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I'm not an expert when it comes to quantum mechanics, so correct me wherever I'm wrong, but:

I've always been a little bit bothered by introductory derivations of the Heisenberg uncertainty relations (for, for example, position and momentum) that involve carrying out an imaginary "observation" of a particle by bonking it with photons. The argument goes (at least the way I've heard it) that the more precision you want in the position measurement, the higher frequency (and so energy) the photon must have. Higher energy bonking leads to more uncertainty in the momentum afterward, and so there's a tradeoff between position and momentum precision. The derivations then manage to work out that the precisions satisfy the Heisenberg uncertainty relations.

The problem I have with those kinds of derivations is that they start out assuming that a particle actually has a position and momentum to speak of, and if you were lucky enough to guess them, you'd be able to predict the particle's motion exactly - it's entirely classical reasoning. The way I understand it, the position and momentum representations of the state are fourier transforms of each other, and so it's just a simple mathematical fact that as one becomes more localized, the other one becomes less localized.

Another way of saying it (and please feel free to comment on the validity/value of this way of thinking about it), that I think makes it qualitatively more obvious, is that each observable corresponds to an eigenbasis. It's pretty intuitive that, in general, if one aligns the state vector closely with any one basis vector in one eigenbasis (corresponding to a nearly definite value of that observable), then the state vector won't be aligned any one basis vector of the other observable, but will instead be a mixture of several basis vectors, and so will have an uncertain value of the other observable. If this sort of reasoning is correct, then it'd seem like the classical derivations of the Heisenberg uncertainty principle are complete gibberish - rather than having to do with a particular way the measurement is made, the limits on measuring simultaneous, definite values of position and momentum are a result of the fact (which the vector space representation makes plain) that a state cannot even have simultaneous definite values.

If it really is true that the classical derivations of the Heisenberg uncertainty relations is whack, is there any value in them at all?


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Looks like you are referring to Heisenberg's microscope. The argument is flawed for many reasons, some of which are described on the Wiki page. I'll leave it to others to give a complete answer. –  Mark Mitchison Jul 20 '14 at 0:33
@MarkMitchison Just to add to that, there is a very long criticism on the talk page of that Wiki article. –  JohnnyMo1 Jul 20 '14 at 0:41
@JohnnyMo1 OK, but the current version looks fine to me, although lacking in detail. The point is that the presumption of a pre-existing value of position is in violation of the Kochen-Specker contextuality theorem; thus it is in disagreement with experiment and with quantum theory in particular. Furthermore, Heisenberg's microscope cannot account for the uncertainty arising from joint measurements on correlated quantum particles. –  Mark Mitchison Jul 20 '14 at 0:50
@MarkMitchison Take your time. Please add a complete answer. –  Cheeku Jul 20 '14 at 1:27
It's with some reluctance that I vote to close (for the "primarily opinion-based" reason) because I have considerable sympathy with the views of the OP. Maybe there's a site this question could be taken forward - is there an SE site for "physics education" type questions? –  John Rennie Jul 20 '14 at 4:58

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