# Bandwidth Theorem

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1. Apr 19, 2015

### bananabandana

1. The problem statement, all variables and given/known data
Consider a propagating wavepacket with initial length $L_{0}$. Use the bandwidth theorem to show that the minimum range of angular frequencies present in the wavepacket is approximately:
$$\Delta{\omega}\approx \frac{v_{g}}{L_{0}}$$

2. Relevant equations
Bandwidth theorem:
$$\Delta f \Delta t > 1$$

3. The attempt at a solution
Use the following approximation for the group velocity $v_{g}$
$$v_{g}=\frac{\Delta \omega}{\Delta k}$$
Using bandwidth theorem:
$$\Delta \omega \Delta t > 2 \pi$$
$$\Delta t =\frac{\Delta v_{g}}{\Delta l}$$
$$\therefore \Delta \omega > \frac{2\pi \Delta v_{g}}{\Delta L}$$
If $\Delta v_{g}$ is small: $$\Delta v_{g} \approx v_{g}, \Delta L \approx L_{0}$$
Therefore:
$$\Delta \omega > \frac{2 \pi v_{g} }{L_{0}}$$
So
$$\Delta \omega \approx \frac{v_{g}}{L_{0}}$$
But this seems a bit of a fudge? Can anyone explain how I might get a more kosher version? Thanks!!

*EDIT: No this is completely wrong - sorry - the limit of $\Delta L$ as $\Delta v_{g} \rightarrow 0$ is zero! How do I fix this? Thanks :)

2. Apr 20, 2015

### collinsmark

Hello bananabandana,

I must ask, did you get the bandwidth theorem, $\Delta f \Delta t \gt 1$ straight from you textbook/coursework, or from some other source?

I ask because I've also seen this uncertainty principle (called the "bandwidth theorem" here) expressed as
$$\Delta \omega \Delta t \approx 1$$
elsewhere. Notice that the $2 \pi$ is folded into it already, the way I've seen it.

Admittedly this definition is an ill-defined version with the "delta" (what is specifically meant by "delta"?) and the approximation symbol. You can make it more precise by working with standard deviations (instead of deltas), and an inequality [and there might be a factor of 2 that fits into that one], but I'm getting off topic. Whatever the case, the above way is a way I have seen it before.

If you use the $\Delta \omega \Delta t \approx 1$ version, you can get the given answer after making some appropriate substitutions. First, you might try to work toward an equivalent form of the relationship in terms of $\Delta x$ and $\Delta k$. You can do this in part by noting that velocity $v = \frac{\Delta x}{\Delta t}$, along with some other substitutions. Once you get there, you can invoke your $v_g = \frac{\Delta \omega}{\Delta k}$ as a near final step. ($\Delta x$ is the same thing as $L_0$ by the way.)

Last edited: Apr 20, 2015
3. Apr 20, 2015

### collinsmark

Okay, I just did the math using Fourier analysis and calculating standard deviations, and, I found that in a more precise form,
$$\sigma_\omega \sigma_t \geq 1$$

And that's assuming I did my math correctly. Here, $\sigma_\omega$ is the standard deviation of the bandwidth in terms of the angular frequency $\omega$, and $\sigma_t$ is the standard deviation of the time based wavepacket. Standard deviations can be precisely calculated and are are less ambiguous than "deltas."

Note that the "$\geq$" represents a hard limit. It is mathematically impossible for the product of the two standard deviations to be less than 1 (the value of '1' assumes I did my math correctly, btw). The special case where they are equal is the situation where the wavepacket's envelope has a Gaussian function shape (like a bell curve). All other wave shapes will have an uncertainty greater than that.

If that's right, you can translate that using $\sigma_\omega = 2 \pi \sigma_f$ which produces $\sigma_f \sigma_t \geq \frac{1}{2 \pi}$ if you wanted to, but I'd just stick with angular frequency $\omega$ for this problem.

Anyhoo, if you define the "Deltas" to be standard deviations, then go with $\Delta \omega \Delta t \geq 1$ and that should give you the given answer.

[Edit: And there's no factor of 2 in there after all (assuming I did my math correctly).]

Last edited: Apr 20, 2015
4. Apr 23, 2015

### bananabandana

Firstly, thanks for all the work, and very sorry for the slow reply on my part. Yes, this is a direct quote from my course notes. It's very strange. This source suggests that the answer is $\sigma_{t}\sigma_{f} = \frac{1}{4\pi}$. As does this one, from University of Toronto. They are using a full $\Delta$ though, so perhaps that makes a difference? However, nowhere can I find the $\Delta t \Delta f \ge 1$ which is in my (handed out) notes though... little bit worrying as I have an exam on Tuesday! I think I will e-mail the lecturer... It was an aside in the course (we have not covered Fourier transforms) but it has come up on previous papers.

However,I am still confused as to why $\Delta x = L_{0}$ though - how come the change in the position of the wavepacket always equals $L_{0}$ ?? I get that $v=\frac{L_{0}}{T}$, if $T$ is the period.. but I don't think that's what you mean - could you explain a bit more? (sorry!)

They also then went on to ask a question, (please see attached) about group velocity dispersion. I thought this was easy to solve - we use the result we are supposed to have just found to get:
$$\Delta v_{g}= \bigg| \frac{dv_{g}}{d\omega}\bigg| \frac{v_{g}}{L_{0}}$$
$$\Delta L=\Delta v_{g}\Delta t =\bigg| \frac{dv_{g}}{d\omega} \bigg|\frac{x}{L_{0}}$$
if $L >> L_{0} \implies \Delta L \approx L$, and result follows.

But now I'm unsure if I cut something out - ?

Thanks!

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5. Apr 24, 2015

### collinsmark

These sources you quoted do seem to include that factor of 2 that I suspected in my first post. However, when I did the math (more on this in a second) that factor of 2 didn't come about.

It's noteworthy that in quantum mechanics (QM), this factor of 2 of is present. The relationship between position and momentum (naming a common example of conjugate variables) is $\sigma_x \sigma_p \geq \frac{\hbar}{2}$. But if I'm not mistaken, the source of this factor of 2 in the denominator comes from that fact that probability distribution functions in QM involve the magnitude squared of the wavefunction -- not the distribution of the wavefunction directly.

However when I did the math for an arbitrary signal $f(t)$ and $F(\omega)$, I was not using the distribution function of the magnitude squared, but rather I assumed that $f(t)$ and $F(\omega)$ (after renormalizing) were distribution functions in their own right, and found the standard deviations of each, directly.

I presume that this treatment of the magnitude squared of the function versus the direct function has something to do with the discrepancy.

Perhaps in a future post I'll show my math regarding my $\sigma_t \sigma_\omega \geq 1$ (same thing as saying $\sigma_t \sigma_f \geq \frac{1}{2 \pi}$), that way if I made a mistake it can be corrected.

[Edit: good luck on your exam Tuesday, by the way. Sorry, but I'm not going to be able to help you on a firm definition of the $\Delta t \Delta \omega$ relationship by the way, since deltas are rather ambiguously defined. It's best if you find it in your textbook or coursework.]

Actually, that was just an assumption I made. If a wavepacket is traveling at some constant group velocity $v_g = \frac{dx}{dt}$ and the "time" it takes for the wavepacket to pass is defined as $\Delta t$ then the length of the wavepacket is $\Delta x = v_g \Delta t$. I assumed that was what was meant in the problem statement as $L_0$. I thought it was pretty straightforward, but if your textbook has some other definition of $L_0$ you'll have to look there. It was just a simple assumption on my part.

Last edited: Apr 24, 2015
6. Apr 24, 2015

### bananabandana

Okay, so long as that's all that it is, that's fine. I understand what you mean by the deltas and also find it really annoying to have all of them flying around with none of them being very precisely defined! To be honest, it's quite odd that it's in the course at all given no Fourier - i.e it was essentially "Oh, just learn this result!" :P Ah, well! Thanks for the help, is much appreciated :)