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As I understand it, Heisenberg's uncertainty principle (HUP) states that the more precisely we know the location of an electron, the less precisely we know its momentum, and vice versa. However, I've looked online, but I haven't found the reason that that is the case.

I'm a first year science student and I haven't taken calculus yet, so a simple answer (even if it sacrifices rigor) would be the most helpful. I'd just like to get some intuition about what makes HUP true.

EDIT

In response to the comments about 'why' being a poor question to ask in physics: It would be equally useful to know how we observed Heisenberg's uncertainty principle.

Suppose a physicist observes the precise location of an electron, what is it about that observation that prevents him from figuring its momentum? If he can calculate it with a less precise idea of its location, why would more information prevent him from calculating it?

I suspect I'm thinking about this the wrong way.

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    $\begingroup$ "Why" is never a very good question in physics, because ultimately, we can't know the answer to that. From a mathematical perspective: The Fourier transform. Momentum and Position are related by a Fourier transform and there we have a theorem that a 'narrow' distribution in real space is a 'wide' distribution in frequency space and vice versa. Have a look here: physics.stackexchange.com/q/47458 $\endgroup$ – Martin May 27 '15 at 12:12
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    $\begingroup$ Equivalently, you could say that it comes from the noncommuting property of operators or the symplectic structure of the canonical commutation relations in companion with the non-squeezing theorem. None of these can be explained in as simple terms as the Fourier transform. $\endgroup$ – Martin May 27 '15 at 12:13
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I'm still tryna get my head around this so please correct me if you find anything that is wrong. It all starts with first of all being able to describe a particle in space with something called the wavefunction. The wavefunction is a probability distribution in space. So essentially, the first step is to see that a particle in space isn't point a point in space with attributes such as energy, mass, momentum, etc... But more rather, there's a probability across the space that you will find the particle in that portion of space with a particular energy, momentum, etc...

Once you have the idea that a particle can be described this way, then we see that as the wavefunction is just a mathematical function and can be described as a sum of orthogonal states (see Fourier analysis), with each of these 'states' having a specific energy/momentum associated with them, the amount each state contributes to the wavefunction can be observed. What we find is that mathematically, the spread in momentum/energy/(whatever observable we are looking at) is related to the spread the wavefunction/probability distribution has in space/time/(whatever the corresponding conjugate variable is).

Basically, everything follows once you accept that a particle isn't actually a point in space, but a probability distribution in space. I don't really know much about why that is the case but hopefully someone else can explain that. Something to do with points in Hilbert space being waves in real space or something? I don't know.

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  • $\begingroup$ this is more or less correct, but doesn't really answer the question IMHO $\endgroup$ – innisfree May 27 '15 at 12:40
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One way to think about it, is in terms of a possible measurement process (but note: the Heisenberg uncertainty principle is not a limitation of our measurement process, but a fundamental restriction!). So keep in mind, this an approximate, possibly misleading argument, but it is intuitive (and was brought forward by Heisenberg himself).

If you want to resolve the position of particle spatially you need to use a wavelength on the scale of the exactness of the position you want to measure. The (lets say electromagnetic) radiation you use, will have a momentum that is inversely proportional to the wavelength. When it interacts with the system you measure, it will transfer some of its momentum, thus changing the particles momentum on the order of its own momentum. Therefore, the more precisely you want to measure the position of your particle, the less precisely you will know its momentum.


Mathematically, the statement can be arrived at in various ways. For momentum and position it simply reduces to the uncertainty principle for the Fourier transform (as momentum eigenstates are plane waves, thus the momentum-representation of the wave function is the Fourier transform of the position representation). More generally (for any pair of operators fulfilling the canoncial commutation relations) it can be proven by using the Cauchy-Schwarz inequality in Hilbert space.

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  • $\begingroup$ As you say yourself, the principle has nothing to do with HUP as usually formulated. Heisenberg's idea of how the measurement disturbance leads to the principle is inuitive, yet we understand today that it is false. Nevertheless, measurements DO disturb the state in a similar way, but today, people understand this in a different way (see e.g. here: physics.stackexchange.com/questions/169730/…) $\endgroup$ – Martin May 27 '15 at 12:55
  • $\begingroup$ I'd not say it is false, rather incomplete. And the energy-time uncertainty relation is still derived similarly in all books I read. And how measurements disturb state can already be found in Landau/Lifshitz III (Quantum mechanics), their description of measurement is eerily close to modern decoherence theory (although they do the partial trace by considering the measurement apparatus to be approximately classical and do not get a density matrix but a state). $\endgroup$ – Sebastian Riese May 27 '15 at 13:01
  • $\begingroup$ With Heisenberg uncertainty relation, I refer exclusively to the (special case of the) Robertson-Schrödinger relations and here, the picture of error-disturbance by measurement is not correct. I'm not saying that the idea of measurement-disturbance is new or that Heisenberg was incorrect in speculating such an effect, I'm just saying that this has nothing to do with the Robertson-Schrödinger uncertainty relation, but is a separate effect. Note also: No such relation can exist for energy-time. $\endgroup$ – Martin May 27 '15 at 14:08
  • $\begingroup$ I know, that for time and energy, there does not exist a uncertainty relation due to the non-commutation of operators as for momentum and position (in this way, it is something completely different), but there exists a formal analogue for energy measurements and time (due to the fact that energies are related to frequencies, which cannot be measured more precisely than "counting the zeros", so this is also due to the properties of the Fourier transform! $\endgroup$ – Sebastian Riese May 27 '15 at 14:10

protected by Qmechanic Jun 5 '17 at 9:34

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