# Why doesn't Gaussian wavepacket broadening in position mean there will be a shortening in momentum?

Many sources that say in free broadening of a Gaussian wavepacket, the momentum uncertainty (I think defined in terms of the range of 'significant' momentum amplitudes) is time invariant even as the Gaussian wavepackets broadens in position.

I'm trying to reconcile how this is consistent with the uncertainty principle.

My idea is that the uncertainty principle only loosely forces the 'bulk area' of the momentum distribution to have a spread that is inversely proportional to the 'bulk area' of the position distriubtion, but in special cases like this maybe the 'bulk area' is rearranged in such a way that the amplitudes of the momenta inside the original 'significant amplitude region' gets much larger, but not much wider?

Any thoughts?

Extras:

The reason I had this thinking about 'bulk areas' was because I was wondering how exactly the fourier transform of the position wavefunction (i.e. giving wavefunction in momentum space) is changed after this wavepacket broadening. Obviously its standard deviation hasn't changed, but I would expect there to be some difference since I though the FT doesn't lose any information from its input and its input has definitely changed. Otherwise you wouldn't be able to get back to this 'broadened' position wavefunction from the momentum wavefunction using an inverse FT.

• Why do you believe uncertainty is a conserved quantity? Sep 18, 2020 at 2:46

the momentum uncertainty (I think defined in terms of the range of 'significant' momentum amplitudes)

I'm not sure what you mean by this "range of significant momentum amplitudes". Uncertainty has an exact definition in QM. If you have a state vector $$|\psi\rangle$$ then the average of some observable $$A$$ is $$\langle A\rangle=\langle\psi|A|\psi\rangle$$, and the uncertainty is just the square root of the variance:

$$(\Delta A)^2=\left\langle \left(A-\langle A\rangle\right)^2\right\rangle$$

My idea is that the uncertainty principle only loosely forces the 'bulk area' of the momentum distribution to have a spread that is inversely proportional to the 'bulk area' of the position distriubtion, but in special cases like this maybe the 'bulk area' is rearranged in such a way that the amplitudes of the momenta inside the original 'significant amplitude region' gets much larger, but not much wider?

Here too, I'm not sure how useful this "bulk area" idea is. You can directly calculate how the uncertainties of position and momentum evolve over time for the free Gaussian wave packet evolving according to the Schrodinger equation. Then you can show that the position uncertainty increases while the momentum uncertainty remains constant.

As pointed out in other answers, this doesn't violate the uncertainty principle. A violation would consist of decreasing uncertainties such that $$\Delta x\Delta p\geq\hbar/2\pi$$ is no longer valid.

To add some intuition here, the free particle Hamiltonian has no position dependence. Therefore, the propagator commutes with the momentum operator, and so any expectation values, variances, etc. for momentum measurements must be time independent.

The reason I had this thinking about 'bulk areas' was because I was wondering how exactly the fourier transform of the position wavefunction (i.e. giving wavefunction in momentum space) is changed after this wavepacket broadening. Obviously its standard deviation hasn't changed, but I would expect there to be some difference since I though the FT doesn't lose any information from its input and its input has definitely changed. Otherwise you wouldn't be able to get back to this 'broadened' position wavefunction from the momentum wavefunction using an inverse FT.

We need to be careful here in distinguishing the wave functions (in terms of position or momentum) to the probability distributions, which involves the squares of the wave functions. Indeed, the wave functions are related by Fourier Transforms, but the probability amplitudes are not.

Assuming an initial wave packet of the form $$\Psi(x,0)=Ae^{-ax^2}$$ one can determine $$\Psi(x,t)$$ and $$\Phi(p,t)$$ for the free particle. I will not go through derivations, but this should be covered in various QM texts, or online. For reference here I was using Griffith's QM book section 2.4 and problem 2.22. Below are the real and imaginary parts of both $$\Psi(x,t)$$ and $$\Phi(p,t)$$ evolving over time (I have set all constants to $$1$$, so the actual numbers are not important):

We can see that both $$\Psi$$ and $$\Phi$$ are changing, and at each time they are related through a Fourier Transform. However, this idea does not carry over into $$|\Psi|^2$$ and $$|\Phi|^2$$; they are not related via direct Fourier transform:

Since the probabilities determine the statistics, we can see that the uncertainty doesn't change for momentum, but it gets larger for position. The uncertainty principle is still followed.

• Ah okay I thought it could be something like this. So am I correct in saying the Uncerainty Principle does not mean that the uncertainties of x and p are inversely proportional, but their bounds? It's just confusing since some sources online make it seem like an increase in uncertainty of one must definitely cause a decrease in uncertainty of the other, but I guess this isn't the case then? Sep 17, 2020 at 9:10
• And I guess will the 'bulk area' idea, I was wondering how exactly the fourier transform of the position wavefunction (i.e. giving wavefunction in momentum space) is changed after this wavepacket broadening. Obviously its standard deviation hasn't changed, but I would expect there to be some difference since I though the FT doesn't lose any information from its input and its input has definitely changed. Otherwise you wouldn't be able to get back to this 'broadened' position wavefunction from the momentum wavefunction using an inverse FT. Sep 17, 2020 at 9:14
• @AlexGower $\Delta x$ and $\Delta p$ do not have to be inversely proportional. The only time you would know anything for sure (before just actually doing the calculations) is if you knew $\Delta x\Delta p=\hbar/2\pi$ and then you made a new state with a smaller $\Delta x$ or smaller $\Delta p$. Then you would know that the other value had to increase. But even then, it wouldn't have to necessarily be an inversely proportional increase. As for the Fourier Transform stuff, I will edit my answer accordingly when I have time. Sep 17, 2020 at 13:47
• @AlexGower I have edited the answer accordingly. Sep 17, 2020 at 15:36
• This is a fantastic answer, but there is a little mistake. You say that '$\Phi$ narrows', but I don't think it does. While it does get higher and higher frequency (because $\Psi$ gains support in larger positions), it is still always contained in the enveloppe $\sqrt{|\Phi|^2}$, which, as you noted, does not change. Sep 17, 2020 at 16:19

The uncertainty principle is simply $$\Delta X\Delta P\geq\frac{\hbar}{2}$$ or $$\Delta P\geq\frac{\hbar}{2\Delta X}$$ If you increase $$\Delta X$$, the bound on $$\Delta P$$ is strictly lowered, not raised. Thus, $$\Delta P$$ staying constant as $$\Delta X$$ increases is a perfectly viable situation according to the uncertainty principle.

• Ah okay I thought it could be something like this. So am I correct in saying the Uncerainty Principle does not mean that the uncertainties of x and p are inversely proportional, but their bounds? It's just confusing since some sources online make it seem like an increase in uncertainty of one must definitely cause a decrease in uncertainty of the other, but I guess this isn't the case then? Sep 17, 2020 at 9:10
• @Alex Gower The two are not inversely proportional; rather, the uncertainty principle just says that their products are greater than $\hbar/2$. This means that I could increase both of them arbitrarily, and since the product is still greater than $\hbar/2$, the uncertainty principle, at the least, allows it. Sep 17, 2020 at 14:08
• Thanks. Now my only final query is what happens to the fourier transform of the position wavefunction (giving the position wavefunction) Sep 17, 2020 at 14:12
• @Alex Gower This was covered in the now accepted answer; if you'd like me to show you how to derive exactly what happens, I can do that. Sep 17, 2020 at 19:35

I’ll just say a word on Fourier transforms, which might help clear up your intuition:

Fourier transforms are complex.

This means, for example, when transforming from position space to momentum space, there is a phase associated with each momentum value, in addition to amplitude.

If the momentum phase is uniform, this means the position spread is “transform limited”, i.e. minimum according to uncertainty principle. If, on the other hand, the momentum phase is non-uniform, then the position is “chirped” or worse. Then the position will be further spread out for the same momentum amplitude spectrum.

Edit to directly answer the question:

“Why doesn’t Gaussian waveform broadening in position mean there will be a shortening in momentum?”

Because the broadening in position can, in certain circumstances, be represented by a change only in the phases of the momentum spectrum (like a chirped wave, for example). Then the amplitudes of the momentum spectrum would remain the same.

• This is interesting, are you suggesting something about how 'momentum uncertainty' in the FT Uncertainty principle is related to its spread in phases too? Sep 17, 2020 at 16:56
• @AlexGower yeah, it’s all just different ways to think about the same thing. The way you’ll get something other than the minimum $\Delta X \Delta P$ product is if the phases of one ($X$ or $P$) are non-uniform so that the other is not transform limited. I’d recommend playing around with Fourier transforms of various waves in mathematical software just to get a feel for it. Sep 17, 2020 at 17:36
• Just to double check, when you say in your original answer 'minimum according to the uncertainty principle', you don't mean satisfying the $\Delta x \Delta p = \hbar /2$ equality, you just mean 'a reduced uncertainty in x than if the momentum phase was non-uniform too' right? Sep 18, 2020 at 9:52
• And secondly to double check, you're not suggesting that momentum wavefunctions with non-unform phases are any 'less spread' or 'less uncertain' than ones with uniform phases i.e. you're not trying to suggest that in the Gaussian wavepacket example, the momentum wavefunction does get less uncertain due to having less unfirom phases, you're just saying that in fourier transforms it can be quite common for the envelope of one variable (e.g. position) to broaden while the envelope for the other (e.g. momentum) doesn't but it just has less uniform phases, right? Sep 18, 2020 at 9:58