Heisenberg uncertainty principle, what is it?

Heisenberg uncertainty principle (HUP), what is it? Before this gets marked as a duplicate, i've went through most of the posts here already, many of them multiple times. I've went through countless resources (youtube, wikipedia, etc.) and for years it seems i've have been trying to get a definitive answer on what the HUP is.

I swear i've seen 5 or so different definitions all from sources that act like they know what's going on. Some will say description X is a misconception while others say it's description Y and so on, nobody seems to be in agreement. Anyway, before someone posts the uncertainty in momentum is inversely proportional to the uncertainty in position equation, i know what that it means mathematically. What i don't understand is why and how that translates to the real world and QM and waves. I was hoping i could get an answer here that might shed some light on what the other answers and videos i've seen have not been able to.

The Heisenberg Uncertainty Principle is a formal property of the Fourier transform. For a function satisfying $\int|f|^2=1$, it is a straightforward exercise (using integration by parts and Cauchy-Schwartz) that $$C\le \int |x f|^2dx \int |\xi \hat{f}|^2 d\xi^2$$ for $C$ a constant (in fact $C=1/4\pi$ if I did this right).

In the context of quantum mechanics, the two integrals on the right of the equal sign can be interpreted in terms of standard deviations of position and momentum, but the inequality holds independent of the interpretation, and conversely, once you commit to the basic formalism of quantum mechanics, the Uncertainty Principle comes down to no more or less than this simple calculation.

• It might be worth saying a few words about how Fourier transforms appear in the formalism of quantum mechanics, and what the relationship is to the non-commutation of observables. – dmckee Feb 17 '18 at 2:43
• @dmckee: I've added a separate answer, sticking to binary variables, which does what I think you're asking for, though without explicitly invoking the language of Fourier transforms. – WillO Feb 17 '18 at 3:40
• I'm sure you know this: it's probably best to leave the things the way you have it as $C$ depends on the FT's definition. If, as in physics, the FT is unitary, then $C=1/4$ (no $\pi$ there; see physics.stackexchange.com/a/91975/26076). (I say this because often you see scale factors in signal processing literature that make the FT nonunitary, weird though that may seem). – WetSavannaAnimal Feb 17 '18 at 5:16
• @WetSavannaAnimalakaRodVance : Yes, rather than saying "If I did this right", I should have said "If you choose the right convention", where "right" doesn't mean right in any absolute sense --- just that there exists a convention that will get you this answer. Thanks for adding this. – WillO Feb 17 '18 at 5:42

The Heisenberg uncertainty principle is a mathematical envelope that describes the behavior of nature in the microcosm of particles,explored in dimensions less than nanometers .

I call it an envelope, because it is a rule of thumb for the way the mathematical model that describes particle behavior, Quantum mechanics, is formulated, and it allows to gauge whether classical mechanics and electrodynamics is sufficient to model the data, or one has to use quantum mechanical models.

The form of the HUP in the mathematical models of quantum mechanics is seen in the commutators (page 31 of the operators of observables.

It is the algebra of operators assumed in the differential equations of quantum mechanics that imposes the uncertainty when two observables do not commute. The expectation value of the commutator operators gives the HUP

The connection with the "wave" comes from the solutions of the QM wave equation . The solutions are sinusoidal and the complex conjugate square gives the probability of the spatial and momentum existence of the particle under consideration. The corresponding expectation values of the momentum and space operators also depend on the probabilistic nature of the QM framework.

QM is a counter intuitive system for intuition built up by everyday observations of particles and waves. Intuition has to be built up by study. This link may help.

It occurs to me that you might find this easier to think about if you start with binary variables, as opposed to continuous variables like position and momentum.

So consider a coin that can be either heads-up ($H$), tails-up ($T$), or in some superposition $aH+bT$ where $a^2+b^2=1$. (I'll stick to real coefficients in order to avoid having to write $||$ signs all the time.) In the latter case, if you observe the coin's orientation, it will be heads-up with probability $p=a^2$ and tails-up with probability $1-p=b^2$.

A reasonable measure of the uncertainty in this observation is $U=p(1-p)$ --- so that whenever you know the outcome of the measurement with certainty (i.e. if $p=0$ or $p=1$) we have $U=0$, and whenever you are maximally uncertain (i.e. if $p=1/2$), the uncertainty achieves its maximum value of $1/4$..

Now consider some other property of your coin --- think of it as, for example, the coin's temperature, which can be either warm ($W$) or cool ($C$). And suppose these states can be written in the form $$W={1\over\sqrt2}(H+T)\qquad C={1\over\sqrt2}(H-T)$$

The state of our coin, which we wrote as $aH+bT$, can be rewritten as $${a+b\over\sqrt2}W+{a-b\over\sqrt2}C$$ If we observe the temperature of the coin, it will be warm or cool with probabilities $q={(a+b)^2\over2}$ and $1-q={(a-b)^2\over 2}$. Then the uncertainty associated to this observation is $V=q(1-q)$, which a little calculation shows is equal to $(p-1/2)^2$.

Here is a graph of the two uncertainties, $U$ and $V$, as a function of $p$:

As you can see, whenever one is large, the other is small. That is the Heisenberg Uncertainty Principle, and as you can also see, once you've committed to the basic setup, it's a purely mathematical phenomenon.

An analogy in classical wave mechanics helps in my opinion. Imagine a single wave packet, for example a single excitation of a rope. It appears as just a single peak moving down the rope. What is important here is that it is straight-forward to see "where" the wave is: just where the crest is, for example the maximum of it. However, what is the frequency of such a wave?

The other extreme is a sine wave in your rope. Each point on the rope oscillates in a sine-fashion in time. We can easily find the frequency of this wave by measuring the wavelength and the propagation speed. But what about the position of the wave? We could again take the maximum, but there are now (ideally) infinitely maxima.

A Gaussian wave packet is somewhat a hybrid case. Here we have some idea where the packet is, but it is not as clear as in the case of a single crest. On the other hand, we can get an estimate for the frequency, but it is not as clearly definied as in the sine case.

We can express this "either-or" by saying that the product of the uncertainty in frequency and in position are fixed and for sure greater than some value. This guarantees that we will never have a wave of which at the same time we can accurately find the frequency and the position.

How is this related to Heisenbergs uncertainty relation? The wave function of quantum particles behaves in a similar fashion as the rope. The position might be clearly defined, but then the momentum (or $k$-vector, which is related to the wavelength) is not. Effectively the HUP gives us an lower bound for how precisely the position and momentum are determined. Wave functions are either exactly this precisely defined, or less precisely (the product of both uncertainties is larger). The Gaussian wave packet happens to be the case where the product of uncertainties is minimal.

I will try -

First of all, many concepts in quantum physics are not understandable by common sense. Therefore, it is not so necessary to understand them that way. Uncertainty principle may be one of such concepts.

Per wikipedia, "it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa"

From definition itself, it is clear that it is about determining position and momentum. Determining is measuring. Therefore it is about measurement at core.

But in a very common language - To determine momentum of a particle, you need to know its velocity. To know the velocity, you need to know at least two positions and time between the positions.

Thus, determining momentum involves two positions, determining position involves exactly one position. Therefore, precise determination of two at the same time is incompatible with one another. This is also true at macroscopic level but is not material, so we do not talk about it.

Per wikipedia, "it arises in quantum mechanics simply due to the matter wave nature of all quantum objects".

In this sense, also, when it has momentum, means it is moving, means it is a wave/particle. Position of wave/particle can not be determined precisely any way as it is not a well defined positional entity.

When it is just a particle, then it is not moving and has no momentum. So we can say we could not measure momentum, or we can say we did, this is the position and momentum is zero. If zero counts as determination.

• That's exactly what i thought it was too. According to the internet though that's not it apparently. – Yogi DMT Jan 20 '17 at 4:03
• @YogiDMT: This type of conceptual descriptions are not always possible to formalize mathematically and so everyone can come up with their own explanations. Therefore when it comes to formalization, math is used. Then at some point people start using math alone and do not bother to grasp the mechanism as it may not have formal value. – kpv Jan 20 '17 at 4:08
• does it also have anything to do with a particle not always being found at it's highest probability location? – Yogi DMT Jan 20 '17 at 4:29
• @YogiDMT: Yes, that seems due to the wave particle duality. – kpv Jan 20 '17 at 5:19

Oka, I will try to be didactic, succesfully or not. Don't call me heretic for recalling Schrödinger's representation, because this is just an example to ease the concepts.

People say QM is not intuitive. Well, Schrödinger's vision is intuitive enough... once you get rid of Classical prejudices. If you think of particles as waves, then things are not only intuitive, but well known results from classical optics.

The uncertainty principle already arises in classical optics, because it is inherent to Fourier's theorem. The width of the spectrum is inversely proportional to the extension of the wave pulse. This is for plain waves, but something very simmilar applies to waves whose group velocity is a half phase velocity, as in spinless QM.

Imagine an harmonic wave. You know that harmonic waves are made by only one frequency, and hence one only $k$, as $\omega=ck$. So, a typical harmonic wave is made by only one $k$. Now I ask: where is this wave? The answer is... everywhere! It is extended from minus infinite to infinite... You know its momentum $(p=\hbar k)$, but you can't tell the position: the amplitude is maximum in so many points, and since probability is proportional to $A^2$, there are so many points where the particle CAN be found with so much probability.

Now imagine a wavefunction concentrated around one point, like a gaussian centered around $x_0$. The probability of finding the particle is now clearer. The particle will be probably found around $x_0$ and it is extremely unlikely to foind it farther than $3\sigma$'s from $x_0$. So now the uncertainty in position is somehow related to $\sigma$.

So we have an example of a wavefunction that is "more localized" around a point. However, if you compute its spectrum, you will find that this function can be written as a superposition of MANY harmonic waves, each one having different frequencies and $k's$. That's Fourier theorem: a continous Square-integrable can always be written as a sum of harmonic waves. The sum might have infinite terms or even be contunous (integral), but it is after all a superposition of harmonic waves.

If that function is equivalent to a sum of harmonic waves, if you try to measure momentum, you can have as many values as different "p's" appear in that superposition. (Example: if you have a beam of red and green photons, you can measure frequency of one, and you can find a red one or a green one, but not a blue one).

So, the more localized position is, the more $k$'s it is made of. So, if you have a narrow wavefunction, you will certainly find it around $x_0$ ($\Delta x$ will be small), but momenta will be so many, so the uncertainty in momenta will be big.

The product of them will always be greater or equal than $\hbar/2$. In the case of a gaussian packet, it wiill be exactly that quantity, but that's not that normal.

Hope this helps.

protected by Qmechanic♦Feb 17 '18 at 3:48

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