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I'm sorry for the long question.

In Brian Greene's book 'The Hidden Reality' he gives a nice illustration of the uncertainty principle. Imagine a butterfly flying around in a room. We would like to measure the speed and position of the butterfly by capturing a photograph of it. We can take a photo with a very fast shutter speed and the resulting sharp photo would give a good measure of the position. But this gives a poor measure of the speed. On the other hand, if we take a photo with a slow shutter speed, the resulting photo with a butterfly trail will give an estimate of the speed, but the position information in this photo will be very coarse. He makes it a point that we cannot measure both the speed and the position accurately at the same time and this illustrates the quantum uncertainty principle.

I understand what he tries to convey. Also I have seen this myself in action when I program the Fourier transform where I have to compromise on the number of frequency bins to get more accurate frequency values.

I was trying to explain this to a friend and I was trying to make a point that this is not a technological limitation and there is no way to improve the accuracy in the butterfly example using a better technology.

Then my friend asked what if we use two different cameras, one with slow shutter speed and one with fast shutter speed? I was bit taken aback. On the face of it, it sounds a smart idea. But I knew that this wouldn't work. However, I was not sure about the minimum argument against it.

I could think about two arguments.

  1. Here the crux is that we need to measure both the parameters simultaneously. Assume that my friend configures both the cameras to fire 'simultaneously' in a simulated manner. If the short exposure time is 10 millisecond and the long exposure time is 1 second, he could program both of his cameras so that the 0.5ms tick of the fast camera coincides with the 0.5s tick of the slow camera. This would mean that both camera would start at different times, but their timing midpoints will coincide and this will make sure that the cameras capture the information somewhat 'at the same instant'. Even though one could argue that this is not quite an 'instant', I believe this is according to the spirit of the experiment.

However, the catch here is that the slow camera still captures a span of 1 second and the resulting estimated velocity couldn't be reliably related to the position captured by the 1ms camera.

  1. This is the macro world equivalent of a quantum measurement. In the quantum world, we cannot make two measurements on the same particle which was in the super position state to get two different parameters. The first measurement disturbs the super position and snaps the particle to a definite eigen state and the second measurement would just measure the value of this eigen state, not of the original super position state.

So, the butterfly experiment is precisely not a quantum one.

My question : Should I necessarily invoke the 2nd argument? Is the 1st one enough to close the deal?

Edit : After learning little bit more about the non-commutative nature of the quantum measurement, it seems like my second argument is not quite accurate as it has nothing to do with the timing, but all about not having a common eigen vector between the two complement quantum operators that contribute to the uncertainty(or the commutator operator not being 0). However, the main question still holds.

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    $\begingroup$ The problem with all attempts to explain the quantum world trough "the lens" of the classical world is that it is the wrong way around. The quantum world is not caused by the classical world but it is the primary (and only) reality that exists and it is the classical world that happens to be a thermodynamic remnant effect. The problem with that for the layman is, that how this "remnant" forms is not only non-trivial, it's also usually explained the wrong way (even in university level physics classes!). $\endgroup$ – CuriousOne Apr 21 '16 at 5:57
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This is a great example of how hard it is to popularize quantum mechanics.

Greene's example is not quite right, because classically, the butterfly does have a definite position and momentum, at all times. We can also measure these values simultaneously to arbitrary accuracy, as your friend says. (As for your concern about exposure time, we could decrease all of those by getting cameras with better resolution.)

In quantum mechanics, the situation is different. Even if you somehow could learn the exact state of the system without disturbing it at all, you still wouldn't know the position and momentum of the object because objects don't have definite positions and momenta. This is a property of quantum states themselves, not of any measurements we perform. (The fact that measurements do disturb the system, as stated in your second argument, is just a red herring, though an important historical one.)

As for why quantum states have this property, I explain it a bit in this answer. You also have likely seen a similar result in Fourier analysis: the smaller the duration of a sample, the wider the frequency distribution is. Up to a factor of $\hbar$, the proof of this is actually the same as the proof of the uncertainty principle.

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  • $\begingroup$ Thanks for the answer. So, you mean that we should necessarily invoke a quantum argument here. However, after reading your other answer I'm puzzled. There you use the case of compass directions to illustrate the principle and categorically states that we cannot determine 'going south-north' and 'going east-west' question simultaneously. It makes sense. This is the same case here. Then do we really need the quantum argument? Isn't this an abstract property? $\endgroup$ – PermanentGuest Apr 21 '16 at 6:06
  • $\begingroup$ Yup! Uncertainty can appear in classical contexts, such as the Fourier transform. The issue is that it doesn't hold for classical position and momentum, so if you specifically want to explain position-momentum uncertainty, you need to do it in QM. $\endgroup$ – knzhou Apr 21 '16 at 6:09
  • $\begingroup$ btw, I understand the 'red herring' part now. In fact, after writing down the question in detail, I knew that the 'order of measurement' is not that important in this case. It is just a timing aspect. $\endgroup$ – PermanentGuest Apr 21 '16 at 6:09

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