The 3B1B YouTube channel has a video The more general uncertainty principle, regarding Fourier transforms which looks at thin peaks in frequency domain corresponding to long-lasting pulses in time domain, and vice versa. The transformation that makes these comparisons possible is the Fourier transform. Unfortunately no equation or mathematical relation is given in the video for this principle. Perhaps there exists a functional equation for this principle, or perhaps it is just something that visual inspection seems to confirm.

The Tom Rocks Maths' YouTube channel has a video Heisenberg's Uncertainty Principle with @Michael Penn that derives the famous $\sigma_x \sigma_p \geq \frac{\hbar}{2}$ using (quantum) expectations and Schrodinger's equation.

I want to learn how closely related these concepts actually are.

On the face of it, the two explanations give me the impression that they are logically independent things because the general uncertainty principle assumes nothing about Schrodinger's equation and could really apply to almost any signal.

But within QM we can think about both of these notions, motivating an ability to distinguish them. Certainly the complex exponential functions involved in solutions to Schrodinger's equation entail a relationship to Fourier series via Euler's formula, so it is natural to suspect that a correspondence between the inverse domains of the Fourier transform should feature somewhere in understanding QM. It isn't clear to me whether this general uncertainty principle is a natural generalization of Heisenberg's uncertainty principle, or only under certain constraints, or even that they are still logically independent considerations within QM.

What relationship, if any, exists between these two principles?


4 Answers 4


3B1B's Youtube video mainly talks about the Fourier uncertainty principle between a function $\psi(x)$ in position space and its Fourier transform $\hat\psi(\xi)$ where $\xi$ is the spatial frequency (i.e. the reciprocal of wavelength $\lambda$). However, the video concentrates on intuitively explaining this principle, but doesn't provide its mathematical formulation. $$\sigma_x \sigma_\xi \geq \frac{1}{4\pi} \tag{1}$$

It basically says that a narrow peak in $x$-space corresponds to a wide peak in $\xi$-space, and vice versa. So far this is a purely mathematical statement. There is no physics involved yet.

You get something physical by introducing de Broglie's relation, which says that every particle (with momentum $p$) behaves like a wave (with wavelength $\lambda=h/p$) propagating in space. This can be written in various ways: $$\begin{align} p&=\frac{h}{\lambda} \\ p&=h \xi \\ p&=\hbar k \end{align} \tag{2}$$ These are all identical because of $\xi=\frac{1}{\lambda}$, $k=\frac{2\pi}{\lambda}$ and $\hbar=\frac{h}{2\pi}$.

Now multiply equation (1) by $h$, and use $p=h\xi$ from equation (2). Voila, you have derived the Heisenberg's uncertainty principle between position $x$ and momentum $p$. $$\sigma_x \sigma_p \geq \frac{\hbar}{2} \tag{3}$$

  • $\begingroup$ IMO a trivial (for a physicist) yet very crucial sentence (the de Brogile axiom) is missing, without which the answer actually makes little sense: connecting the variables in eqs. (1) and (2), or in other words, explaining why can we use eq. (1) with the quantities in eq. (2)! And this sentence is the key to quantum mechanics: "The de Broglie picture expresses particles as waves with a wavelength $\lambda = h/p$ propagating in space." $\endgroup$
    – Neinstein
    Jul 18, 2022 at 12:49
  • $\begingroup$ @Neinstein You're right. I have added this crucial bit now. $\endgroup$ Jul 18, 2022 at 14:45

It is very easy to be seduced by the analogy between Fourier pairs like $x$ and $p$ and the HUR but the relation is basically coincidental.

It is true that $x$ and $p$ satisfy Fourier-type relations but these don’t have much to do with HUR. The HUR is a relation between variances of operators and the average value of their commutator, and there is no average value or commutator in Fourier pairs. (Although one can substitute the commutator (up to $i\hbar$) by a Poisson bracket.)

If one attempts to continue the analogy of $x$ and $p$, one obtains a Fourier-type relation between $\omega$ and $t$, and naively subbing $E=\hbar \omega$ one can obtain a Fourier relation between $E$ and $t$, which has no quantum analogue since $t$ is not an operator so there is no sense to $\Delta t$ as the variance of an operator in the quantum HUR. (There are multiple questions on this site on the pitfalls of $\Delta E\Delta t$, tied to the problem of making sense of $\Delta t$; see this post for a precise definition of the sense of $\Delta t$.) Moreover, continuing extending the analogy $L_z$ and $\theta$, one runs into even more problem as there is no good angle operator in quantum mechanics. There are also multiple questions on this site about quantization of the angle variable and its own set of problems, such as this one or this one .

Finally there is an HUR for any two non-commuting operators, irrespective of the existence of Fourier relations between these operator. The simplest example is $L_x$ and $L_y$ where we have $$ \Delta L_x\Delta L_y=\frac{1}{2}\vert \langle i\hbar L_z\rangle\vert\,. $$ Never mind the fact that $L_x$ and $L_y$ are not Fourier pairs (in the usual sense), it turns out the RHS can be $0$ even if neither $\Delta L_x$ or $\Delta L_y$ is $0$ because the RHS is state dependent. This cannot happen under any Fourier scenario.

The uncertainty principle is usually taken as slightly more general of Heisenberg’s uncertainty relations, which are often introduced as a special case of the general uncertainty relations specialized to $x$ and $p$. The connection with Fourier analysis breaks beyond $x$ and $p$ so is not terribly useful to analyse or get insight into the UR.

  • $\begingroup$ It might be instructive to include the derivation of the general uncertainty relation, though that probably already exists elsewhere on this site. $\endgroup$ Jul 18, 2022 at 17:01
  • $\begingroup$ @CharlesHudgins a simple wiki search is enough: en.m.wikipedia.org/wiki/… $\endgroup$ Jul 19, 2022 at 15:59

The uncertainty principle as a mathematical theorem says, in loose words, that the broader the function, the narrower its Fourier transform.

In QM of a single particle, the state can be described by a wave-function in $x$-representation $\psi(x)$ or in the momentum representation $\tilde\psi(p)$, and these two are related via Fourier transform. Furthermore, $\sigma_x$ and $\sigma_p$ are measures of the broadness of the wavefunction in the respective representations. The Heisenberg formula $\sigma_x \sigma_p \geq \hbar /2$ is the general mathematical theorem interpreted for the specific case of quantum mechanics.

I think Schrödinger equation is not needed to derive the Heisenberg uncertainty principle.

  • 5
    $\begingroup$ Right! Various more-or-less precise versions of "the broader a function, the narrower its Fourier transform" are mathematical assertions, which apply to any function/signal. The connection to physics is interpreting/postulating position/momentum as Fourier transform pairs. :) $\endgroup$ Jul 16, 2022 at 18:55
  • 3
    $\begingroup$ "Schrödinger equation is not needed" A quantum mechanical wave equation is needed in order to have the quantum mechanical HUP. The wavefunction postulate that connects the mathematical wave solution with the probability is necessary in deriving the HUP. $\endgroup$
    – anna v
    Jul 16, 2022 at 19:00
  • 1
    $\begingroup$ @annav You need to know the interpretation of $\psi(x)$, but it is not the Schrödinger equation. Schrödinger equation only gives you its time evolution. $\endgroup$
    – Hoody
    Jul 17, 2022 at 19:51

Let me make a comparison.

There is the general principle of conservation of angular momentum, and there are the many instances where this conservation of angular momentum manifests itself.

For instance, in the case celestial mechanics: in our solar system all the celestial motions of the planets and asteroids and comets is dominated by the gravity of our Sun, which means angular momentum is applicable. The effect is not vivid for the planets and asteroids, as they move in close to circular orbit; their angular velocity with respect to the Sun barely changes. But take the case of Halley's Comet: at closest approach to the Sun the angular velocity of Halley's comment is far larger than at the point of furthest distance to the Sun.

At human scale there is the example given in every physics textbook: the spinning skater. She pulls her arms closer to her torso, and the increase in angular velocity is in accordance with conservation of angular momentum.

It would be ludicrous to suggest:
"The general principle of conservation of angular momentum assumes nothing about the inverse square law of gravity, so I don't see why celestial motion would be subject to conservation of angular momentum."

The whole point of recognizing a general principle is that its validity is independent of the details of the various instances where the principle manifests itself.

The point that Grant Sanderson (3B1B) is making is that the phenomenon that is being described with the Schrödinger equation is an instance of wave mechanics, and as such Heisenberg uncertainty is an instance of the general uncertainty inherent in wave mechanics.

Now, of course the Schrödinger equation is very different from the classical wave equation. Caution is warrented, that I agree with. Overall: in my opinion the physics described with the Schrödinger equation does count as an form of physics taking place where the general uncertainty principle manifests itself. That is: I concur with Grant Sanderson.

In the specific instance of the physics described with Schrödinger's equation the manifestation of uncertainty is referred to as Heisenberg uncertainty.


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