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

  • #1
blue_leaf77
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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.
 

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  • #2
DrClaude
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Usually, a wave packet is a single-particle wave function.
 
  • #3
jfizzix
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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
blue_leaf77
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A wave function for a single particle can be represented as a sum (or integral) over many plane waves.
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
DrClaude
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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
jfizzix
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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
DrClaude
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Wavepackets are reasonably well-localized (minimum uncertainty in position and momentum) wavefunctions (often Gaussian), travelling either freely or inside a potential well.
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
jfizzix
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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
DrClaude
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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.
 
  • #11
blue_leaf77
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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
vanhees71
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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
 
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  • #13
blue_leaf77
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A plane wave (momentum eigenmode) is NOT representing a physical state, because only square-integrable wave functions do so.
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
DrClaude
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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 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.

I don't know if we can determine the product between zero and infinity in the case of uncertainty relation of plane wave.
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
blue_leaf77
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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
Nugatory
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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 = ? ##).
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
Nugatory
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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".
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 :smile:

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
blue_leaf77
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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
vanhees71
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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
blue_leaf77
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t's good advice to warn beginners of quantum theory against to look too early on the socalled "philsophical foundations"
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.
 
  • #21
Nugatory
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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.
I expect that that warning was directed at me, not you :smile:
 
  • #22
vanhees71
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I was referring to Nugatories posting saying something about the socalled "philosophical foundations of quantum theory". Those mutterings however, usually tend to confuse the beginner in quantum theory more than they help them. It's far more important to learn about the physical foundations first. Of course, there's no other way to express those foundations without a certain degree of learning about the mathematics too. Also here, I recommend to first read good physicists' quantum-theory books. My favorites are

J. J. Sakurai, Modern Quantum Mechanics (the >=2nd edition is somewhat better than the 1st, because it contains some gems the 1st edition is missing); this is a great book to begin with. It uses the Dirac formalism from the very beginning without overemphesizing wave mechanics too much. It lacks somewhat the applications. E.g., the hydrogen atom is somewhat brief.

L. Ballentine, Quantum Mechanics, A modern development; delves somewhat deeper into the modern mathematical tools making the physicists' hand-waving arguments a bit more rigorous but still avoids to be too rigorous. It also provides a very clear exposition of the very important symmetry principles for non-relativistic quantum theory. Last but not least it uses the minimal statistical interpretation througout, avoiding to much philosophical balast.

S. Weinberg, Lectures on Quantum Mechanics; as all textbooks by Weinberg, also this one provides a lot of deep insights; it's brief in the foundations and thus should be read by the somewhat more experienced reader

Landau/Lifshitz vol. III: a more traditional approach, based on wave mechanics; the biggest advantage against more modern books is the huge collection of applications in atomic and molecular physics.

For the "interpretational/philosophical issues" I recommend the no-nonsense book

A. Peres, Quantum Theory: Concepts and Applications
 
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  • #23
blue_leaf77
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Thanks vanhees71, that's quite a describing review. Actually I'm also starting with Sakurai's, it's indeed a good book even though now I'm having a hard time with how the book explains the Clebsch-Gordan coefficients diagram, that's why I switch to Griffith's and Ballentine's only on that part.
It lacks somewhat the applications. E.g., the hydrogen atom is somewhat brief.
This is indeed rather ironic though, Sakurai said himself in the foreword section that the motivation behind him writing this book was because there were many physicists who were good theoretician but not very capable of dealing with more applied sides of QM. And it seems like there is no discussion on the hyperfine splitting phenomena, given relatively long pages on the chapter of angular momentum alone.
 
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  • #24
blue_leaf77
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I'm sorry for bringing up this topic again, but I have found something I couldn't understand in my textbook. It's written there that the issue of identical particles becomes less important either when we have large number of identical particles or when the particles are very far away apart. I can understand the latter but not the former. I ask this question here because in my original post I wrote that when there are many electrons we might have to take their identity into account, which seems to counteract that statement in my textbook.
 
  • #25
vanhees71
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The former is wrong. It's the particle density, which makes the difference. In quantum statistics you have the phase-space distribution function in thermal equilibrium given by Fermi-Dirac or Bose-Einstein statistics
$$f(E)=\frac{1}{\exp[\beta(E-\mu)] \pm 1}.$$
This becomes the classical Maxwell-Boltzmann distribution in cases, where you can neglect the ##\pm 1##, i.e., if for typical energies (given by the average energy, which is of order ##T##) the exponential is much larger than 1, i.e., the phase-space density much less than 1.
 
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