Heisenberg's Uncertainty Principle states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa.

What I am getting out of this is that we as humans cannot know both the position and momentum of a particle at the same time to 100% certainty. While we as humans cannot know both the position and momentum of a particle at the same time to 100% uncertainty, these values do exist in nature at any instant of time, right? While we do not (or perhaps cannot) know what these values are, there are numerical values to these variables at any given time, correct?

I've heard arguments about Heisenberg's uncertainty principle disqualifying the hard-deterministic theory that if all particles are simply obeying the laws of physics, everything is pre-determined. I don't see how Heisenberg's uncertainty principle disqualifies this theory as predictable is not the same thing as pre-determined -- if we, as humans, cannot know the position and momentum of any particle at any instant of time with 100% accuracy, we cannot predict the future as human beings. However, if these values still exist as finite numerical values, and assuming they obey the laws of physics, everything can be pre-determined without it being predictable, correct?

I'm sorry for taking this off on a philosophical tangent. Please answer the physics questions (in the 2nd body of text), and feel free to ignore the third paragraph.


The idea that position and momentum exist and we do not know them is untenable if one also assumes that the particle evolves with the laws of classical physics.

If (classical) position and momentum existed independently from our knowledge, the evolution of the wavefunction and the probabilities would be different from what we actually observe in physical experiments.

I mean, for instance, a free particle, independently from my knowledge of its initial position and momentum, moves along a straight line with constant speed. This would produce a certain law of evolution of probabilities to find the particle somewhere (probabilities simply due to my initial incomplete knowledge of the state of the particle) similar to that of classical statistical mechanics.

This picture is not corroborated from facts when analyzing elementary quantum phenomenology.

If one tries to insist on the epistemic intepretation of Heisenberg principle (as stated in the second block of your question), one must also assume that the evolution of the particle is subjected to laws different from those of classical mechanics. In particular some sort of quantum force enters the picture with quite classically weird peculiarities (e.g. non-locality). This is the idea behind Bohm's interpretation of quantum mechanics.

The standard formulation of QM assumes instead that

position and momentum generally do not exist and they cannot exist simultaneously in any case.

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Here's a 2-minute, mathematically accurate explanation of the Uncertainty Principle:

Particles are singular points, which means that they have one and only one location is space (a Dirac delta function, if you know about that sort of thing). Idealized waves are infinite in extent (think about the graph of a sine function carrying on towards infinity). In quantum mechanics, all of the information about the system is contained within its wave function (which is usually more like a sine wave than a Dirac delta function).

However, the wave function can take on a range of shapes between a sine wave and a Dirac delta function. The more DDF-like the wave function is the better we "know" its position. The more sine-like the wave function is the better we "know" its momentum. Now, in truth, we don't actually know the particle's position or momentum in either case, and the only way to find out is to do a measurement and thus collapse the wave function. But in the more localized cases (DDF-like) the range of possible positions is small while the range of possible momenta is big. Conversely, in the sine-like cases the range of possible momenta is small while the range of possible positions is big.

In fact, these ranges are closely linked through a process called the Fourier Transform of the wave function. And from the mathematical analysis of Fourier Transforms it can be deduced that the "width" of the position wave function times the "width" of its Fourier Transform (i.e. the momentum wave function) must be no less than $\frac{\hbar}{2}$. Thus,

$$ \sigma_x\sigma_p\geq\frac{\hbar}{2} $$

where the sigmas represent the standard deviations of their respective wave functions.

EDIT: To address @ZeroTheHero from the comments. I agree that the way I've presented this material is fairly tightly focused towards the position-momentum uncertainty relation, but that's what OP was asking about and it is the best example for developing a meaningful intuition about this property of quantum mechanics.

That said, there is nothing analogical about the Fourier Transform explanation. It is exact, and it is easy to show its exactness. Here is the proof:

To show the inequality, it suffices to show that the minimum uncertainty state satisfies the equality (i.e. $\sigma_x\sigma_p=\frac {\hbar}{2}$). The minimum uncertainty state is known to be a Gaussian with arbitrary mean and standard deviation (see Shankar or Griffiths for that proof). Without loss of generality, let's examine the normalized position-space state

$$\psi (x)=\frac{1}{\pi^{\frac{1}{4}}}e^{-\frac{x^2}{2}} $$

which has a corresponding Fourier Transform (that also happens to be its momentum-space wave function)

$$\phi (p)=\frac{1}{\pi^{\frac{1}{4}} \sqrt{\hbar}} e^{-\frac{p^2}{2\hbar^2}} $$

From here,

$$\sigma_x^2=\frac{1}{\sqrt {\pi}}\int^{\infty}_{-\infty} x^2 e^{-x^2} dx =\frac {1}{2}\\ \ \ \\ \sigma_p^2=\frac{1}{\hbar\sqrt {\pi}}\int^{\infty}_{-\infty} p^2 e^{-\frac{p^2}{\hbar^2}} dp=\frac{\hbar^2}{2} \\ \ \ \\ \therefore \sigma_x\sigma_p=\frac {\hbar}{2} $$

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  • $\begingroup$ You might want to be careful in suggesting the Fourier relations are a mathematically accurate explanation of the HUP. They do not provide you with the $1/2 $ factor on the right hand side and are not an inequality. Also, it is difficult to make a similar Fourier-type argument for anything beyond $x$ and $p$: for angular momenta for instance the right hand side is state-dependent. $\endgroup$ – ZeroTheHero Oct 7 '17 at 17:50
  • $\begingroup$ @ZeroTheHero I worked through some responses to what you said. I hope this clears it up. $\endgroup$ – Geoffrey Oct 8 '17 at 16:47
  • $\begingroup$ Right... I just wanted to point out that the Fourier relations would be more like $\Delta x\Delta k\approx 1$, not a strict inequality, and no factor of $1/2$. Of course for Gaussian states the FT approach works because the FT of a Gaussian is also a Gaussian, but this is exceptional rather than normal. $\endgroup$ – ZeroTheHero Oct 8 '17 at 16:53
  • $\begingroup$ @ZeroTheHero I'd say Gaussians are exceptional precisely because they ARE normal! rimshot Look, you can't argue that the FT isn't the p-space wavefunction because it is. That's a fact. Another fact is that the HUR holds regardless of whether you evaluate the expectation values both in x-space, both in p-space, or one in each. That means - like it or not - the relationship between the variance of the x-space function and its FT is hard and fast. I think whoever showed you this before didn't finish the proof. Also, your missing 1/2 comes from squaring the function before finding the variance. $\endgroup$ – Geoffrey Oct 9 '17 at 0:02
  • $\begingroup$ See the excellent discussion in Berkeley Physics on Waves amazon.ca/Waves-Berkeley-Physics-Laboratory/dp/0070048606 (which sort of shows my age). $\endgroup$ – ZeroTheHero Oct 9 '17 at 0:09

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