremnt problem are distinct. It's no inconceiveable that the measuremnt prolem could be solved, but this would not change the uncertainty principle at all.
  9. Aug 14, 2004 #8
    dear willem;
    can you tell me the exact part of the book that the fact was discussed;
    thank you
  10. Aug 14, 2004 #9
    He states the problem in paragraph 1.2 (The statistical interpretation) and returns to it at the end of the book.

    At A.3 (What is a measurement) of his afterword he goes deeper into the problem (read also footnote 10 and 12 of the same paragraph).
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