De Broglie Wavelengths & Electron Orbitals

  • Thread starter SSSUNNN
  • Start date
  • Tags
    De broglie
In summary: It is, but in the dual of the nuclear subspace, so in a rigged Hilbert space, the natural mathematical "environment" of any spectral problem." In summary, the electron can be described by a summation of many different waves according to de Broglie. Each wave corresponds to a certain orbital and the electron is described by their summation because for each orbital there will be only one wave.
  • #1
SSSUNNN
3
0
Hi everyone,

I read that: The electron (according to de broglie) can be described by a summation of many different waves according to

PSI(X,t)= integration (A e^i(kx-wt))

Does each wave correspond to a certain orbital? and if each of them is for an orbital, why the electron is described by their summation? I expected that for each orbital there will be only one wave (not a summation)

Thank you for your help
 
Physics news on Phys.org
  • #2
SSSUNNN said:
Hi everyone,

I read that: The electron (according to de broglie) can be described by a summation of many different waves according to

PSI(X,t)= integration (A e^i(kx-wt))

Does each wave correspond to a certain orbital? and if each of them is for an orbital, why the electron is described by their summation? I expected that for each orbital there will be only one wave (not a summation)

The de Broglie relationship [itex]p = h/\lamba[/itex] associates a definite de Boglie wavelength [itex]\lambda[/itex] with a definite momentum [itex]p[/itex]. Wavenumber [itex]k = 2 \pi/\lambda[/itex] is associate with wavelength [itex]\lambda[/itex], so de Brgolie's relation can be written as [itex]p = \hbar k[/itex]. Consequently, [itex]e^{ikx - \omega t}[/itex] represents a matter-wave with definite momentum, and

[tex]
\psi(x,t) = \int_{-\infty}^{\infty} e^{i(kx-\omega t)} dk
[/tex]

is the summation (with equal weights) of all possible momenta. In quantum theory, [itex]\psi[/itex] represents the state of a free particle that is at the definite position [itex]x[/itex]. By Heisenberg's uncertainty principle, a particle with definite position has completely uncertain momentum. Thus, the summation of all possible momenta.

Regards,
George
 
  • #3
There are no orbitals for a free particle.

Daniel.
 
  • #4
SSSUNNN said:
Does each wave correspond to a certain orbital? and if each of them is for an orbital, why the electron is described by their summation? I expected that for each orbital there will be only one wave (not a summation)

The big picture: for a free electron (i.e. in the presence of no external potential) the wave function is given by the integral because it is a sum of different momentum modes. (Alternatively you can Fourier transform and say it is a sum of position modes.) Intuitively this means that an plane-wave electrons can have a range of momenta (positions)... then when you actually check to observe the momentum (position) of the electron, the wavefunction collapses into a particular momentum (position) eigenstate. The fact that momentum eigenstates are different from position eigenstates is Heisenberg's position-momentum uncertainty principle.

The orbitals that you're thinking of (the ones that are talked about in chemistry, among other places) arise from the wavefunction of an electron in a central potential--that caused by the nucleus of a Hydrogen atom, for example. In this case, Schrodinger's equation has an extra potential term, and one has to do some mathematical tricks to find closed-form solutions for the position-space wavefunction. It turns out that these position-space wavefunctions take the form of the orbitals that you're thinking of. You can check out the details in any standard introductory QM book.

Hope that helps,
Flip
 
  • #5
But isn't Aexp[kx-wt] a solution of the schrödinger equation? So the wave function must not be given by an integral, correct?
 
  • #6
Yes, true, but quantum states, in agreement with the I-st principle, must be described by normalizable (wrt the scalar product on the Hilbert space [itex] \mathbb{L}_{2}\left(\mathbb{R}\right) [/itex] ) wave function. A plane wave is not normalizable in the Hilbert space aforementioned.

Daniel.
 
  • #7
I do not think that the other integral "formula" is normalizable either.
 
  • #8
It is, but in the dual of the nuclear subspace, so in a rigged Hilbert space, the natural mathematical "environment" of any spectral problem.

Daniel.
 
  • #9
Kruger said:
I do not think that the other integral "formula" is normalizable either.

As a multiple of the Dirac delta function, the "integral formula" (which can be expressed as a formula in the theory of tempered distributions) that I gave is not normalizable. However, going back to the original post, the wave packet

[tex]
\psi(x,t) = \int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)} dk
[/tex]

is normaiizable for appropriate [itex]A(k)[/itex]. In particluar, [itex]A(k)[/itex] can be found such that the wave packet is both normalizable and "almost" a position eigenstate.

Regards,
George
 
  • #10
dextercioby said:
There are no orbitals for a free particle.

Daniel.

Are you sure? really sure?


Seratend.
 
  • #11
dextercioby said:
... the natural mathematical "environment" of any spectral problem.Daniel.

Not according to Reed and Simon. :smile: In the endnotes to one of the chapters of Vol. I, they say something like "We recommend the rigged Hilbert space only to those that have a strong emotional attachment to Dirac notation." I'm quoting form memory, so I might have it a bit wrong.

Even though I'm not that familiar with the rigged Hilbert space approach, I kind of like it. There is a nice physicist's synopsis of it in the undergrad/grad book by Anton Capri.

Regards,
George
 
Last edited by a moderator:
  • #12
seratend said:
Are you sure? really sure?


Seratend.

Depends on your acception of orbital.A plane wave can be decomposed in a sum over a product of spherical harmonics and spherical bessel functions, it's true;however, the notion of "orbital" appears only in the context of atomic physics, where one has atomic and molecular orbitals.

Daniel.
 
  • #13
George Jones said:
Not according to Reed and Simon. :smile: In the endnotes to one of the chapters of Vol. I, they say something like "We recommend the rigged Hilbert space only to those that have a strong emotional attachment to Dirac notation." I'm quoting form memory, so I might have it a bit wrong.

Even though I'm not that familiar with the rigged Hilbert space approach, I kind of like it. There is a nice physicist's synopsis of it in the undergrad/grad book by Anton Capri.

Regards,
George

Bogoliubov et al. [1] reccomend using the rigged Hilbert space for any quantum theory (field or nonrelativistic) dealing with unbounded linear operators on separable Hilbert spaces.

Incidentally, the course on QM I've taken back home reccomends the same thing. :smile:

Daniel.

[1]N.N.Bogoliubov et al., "Introduction to Axiomatic Quantum Field Theory", Benjamin/Cummings, NY, 1975.
 
Last edited by a moderator:
  • #14
seratend said:
Daniel said:
Originally Posted by dextercioby
There are no orbitals for a free particle.

Daniel.
Are you sure? really sure?

Seratend.
That would seem to be the definition of a 'free' particle.

In the OP, it seems that the question inferred an electron in an orbital.

An electron has a deBroglie wavelength whether it is free or in an orbital.
 
  • #15
Kruger said:
But isn't Aexp[kx-wt] a solution of the schrödinger equation?

Yes.

Kruger said:
So the wave function must not be given by an integral, correct?

Incorrect. The Schrödinger equation is linear, so the sum of any two solutions is also a solution. So is the sum of three, four,... even a countably infinite number of solutions. So is the integral of an uncountably infinite number of solutions, such as George's integral, which you will find in every QM text that discusses "wave packets."

dextercioby said:
A plane wave is not normalizable in the Hilbert space aforementioned.

Kruger said:
I do not think that the other integral "formula" is normalizable either.

That depends on the particular [itex]A(k)[/itex] used. If [itex]A(k)[/itex] is normalizable, i.e. if

[tex] \int_{-\infty}^{\infty} A^{\star}(k) A(k) dk [/tex]

is finite, then so is [itex]\psi (x)[/itex].
 
  • #16
dextercioby said:
Depends on your acception of orbital.A plane wave can be decomposed in a sum over a product of spherical harmonics and spherical bessel functions, it's true;however, the notion of "orbital" appears only in the context of atomic physics, where one has atomic and molecular orbitals.

Daniel.

As you said *atomic* orbitals and *molecular* orbitals and not orbital alone.

However, we have the choice to describe free particles by "plane" waves or by "spherical" waves that are both eigen vectors of the free hamitonian p^2/2m.


Seratend.
 
  • #17
Well, i'd still not use the word orbital in connection with a free particle. Actually, it's the chemists that use this word a lot, and I'm sure they're workking only with bound systems.

Daniel.
 
  • #18
dextercioby said:
Bogoliubov et al. [1] reccomend using the rigged Hilbert space for any quantum theory (field or nonrelativistic) dealing with unbounded linear operators on separable Hilbert spaces.

[1]N.N.Bogoliubov et al., "Introduction to Axiomatic Quantum Field Theory", Benjamin/Cummings, NY, 1975.

Whenever, I look at [1], I like what I see. I've always meant to buy my own copy of it, but I never manage to get around to doing so.

Regards,
George
 
  • #19
I too don't have a book for myself. I've had to borrow it from the library and needed to return it, of course. I could say i compensate, in a way, with the four volumes of Reed and Simon and Lopuszanski's little book, but of course, the first two are mathematicians and they don't really attack each physical issue (like coherent subspaces and supraselection rules, rigged Hilbert spaces and operators acting on them and other issues), while the Polish guy wrote a physics book too short and unfortunately missing a lotta stuff.

Daniel.
 
  • #20
question on de Broglie wavelength

I have a question on the fundamental physical interpretation of the de Broglie wavelength. The physical situation I am trying to understand is ions trapped in a low temperature environment such as electrons in liquid helium

In QM a particle is correctly described as a superposition of waves which, when summed together, results in a wavepacket.

Question 1: For a free particle this wavepacket is a Gaussian. Is the DB wavelength of the particle the full width at half maximum (FWHM) of the Gaussian?

Question 2: What is the physical interpretation of the DB wavelength? Is it the uncertainty in the position of the particle?

Question 3: As is well known the DB wavelength of a particle is inversely related to temperature, growing larger as the square root of the temperature. Is the uncertainty in a free particle's position growing as the particle gets colder?

What is not clear to me is how the uncertainty in an electron's position is changing as I cool its environment. It seems to me that the uncertainty in its position--I know it is somewhere in the liquid helium, but that's all I know--remains unchanged as i cool the liquid helium. But this seems in contradiction to the idea that the DB wavelength is growing as I cool the electron.

Thanks for the help
 
  • #21
Unfortunately, this does not answer all your questions, because I'm not familiar with your particular situation. I can address only the general QM aspects.

japeth said:
In QM a particle is correctly described as a superposition of waves which, when summed together, results in a wavepacket.

Correct.

Question 1: For a free particle this wavepacket is a Gaussian. Is the DB wavelength of the particle the full width at half maximum (FWHM) of the Gaussian?

No. When you measure the wavelength of the particle, or equivalently its momentum, you get a random value selected according to a Gaussian probability distribution for the momentum. The most likely value for the wavelength is the one that corresponds to the momentum at the centroid of the momentum probability distribution.

Question 2: What is the physical interpretation of the DB wavelength? Is it the uncertainty in the position of the particle?

The de Broglie wavelength determines how the particle will diffract, or behave in other wave-like contexts. It is not the position uncertainty.

The position uncertainty is related to the momentum uncertainty by the Heisenberg uncertainty principle. The momentum uncertainty is in turn related to the wavelength uncertainty by a simple mathematical transformation that follows from de Broglie's formula.
 
  • #22


The de Broglie wavelength determines how the particle will diffract, or behave in other wave-like contexts. It is not the position uncertainty.

Question: Is the DB wavelength related to the scattering cross section of a particle then?

Consider a system of interacting particles like an ideal gas (electrons on liquid helium can be modeled this way). I'd be interested in feedback on the following: does it follow from the above that as the DB wavelength of the electrons gets larger (with decreasing temperature as I cool the liquid helium environment in which they are suspended) their equation of state changes as well?
 

1. What is the de Broglie wavelength?

The de Broglie wavelength is a concept in quantum mechanics that describes the wavelength of matter particles, such as electrons. It is named after French physicist Louis de Broglie, who proposed that all matter particles have both wave-like and particle-like properties.

2. How is the de Broglie wavelength related to the momentum of a particle?

The de Broglie wavelength is inversely proportional to the momentum of a particle. This means that as the momentum of a particle increases, its de Broglie wavelength decreases. This relationship is described by the de Broglie equation: λ = h/mv, where λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity.

3. What is the significance of the de Broglie wavelength in the study of electron orbitals?

The de Broglie wavelength is important in understanding the behavior of electrons in an atom. It helps to explain why electrons can only exist in certain energy levels or orbitals, as their de Broglie wavelengths must fit within the circumference of the orbital. This concept is key in the development of the quantum mechanical model of the atom.

4. Can the de Broglie wavelength of an electron be measured?

Yes, the de Broglie wavelength of an electron can be measured using a technique called electron diffraction. This involves firing a beam of electrons at a crystalline material and measuring the diffraction pattern that is produced. The de Broglie wavelength can then be calculated from the spacing of the diffraction pattern.

5. How does the de Broglie wavelength of an electron change in different energy states?

The de Broglie wavelength of an electron changes as it moves between different energy states. When an electron gains energy, its velocity increases and its de Broglie wavelength decreases. Similarly, when an electron loses energy, its velocity decreases and its de Broglie wavelength increases. This is consistent with the de Broglie equation, as the mass and Planck's constant remain constant.

Similar threads

  • Quantum Interpretations and Foundations
Replies
3
Views
826
  • Quantum Interpretations and Foundations
Replies
6
Views
2K
  • Quantum Interpretations and Foundations
Replies
1
Views
1K
  • Quantum Interpretations and Foundations
Replies
14
Views
2K
  • Quantum Interpretations and Foundations
Replies
28
Views
7K
  • Quantum Interpretations and Foundations
Replies
4
Views
1K
  • Quantum Interpretations and Foundations
Replies
1
Views
1K
  • Quantum Interpretations and Foundations
Replies
1
Views
1K
  • Quantum Interpretations and Foundations
Replies
2
Views
2K
  • Quantum Interpretations and Foundations
Replies
17
Views
3K
Back
Top