# "Wavepacket of electrons" or "wavepacket of electron"?

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1. Mar 25, 2015

### blue_leaf77

For simplicity let's assume Gaussian shape. I just want to verify my understanding about wavepacket. As the thread title says, when we have a wavepacket, is it understood as the superposition of many plane waves corresponding to many free electrons, or as the wavefunction of only one electron, only that this electron was somehow produced with momentum distribution of nonzero uncertainty ($\Delta p \neq 0$)?
I prefer the latter because for the former, we must take indistinguishability nature of electrons into account. In this case each electron must be assigned with its own degree of freedom $\mathbf{r_n}$ and the total wavefunction should look much more complicated.

2. Mar 25, 2015

### Staff: Mentor

Usually, a wave packet is a single-particle wave function.

3. Mar 25, 2015

### jfizzix

A wave function for a single particle can be represented as a sum (or integral) over many plane waves. The amplitude associated to each plane wave is the amplitude for that particle to have a particular momentum value.

4. Mar 25, 2015

### blue_leaf77

I see what you mean is $\langle\mathbf{r}|\psi\rangle = \int \langle\mathbf{r}|\mathbf{p}\rangle \langle\mathbf{p}|\psi\rangle d\mathbf{p}^3$, right?
But this makes me thing that the term wavepacket is a synonim of wavefunction, but if it's true why people define the wavepacket at all?

5. Mar 25, 2015

### Staff: Mentor

The wave function is what describes a quantum system. A wave packet is a particular state where the particle is localized on a scale smaller that than what would be normally expected from the potential. It is by looking at the wave function that you can see if you have a wave packet or not.

6. Mar 25, 2015

### jfizzix

I not sure if there is an ironclad distinction between wavepackets and wavefunctions, but the usage as I see it is:

Wavepackets are reasonably well-localized (minimum uncertainty in position and momentum) wavefunctions (often Gaussian), travelling either freely or inside a potential well.

When we want to describe how an otherwise well-defined classical particle might behave in some quantum potential, we see how a wavepacket wavefunction behaves.

As an example, you can have the wavefunction of an electron in a hydrogen atom be a wavepacket (i.e., a well-localized blob of probability orbiting the nucleus in a nearly classical fashion). However, expressing this wavepacket in terms of the eigenfunctions of the hydrogen atom shows us that such wavepackets are superpositions of a very large number of orbitals.

So, people use wavepackets because they are convenient to work with when trying to understand the limit in which quantum systems begin to behave classically.

Long story short, all wavepackets are wavefunctions, but not all wavefunctions are wavepakcets.

7. Mar 25, 2015

### Staff: Mentor

If you have minimum uncertainty, then you have a coherent state which must be Gaussian. A general wave packet does not have that characteristic.

8. Mar 25, 2015

### jfizzix

yes, if they are truly of minimum uncertainty, then they must be Gaussian (a coherent state).

What other shapes of wavepackets are used in physics?

9. Mar 25, 2015

### Staff: Mentor

I don't know of particular shapes, but the physics can dictate that the system ends up in a localized wave packet that is not Gaussian. For instance, in pump-probe spectroscopy, you can have an initial state that is excited, then evolves as a wave packet, before being probed. That initial state is not always Gaussian.

10. Mar 25, 2015

### jfizzix

Good point.

11. Mar 25, 2015

### blue_leaf77

Ok I think I kind of get the idea behind this. But I agree that there is a very transparent curtain between wavepacket and wavefunction. Wavefunction is the space/momentum representation (at least as far as I know) of a general ket vector in a ket vector space, I can actually also think that a wavepacket be a general ket in ket space. Only that in position and momentum space, it's more localized. Well it seems to me that the difference is just a matter of "size".

12. Mar 26, 2015

### vanhees71

It is often not clearly stated in undergrad introductory quantum theory lectures, but it's a very crucial point! A plane wave (momentum eigenmode) is NOT representing a physical state, because only square-integrable wave functions do so. This is clear from Born's rule, according to which
$$P_{\psi}(t,\vec{x})=|\psi(t,\vec{x})|^2$$
is the probability density for the position of a single particle, provided that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(t,\vec{x}|^2=1.$$
Consequently, since the plane waves do not belong to the Hilbert space of square-integrable functions and thus do not represent a quantum state. They belong to the dual space of the domain of the position and momentum operators.

For a good introduction into the modern treatment of these somewhat delicate issues in terms of the rigged-Hilbert space formalism, see

L. Ballentine, Quantum Mechanics, Addison Wesley

or the various freely from the arXiv available reviews by de la Madrid, e.g.,

http://arxiv.org/abs/quant-ph/0109154v2

13. Mar 26, 2015

### blue_leaf77

Thanks for pointing this out vanhees71, and I have also been aware of this issue with plane wave. To me a wavepacket propagating in free space is a single traveling electron whose momentum is not well defined, and in this sense wavepackets satisfy the uncertainty relation between momentum and space. I don't know if we can determine the product between zero and infinity in the case of uncertainty relation of plane wave.

14. Mar 26, 2015

### Staff: Mentor

I wouldn't say that the momentum is not well defined, but rather that there is a spread in momentum, the same way as there is a spread in space.

When you take the Fourier transform of a single plane wave, you have a Dirac delta in momentum, and hence the wave function is constant over all space.

15. Mar 26, 2015

### blue_leaf77

But the spread of probability means when a measurement is performed, one will not get the same values for all the time assuming the state is brought back to its original after each measurement. That means it's not well defined, I think it's just a matter of the term usage.
What I meant in my last sentence is that I am not sure if plane waves satisfy uncertainty relation for coordinate-momentum ($\Delta p\Delta x = 0\times \infty = ?$).

16. Mar 26, 2015

### Staff: Mentor

You can save yourself some grief by considering an ensemble of identically prepared systems. We make one position or one momentum measurement on each member of the ensemble and never look at that member again, so we don't have to worry about restoring the system to its original state. Each measurement yields a number as a result, never infinity, so we can make sensible statements about the distribution of momentum results and position results, without having to deal with the $0\times \infty = ?$ mess.

Of course we cannot prepare a system in a perfect plane wave, but this procedure allows us to assign a meaning to the uncertainty relation as applied to plane waves.

17. Mar 26, 2015

### Staff: Mentor

You're making this question harder than it has to be... although to be fair, the philosophical foundations of quantum mechanics are such that if you if aren't doing this from time to time you aren't thinking about it deeply enough keep your curiosity under superhuman restraint

Any physical system is described by a solution to the time-dependent Schrodinger equation. We call these solutions "wavefunctions", especially when we're working in the position and momentum bases. A subset of these solutions, whose members we call "wave packets", describe more or less localized single particles moving in a potential.

18. Mar 26, 2015

### blue_leaf77

Maybe I should have written a wavepacket to be a general ket represented in position/momentum space whose spread is more localized among the position/momentum-represented kets. Other than that, I don't see anything else that made the question harder.

19. Mar 27, 2015

### vanhees71

It's good advice to warn beginners of quantum theory against to look too early on the socalled "philsophical foundations". There are no philosophical foundations necessary. First one should understand the science. After retirement scientists might muse about "philosophical foundations" although it's a waste of their true expertise ;-)). SCNR.

20. Mar 27, 2015

### blue_leaf77

I appreciate this advice of yours, but would you please explicitly mention which part was too philosophical? I guess about the ket thing?
By the way I have actually taken and passed QM course few years ago, but I just realized that the taught QM in my undergrad was slightly different than you guys, it was too focused on the math and rarely discussed the physical arguments which should take more priority. Actually the math itself was still slightly low leveled than how it should have been, that's why now I'm trying to compensate all those gaps myself.