# Understanding the statement of the bandwidth theorem

I know that the Bandwidth Theorem (BT) and the Heisenberg Uncertainty Principle (HUP) are basically the same thing, and stem from the fact that for operators $$A,B$$, we have:

$$\Delta A \Delta B \geq \frac{1}{2} | \langle \psi |[A,B]|\psi\rangle.$$

The HUP is obtained using the fact that $$[x,p] = i\hbar$$. In my class, we were asked to prove this inequality, but were given without proof the value of $$[x,p]$$.

I've now been asked to prove the bandwidth theorem, which states that

$$\Delta f \Delta t \geq \text{some positive constant}$$

I assume that $$f$$ is frequency and $$t$$ is time. It seems that this theorem is folklore, because I haven't been able to find any rigorous proofs. It seems intuitive, especially after watching this 3B1B video, however I have some questions about how to prove this:

1. What even are $$f$$ and $$t$$? Functions? I know that the formula for $$\Delta A^2$$ is given by $$\langle \psi | A^2 |\psi\rangle -\langle \psi | A |\psi\rangle^2$$. It's unclear how to compute this for $$f$$ or $$t$$, since I don't see how they can "operate" on $$\psi$$. I'm aware that we can use $$\langle f_1(x)|f_2(x)\rangle := \int_\mathbb{R} \bar{f_1}(x)f_2(x) dx$$ as an inner product.
2. Similarly, how was $$[x,p] = i\hbar$$ obtained?

I'm sure that the answer has something to do with a Fourier series, but my knowledge of them is shaky (basically just know the definitions), and most discussions of this theorem that I've found have been handwavy to say the least.

For what it's worth, I would still like to figure this problem out for myself (I'm not asking for a full solution), but I need to understand more about what the BT is actually stating before I can try to prove it.

The theorem is a fully rigorous one.

For a signal $$s(t)$$ (that is, a function from $$\mathbb{R}$$ to $$\mathbb{R}$$ or $$\mathbb{C}$$, representing some physical quantity as voltage, current, heat, etc as a function of time) define the real number $$\Delta t$$ as $$\Delta t\equiv \left(\int_{-\infty}^\infty(t-\bar t)^2 \frac{|s(t)|^2}{\left\lVert s\right\rVert^2}\,dt\right)^{1/2}$$ where $$\left\lVert s\right\rVert \equiv \left(\int_{-\infty}^\infty |s(t)|^2\,dt\right)^{1/2} \qquad\text{and} \qquad \bar t \equiv \int_{-\infty}^\infty t\, \frac{|s(t)|^2}{\left\lVert s\right\rVert^2}\,dt.$$ You can interpret $$\bar t$$ as a temporal mean of the signal and $$\Delta t$$ as a scale of its duration (i.e. for how long it is significantly different from zero).

$$\Delta f$$ is defined in exactly the same way, but with with the signal Fourier transform $$\hat s(f)$$ in place of $$s(t)$$. Explictly $$\Delta f\equiv \left(\int_{-\infty}^\infty(f-\bar f)^2 \frac{|\hat s(f)|^2}{\left\lVert \hat s\right\rVert^2}\,d\!f\right)^{1/2}$$ where $$\left\lVert \hat s\right\rVert \equiv \left(\int_{-\infty}^\infty |\hat s(f)|^2\,d\!f\right)^{1/2} \qquad\text{and} \qquad \bar f \equiv \int_{-\infty}^\infty f\, \frac{|\hat s(f)|^2}{\left\lVert \hat s\right\rVert^2}\,dt.$$

Proving the BT is a matter of tedious calculations together with some properties of the Fourier transform, nothing else, and can be found in any textbook on signal processing. Note that this theorem is not related to operators or any machinery of QM. It is only a property of the Fourier transform.

The connection to QM comes from considering instead of $$s(t)$$ a function $$\psi(x)$$, the wavefunction. This has a completely different physical meaning but mathematically they are both just functions from the reals to the complexes. So, in the same way as for the signal, you can define for the wavefunction $$\Delta x$$ and $$\Delta k$$ in place of $$\Delta t$$ and $$\Delta f$$. From De Broglie relation $$p=\hbar k$$ you then get the HUP. The commutation relation is a bit subtler, but boils down to the fact that multiplying the Fourier transform $$\hat\psi(k)$$ by $$k$$ is "equivalent" to taking the derivative of the wavefunction $$\psi(x)$$.

I know that I was very succinct, but the complete story would get too long. You can find a beautiful discussion of all of this on the first chapter of Griffith's book, especially paragraph 1.6. That is were I learned it first.

• I'm giving this answer the bounty because I think it's the most concise and self-contained. My only complaint is that it doesn't give any hints about how to actually prove the BT, other than that it's tedious, which I assumed to be the case. Otherwise, thank you for a well-written response. – Alex Feb 22 at 0:49

The bandwidth theorem was first fully understood in an all-time fantastic paper: Gabor's Theory of Communication (1946). I'd recommend checking it out. Not the easiest read if you're new to Fourier analysis, but it's really nice, and Part 2 even gives a detailed analysis of human hearing and how it relates.

The bandwidth theorem is a statement about (Fourier) signal analysis. Forget QM for now. Let's just talk about any function $$\psi(x)$$. Could be complex or real valued.

(Note: I'd call this function $$f(x)$$, but then $$f$$ could be confused for frequency. In signal analysis they usually talk about a real-valued signal $$s(t)$$ which is a function of time, and its spectrum $$S(f)$$ which is complex.)

Fourier analysis says that every function $$\psi(x)$$ has a unique frequency spectrum $$\phi(f)$$.

The meaning of this spectrum is that $$\psi(x)$$ can be written as a superposition of the complex functions $$e^{2\pi i f x}$$ (in other words, as a superposition of sines and cosines of frequency $$f$$). The spectrum $$\phi(f)$$, also called the "Fourier Transform", tells you how much of each sinusoidal component you need to use.

It turns out that "narrow" functions have a "broad" spectrum, and "broad" functions have a "narrow" spectrum:

• The spectrum of a localized spike (Dirac delta function) contains all frequencies equally.
• The spectrum of a perfect sinusoidal wave (totally spread out through space) has just a single frequency.
• The spectrum of a Gaussian distribution with width $$\sigma$$ is a Gaussian distribution with width $$1/\sigma$$.

The bandwidth theorem makes this observation quantitative: The product of the width $$\sigma_x$$ of a function times the bandwidth $$\sigma_f$$ of its spectrum obeys $$\sigma_x \, \sigma_f \geq \frac{1}{2}$$.

The width and bandwidth of the function and spectrum are simply defined as $$\sqrt{2\pi}$$ times their standard deviations (i.e. times their root-mean-square deviations from the mean): $${\sigma_x }^2 = 2\pi \int (x - \bar{x})^2 \, |\psi(x)|^2 \,dx \qquad \textrm{and} \qquad {\sigma_f }^2 = 2\pi \int (f - \bar{f})^2 \, |\phi(f)|^2 \,df$$ where $$\bar{x}$$ and $$\bar{f}$$ are the mean (expected) values of the position and frequency. This assumes $$\psi(x)$$ is normalized, if not then both sides get divided by a normalization factor. The factor of $$2\pi$$ is conventional and relates to information content, as explained by Gabor.

If you change notation so that $$\sigma_x = \Delta x$$ and change $$x$$ to a time variable $$t$$, nothing at all changes, and you can write the theorem as $$\Delta t \, \Delta f \geq \frac{1}{2}$$. This is a property of functions, not of energies, quantum mechanics, momentum, or anything else.

This is related to (but not caused by) the fact that $$[x,\frac{\partial}{\partial x}]=-1$$ (to check this, remember to apply a test function). The relation to quantum operators is obtained by simply plugging in $$\hat{x} = x$$ and $$\hat{p} = -i\hbar \frac{\partial}{\partial x}$$. The role of QM in obtaining the uncertainty principle is usually overblown: all the interesting generalized uncertainty relations boil down to the bandwidth theorem. For finite-dimensional observables, there's always a state with zero uncertainty in one operator and finite uncertainty in the other. In other words, the reason for the Heisenberg uncertainty principle in QM is not some magic of operators $$-$$ it is simply that the state is represented by a wavefunction, and any function can either have a well-defined position, or a well-defined frequency, but not both (spatial frequency corresponds to momentum in QM). If anyone tells you taking $$\hbar \to 0$$ makes the uncertainty principle "go away in the classical limit", don't listen.

An amazing consequence of this theorem is that, if allowed to communicate using a finite bandwidth for a finite amount of time, you can only transmit a finite amount of information... and Gabor's analysis tells you how much information that is.

# Where $$t$$ and $$f$$ enter

You are correct: $$t$$ in quantum mechanics is not clearly an observable and $$f$$ is mostly intelligible only by the Einstein relation that $$E = h f$$ as the observable $$\hat H/h,$$ which is somewhat unsatisfying in trying to derive an uncertainty principle between them. This is fundamentally a Fourier-space argument about a time signal but wavefunctions are not time signals, they are distributions that may evolve over time.

As you might expect in terms of teachers who pay considerable attention to detail, Griffiths's QM textbook stresses this point considerably in section 3.5.3 where he derives this new point, saying,

My purpose now is to derive the energy-time uncertainty principle, and in the course of that derivation to persuade you that it is really an altogether different beast, whose superficial resemblance to the position-momentum uncertainty principle is actually quite misleading.

Indeed the general uncertainty principle that you have already come to contains the thrust of his derivation, which is that if the system is going to be observed to change, this change must be witnessed by some observable $$\hat A$$. It changes at a rate $$\frac{d\langle A\rangle}{dt}$$ and to be sure of the change you want it to have changed by an amount proportional to $$\Delta A$$, which motivates a definition of a characteristic time length over which the system has changed,$$\Delta t = \frac{\Delta A}{d\langle A\rangle/dt}.$$

Once you have this definition you can directly see the application to your case because by the chain rule, $$\frac{d\langle A \rangle}{dt} = \langle \partial_t \psi |\hat A|\psi\rangle + \langle \psi |\partial_t \hat A|\psi\rangle + \langle \psi |\hat A|\partial_t \psi\rangle.$$ Since $$i\hbar |\partial_t \psi\rangle = \hat H |\psi\rangle$$ and we generally would want $$\partial_t \hat A = 0$$ so that the thing we're measuring to witness the change is not itself changing with time, we are left with $$i\hbar~\frac{d\langle A \rangle}{dt} = \langle A H - H A\rangle.$$ I like to call this the "ah-ha" relation because I mostly just tutor individual students and occasionally they groan, but they remember it. It's terribly useful especially when one gets a little further and starts to do all of quantum mechanics in this way, the so-called "Heisenberg picture" of quantum mechanics. But anyway, this is just your commutator above, so one rather directly has $$\Delta E~\Delta A \ge \frac12 ~\hbar~ \left| \frac{d\langle A\rangle}{dt} \right|,$$ and with $$\Delta f = \Delta E/\hbar$$ one then has the desired relation $$\Delta f ~\Delta t \ge 1/2.$$ It says that wavefunctions with very precisely defined energies and therefore very precise frequencies cannot change in any observable way, except as the proof indicates above -- they can be "seen to change" by observables that themselves are not constant in time.

With that said there is also a general Fourier relation between $$t$$ and $$f$$ for signals just like there is between $$x$$ and $$1/\lambda$$ for spatial signals: as you squash a signal into a smaller and smaller time/space "domain", you necessarily make it harder to resolve anything but the highest frequencies as those can still have many repetitions across the domain. As a result the distribution in the conjugate Fourier space widens whenever you do anything to try and narrow the distribution in the base space, and vice versa -- if you want a signal that has a very well-defined wavelength then it needs to consist of many, many repetitions of that wavelength, the more the better; and so it needs to occupy a very large amount of your base space. You can even view the $$x/p$$ relation and the $$t/E$$ relation as special relativistic analogues of each other. But in nonrelativistic quantum mechanics this relation persists -- but it is not meaning quite the same thing in the same way.

# The canonical commutation relation

In many contexts we would say that $$[x, p]=i\hbar$$ is just a fundamental axiom of the theory. What it means is that $$p$$ as an operator takes the function of $$-i\hbar\partial_x$$, since $$[x, \partial_x] = -1.$$ In turn this is a deep way of trying to induce the de Broglie relation that $$p = h/\lambda.$$

That is made more clear when one realizes that one is dealing with Fourier transforms of probability distributions over space. So one starts from some $$\psi(x)$$ and constructs its Fourier transform, $$\psi[K] = \int_{-\infty}^{\infty} e^{-2\pi i~K~x}~dx~\psi(x).$$Note that when we use this $$2\pi i$$ factor in the Fourier transform we're automatically normalized, but we do have to remember that $$K = 1/\lambda$$ (as opposed to how physicists normally think about $$K$$-space which is that $$k=2\pi/\lambda$$). Normalization just means that: $$\int_{-\infty}^\infty dK~ \psi^*[K]~\psi[K]=\iint_{\mathbb R^2}dx~dx'~\psi^*(x')~\psi(x)~\delta(x-x')=1,$$because $$\int_{\mathbb R} dK~e^{-2\pi i ~K~u} = \delta(u),$$ where $$\delta$$ is Dirac's delta-function.

But when we want to perform the multiply-by-$$K$$ operation in Fourier space that now has a direct interpretation in position space because we can just write it using the inverse Fourier transform as,$$\hat K \psi(x) = \int_{-\infty}^{\infty}~e^{2\pi i~K~x}~dK~K~\int_{-\infty}^{\infty} e^{-2\pi i~K~x'} ~dx'~\psi(x'),$$ and one can simply rewrite that internal $$K$$ as an operation acting upon the exponential, which is the only thing that depends on $$x$$, $$\hat K \psi(x) = \int_{-\infty}^{\infty}~dK~\frac{1}{2\pi i}\partial_x ~e^{2\pi i~K~x}~dK~\int_{-\infty}^{\infty} e^{-2\pi i~K~x'} ~dx'~\psi(x').$$ Commuting this operation on $$x$$ out of the integral on $$K$$ and using $$\hat p_x = h~\hat K$$ as our de Broglie relation gives the formula $$\hat p_x = -i\hbar \partial_x$$ more directly. So if one looks in Fourier space it is obvious that $$\hat p = h~\hat K$$ as suggested by de Broglie must map to the equivalent $$\hat p = -i\hbar \nabla.$$