I am trying to understand logic behind the uncertainity principle. And as far as I understand, it follows mathematically if we assume that wave function in momentum space is Fourier transform of the wave function in position space. I tried to dig in and find out why they should be related so, and the only explanation I could find out was Pontryagin duality.

  • $\begingroup$ Reminder: Comments are for critiquing or clarifying the question, not for giving half-baked answers. Several comments deleted. $\endgroup$ – ACuriousMind Oct 24 '17 at 11:16

Seems that one may have a bit of a chicken-and-egg situation here. What comes first, quantum mechanics or the Fourier transform? According to tparker's answer it seem that one should take quantum mechanics as more fundamental and the Fourier transform then follows from it. However, I suspect it is the other way around.

The Fourier properties were more or less imposed at the point where Planck discovered the relationship between energy and frequency $E=\hbar\omega$, which was later extended to a relationship between momentum and the propagation vector ${\bf p}=\hbar{\bf k}$. Due to these relationships, the fathers of quantum mechanics expanded everything in terms of plane waves. Well, plane waves form an orthogonal basis. Hence, such an expansion comes down to a Fourier analysis. As a consequence, one then obtains the Heisenberg uncertainty relationship.

However, one also finds in quantum mechanics some Heisenberg-type uncertainty relations that does not seem to follow directly from a Fourier relationship. For example, consider the uncertainty relationship associated with spin. This begs the question, what is the underlying principle that leads to an uncertainty relationship, which is shared by Fourier analysis?

This underlying principle, in my view, is the notion of mutually unbiased bases. Any inner product between elements from the respective mutually unbiased bases $\langle x|k\rangle$ gives a constant magnitude, independent of the choice of elements (the phase could be different). Any state with a particular representation in one basis will have a representation in a mutually unbiased basis that obeys a Heisenberg-type uncertainty relationship; the width in terms of one representation would be inverse proportional to the width in the other representation.

What does this have to do with Fourier analysis? Well, a Fourier transform is a link between representations in two mutually unbiased bases. This follows from the fact that for these bases $\langle x|k\rangle=\exp(-ixk)$, which means that $|\langle x|k\rangle|=constant$. This property, ultimately leads to the uncertainty relationship as we know it.

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  • $\begingroup$ Perhaps your answer hinges on the phrase "as we know it", but to me the uncertainty principle is much more general than just applying to mutually unbiased bases. For any two operators that don't commute, we have the uncertainty relation $\sigma_A \sigma_B \geq |\langle [A, B] \rangle| / 2$. This statement only refers to the two operators themselves, but to talk about MUBs you need to specify two complete (and often infinite) sets of operators, the large majority of which you never use. This seems like a huge amount of unnecessary baggage. $\endgroup$ – tparker Oct 23 '17 at 6:09
  • $\begingroup$ There's still an uncertainty relation between $\sigma_x$ and $\sigma_x + (1/2) \sigma_y$, even though they don't belong to any MUBs. $\endgroup$ – tparker Oct 23 '17 at 6:11
  • $\begingroup$ Thanks for your comments @tparker. The way I understand it is that the operators $A$ and $B$ are associated with eigenbases, which are mutually unbiased if these operators do not commute. Part of $\sigma_x+\sigma_y/2$ obviously commutes with $\sigma_x$. I'd say the idea is to determine which parts actually do not commute, which in this case would be $\sigma_x$ and $\sigma_y$. $\endgroup$ – flippiefanus Oct 23 '17 at 12:18
  • $\begingroup$ Can you please explain the chat above and your answer for less mature audience. Specifically, the example by @tparker, I don't know how to go about understanding it. The notion of mutually unbiased bases is very helpful. Thanks. $\endgroup$ – Pratyush Rathore Oct 23 '17 at 17:04
  • $\begingroup$ So you're saying that starting from any collection of operators, you can do something like the Gram-Schmidt orthogonalization procedure, but instead of getting rid of parallel components of vectors, you get rid of parts of operators that commute with previously considered operators? And the end result of this process will always be a mutually unbiased basis of some Hilbert space? That's possible, but I'm skeptical that there's an equivalent of Gram-Schmidt's inner product that allows us to uniquely pick out "the part of $B$ that commutes with $A$". $\endgroup$ – tparker Oct 23 '17 at 17:16

Practically speaking, the full machinery of Pontryagin duality is way more advanced than physicists need to understand the uncertainty principle. There are several ways to "derive" that the momentum-space wavefunction is the Fourier transform of the position-space wavefunction, which depend somewhat on your choice of starting postulates. Here's one common path:

One common starting fundamental postulate is the commutation relation $[\hat{x}, \hat{p}] = i \hbar.$ The most common position-space representation of this commutation relation is $\hat{x} \to x,\ \hat{p} \to -i \hbar \frac{\partial}{\partial x}$. In this representation, taking the inner product of $\langle x |$ and the eigenvalue equation $\hat{p} |p\rangle = p | p \rangle$ gives the differential equation $$-i \hbar \frac{d\, \psi_p(x)}{dx} = p\, \psi_p(x),$$ which has solution $\psi_p(x) = \langle x | p \rangle \propto e^{(i p x)/\hbar}$. Then to express an arbitrary state $| \psi \rangle$ in the momentum basis, we can use the resolution of the identity $$ \psi(p) = \langle p | \psi \rangle = \int dx\ \langle p | x \rangle \langle x | \psi \rangle \propto \int dx\ e^{-ipx/\hbar} \psi(x),$$ which is just the Fourier transform. This generalizes straightforwardly into higher dimensions.

BTW, the fact that position-space and momentum-space wavefunctions are Fourier transforms of each other (or more precisely, can be chosen to be Fourier transforms of each other) gives some nice intuition for the uncertainty relation but isn't actually necessary to derive it. All you need is the commutation relation, as I explain here.

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  • $\begingroup$ Can you please explain why we believe in the commutation relation and what does it convey? I read about the commutation for the first time today and although I tried reading online, I haven't been able to grasp it. Also, if I want to learn the mathematical notation you are using, [math]\hat{p}|p>[/math], what topics should I search for? Thanks. $\endgroup$ – Pratyush Rathore Oct 23 '17 at 16:42
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    $\begingroup$ @PratyushRathore en.wikipedia.org/wiki/Bra%E2%80%93ket_notation $\endgroup$ – tparker Oct 23 '17 at 17:09
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    $\begingroup$ @PratyushRathore Asking why we "believe in" the commutation relations is essentially asking: Why quantum mechanics? $\endgroup$ – ACuriousMind Oct 24 '17 at 11:17

I will try to adress this question from a point of view that is useful when building intuition. Thus, see other peoples answers, and references in Wikipedia, to fill in technical details.

First a comment on theory-building. Much of theoretical physics begins as someones hunch, or educated guess. You then see if you can formalise what you feel should be the case, and use the new framework to make predictions that are tested with experiment (ideally). Iterations will then strengthen your framework, or prove it to be unfeasible. Inevitably, theories will contain some axioms (statements that you assume are true without proving them). Note that these axioms are not necessarily unique, and what you choose to be an axiom, and what is a theorem (consequence of the axioms) is somewhat ambiguous; the distinction can dependent on your personal tastes (though there is commonly some consensus in a field).

One possible axiom for Quantum Mechanics is that: a state of a system can be described as a "wave-function" over the spacial coordinates, like x y z. (This axiom is usually generalised and expressed in terms of Hilbert spaces instead.)

Then, if a state of a physical system (like a particle) is modelled with a function you might wonder what you can do with this function, and what that can tell you. One obvious thing you can do with functions is decomposing it as sums. Like consider for instance the functions $f$, $g$ and $h$.

$$ f(x) = x^2 + 5x \hspace{10pt};\hspace{10pt} g(x) = x \hspace{10pt};\hspace{10pt} h(x) = x^2 $$

Clearly, $f$ can be expressed was a sum of $g$ and $h$; if you take 5 of $g$ and 1 of $h$. But there are much more sophisticated ways to do the exact same thing. Like on a interval $x \in[a,b]$, any function can be expressed as an infinite sum of sines and cosines. This is the idea behind Fourier series. Then you can ask yourself if the same thing is possible on an infinite interval, $x\in[- \infty, \infty]$, and it turns out that it is. This is the idea of Fourier transforms. Clearly when you do this you will pick up at lot of technicalities, which are important when actually dealing with Fourier transforms. But in spirit you're doing the same thing as in the trivial example above.

When you do a Fourier transform, how to mix the functions (like 1 of $g$ and 5 of $h$ in the example) is summarised in a second function, called the "transform". This function cannot be a function of $x$ (clearly), so it is a function of something else, let's call the variable $k$ and the function $\tilde{f}(k)$. You can then go back to your physical theory and ask if $\tilde{f}(k)$ has any kind of physical significance. I mean, if it tells you anything useful abut your system.

It turns out that it does; $k$ is actually associated with the momentum $p$ of your system. Where $p = \hbar k$ gives you the correct predictions when compared to experiment. On top of that it ties nicely into other existing theories, and other ideas you hold for true.

Your friends doing the experiments then come back and tell you that they cannot seem to simultaneously nail down the momentum and position of a particle. And you respond that you know why. That it is a consequence of the fact that particles are best modelled by functions, and a well localised function in $x$ becomes a non localised function in $k$ (which is basically $p$), and vise versa. Then doing some further investigations into the theory of Fourier transforms you find that the function that is simultaneously "the most localised" in both $x$ and $k$ is the gaussian function, and from that function you conclude that $\Delta x \Delta p \geq \frac{\hbar}{2}$, where $\Delta x$ is the standard deviation in $x$, and $\Delta p$ is the standard deviation in $p$.

You then make a cup of coffee, and wait for a phone call from Stockholm.

[Edit below:]

The idea that $p = \hbar k$ is called the de Broglie hypothesis, and an intuition about why anyone (like de Broglie) would suggest this hypothesis can be gained by considering special relativity. Basically: If you have a stationary wave alternating in sync, and you travel by it at some (high) speed, the loss of simultaneity will mean that the wave gains a spacial frequency, as well as momentum (relative to you).

See this resource for some pretty animations and further explanations: 3blue1brown about the uncertainty principle.

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Each uncertainty is a reflection of an existing fundamental symmetry of nature. For example, per the Noether theorem, the time symmetry leads to energy conservation while the space translation symmetry leads to conservation of momentum. The quantum reflection of these symmetries is the time/energy and position/momentum uncertainty. While the Noether theorem is limited to only a few symmetries for technical reasons, it reveals the connection between symmetries and conservation laws and gives an insight on the nature of uncertainty.

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    $\begingroup$ Continuous symmetries and their generators always have a commutator equal to $i \hbar$, but the uncertainty principle holds more generally for operators with an arbitrary nonzero commutator. $\endgroup$ – tparker Oct 23 '17 at 16:15
  • $\begingroup$ @tparker I don't disagree with your comment at all, except for one very important point: it doesn't apply to my answer. Please read my answer carefully and tell me where exactly it states what you are objecting. I can see how one may make this connection by reading without attention and making mental assumptions that the answer states what it actually doesn't. For example, where do you see me connecting uncertainty to specifically continuous symmetries??? Please remove the undeserved downvote, I'd appreciate it. My answer is complimentary and doesn't contradict yours. $\endgroup$ – safesphere Oct 23 '17 at 16:38
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    $\begingroup$ Your first sentence clearly states that you think that each uncertainty is related to an "existing fundamental symmetry". This is precisely what tparker objects to - this is often true for uncertainty principles of the form $...\geq \mathrm{i}\hbar$ but need not be(!), and simply false for most pairs of operators one might look at. Secondly, the "time/energy" uncertainty should not be treated as being of the same kind as the position/momentum uncertainty, cf. physics.stackexchange.com/a/53804/50583 for the rather different meaning of "$\Delta t$" in there. $\endgroup$ – ACuriousMind Oct 24 '17 at 11:21
  • $\begingroup$ Yes, each uncertainty represents a symmetry, but not necessarily continuous. This is rather obvious, as any uncertainty is for a pair of Fourier conjugates while a Fourier relation can be seen as representing a symmetry. Secondly, time/energy symmetry and uncertainty is absolutely of the same kind as position/momentum, if you look at them the right way. Time is a coordinate of spacetime along with the space coordinates. Energy and mimentun are components of 4-momentum. Both symmetries are between a coordinate and the corresponding component of 4-momentum, exactly the same type of symmetry. $\endgroup$ – safesphere Oct 24 '17 at 13:32
  • $\begingroup$ I see where the confusion may be. The symmetry used in the Noether theorem is not the symmetry I am talking about. For example, per Noether, momentum conserves if space has a continuos translation symmetry. The symmetry in the theorem is for position only while the momentum is constant. This is a limited special case of the general position/momentum symmetry where space is not uniform and the space translation symmetry is not present, but the position/momentum symmetry still holds. This allows us to represent the space curvature as a "force field" In a flat space and restore the conservation. $\endgroup$ – safesphere Oct 24 '17 at 14:11

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