De Broglie wavelenghts

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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
 

George Jones

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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
 

dextercioby

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There are no orbitals for a free particle.

Daniel.
 
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
 
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But isn't Aexp[kx-wt] a solution of the schrödinger equation? So the wave function must not be given by an integral, correct?
 

dextercioby

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

dextercioby

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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.
 

George Jones

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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
 
dextercioby said:
There are no orbitals for a free particle.

Daniel.
Are you sure? really sure?


Seratend.
 

George Jones

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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 https://www.amazon.com/exec/obidos/tg/detail/-/981024651X/qid=1120420446/sr=1-2/ref=sr_1_2/104-6066677-3407149?v=glance&s=books by Anton Capri.

Regards,
George
 
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dextercioby

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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.
 

dextercioby

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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 https://www.amazon.com/exec/obidos/tg/detail/-/981024651X/qid=1120420446/sr=1-2/ref=sr_1_2/104-6066677-3407149?v=glance&s=books 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.
 
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Astronuc

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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.
 

jtbell

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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].
 
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.
 

dextercioby

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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.
 

George Jones

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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
 

dextercioby

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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.
 
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
 

jtbell

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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.
 
re: question on DB wavelength

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 modelled 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?
 

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