View Full Version : A measurement in QM.
da_willem
Aug12-04, 10:25 AM
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?
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?
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.
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...
regards
marlon
Eye_in_the_Sky
Aug12-04, 05:41 PM
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".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:
This means that the equation of motion for the probability function does now contain the influence of the interaction with the measuring device. This influence introduces a new element of uncertainty, since the measuring device is necessarily described in the terms of classical physics; such a description contains all the uncertainties concerning the microscopic structure of the device which we know from thermodynamics, and since the device is connected with the rest of the world, it contains in fact the uncertainties of the microscopic structure of the whole world.D'Espagnat responds to this argument with:
As Barman and Shimony have pointed out, the obvious interpretation of these passages is that the pointer reading is definite though unknown when the final state of object plus apparatus is reached (but before registration upon the consciousness of the observer), and that this comes about in spite of the initial indefiniteness of the measured quantity, because of the initial uncertainties concerning the state of the apparatus. But if instruments are finite, the fallaciousness of this analysis is essentially the content of the results reported in Chapters 17 and 18.---------------------
What constitutes as a measurement?
... 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?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.
---------------------
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?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.
vanesch
Aug12-04, 11:53 PM
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.
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.
cheers,
Patrick.
isn't this Hisenbergs Uncertany principle. The more you want to know abou position, the less you know about momentum and vice versa.
isn't this Hisenbergs Uncertany principle. The more you want to know abou position, the less you know about momentum and vice versa.
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.
dear willem;
can you tell me the exact part of the book that the fact was discussed;
thank you
da_willem
Aug14-04, 09:08 AM
dear willem;
can you tell me the exact part of the book that the fact was discussed;
thank you
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).
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.