# Free particle spectrum?

1. Oct 5, 2004

### altered-gravity

Hi, this is a silly question

Energy of a free particle is not quantized. Does it mean that it should have a continuous absorption spectrum?. When a cloud of free electrons in some type of plasmas is irradiated with light, theese "hot electrons" are accelerated as they absorb light.

So, imagine water. We want to measure the IR spectrum of water, but instead of doing it in solution or in gas phase, we do it in a molecular beam (without collisions and interaccions), supposing that water molecules in the beam are completely free and cold would they absorb radiation in all the wavelengths, increasing the energy in traslation degrees???

It´s obvious I´m wrong, but where?

Last edited: Oct 5, 2004
2. Oct 5, 2004

### Gonzolo

Unlike a monoatomic particle, water particles have the extra degrees of freedom of rotation and vibration. The associated energy levels are quantized, and I believe that it them that allow IR absorption in molecules with 2 or more atoms

Single atoms absorb IR by having their electrons jump up levels, still quantized.

So IMO, besides for particles with a net charge, such as electrons, protons, alphas, and ions, absorption is necessarily quantized. Neutrons should strictly not be aware of any IR.

Would IR make an ion translate continuously or have its electrons jump levels? I suppose it would depend on whether the IR wavlength matches a level jump or not.

Last edited by a moderator: Oct 5, 2004
3. Oct 5, 2004

### altered-gravity

Hi Gonzolo, thanks for your time and excuse me if my question is nonsense.

That´s true, but it has also translation degrees. If a molecule of water is "completely free" (not influenced by any potential) the energies of theese degrees are continuous. So translation degrees would absorb themselves all the radiation leaving no possibility for measuring other degrees.

If you prepare a beam of any type of particle propagating coherently and free, wolud it be "black" and opaque to light? NO but why not?

Perhaps the reason is that it´s impossible to get a molecule "completely free", but that wolud quantize the translation energy thus creating "translation spectrums"... of course not, but I can´t explain why not.

4. Oct 5, 2004

### Claude Bile

You are trying to make a comparison between electrons and molecules. Electrons are free particles, yes, but molecules are not free particles in a Spectroscopic sense. They consist of electrons bound to a nucleus and two atoms bound together.

Molecules will still exhibit electronic, vibrational and rotational spectra. Translational motion simply doppler shifts the spectrum one way or the other, is does not contribute to the spectral features and is usually disregarded in spectroscopy for this reason.

Claude.

5. Oct 6, 2004

### altered-gravity

Ok, Born-Oppenheimer approx. says that electronic movements are much quicker than nucleus movements, so in VIS radiation-matter interaccions we can consider total wavefunction as electronic wavefuncion.

Even rotational and vibrational movements are much quicker than translational movements. So if we irradiate the beam with MW or IR (resonant with any transition) it would absorb as if the molecule hadn´t translational energy.

But if we irradiate with non-resonant ligth? What is the reason for not to think that it wolud use that energy to increase translational energy (as if it was a photon-molecule collision)?

6. Oct 6, 2004

### Gonzolo

First, there are many more photons than water molecules in the beam, so many get through.

And in a case where you have exactly the same number of photons than molecules, there is some distance between the molecules, so that light can scatter through, such that it does through any gas. There is a limited cross-section to a molecule that might be smaller for translation absorbtion than for vibrationnal or rotationnal absorption.

In fact, the closest thing to the beam you envision might be very dense steam. It is not black indeed, but rather white-ish, due to scattering. This white is indeed continous. You might want to think of this white as what happens "after a translationnal absorption" in the VIS. But remember that a water molecule is a dipole, so that if it is accelerated by EM (such as a photon), it can also emit. Such "emission and absorption", which I believe is better termed "scattering" is quite symetrical - what the dipole (or whatever charge) receives doesn't necessarily have a reason to stay, such translations would cause molec.-molec. collisions emitting further photons and some of the molecules might return to their original state. And beam temperature would rise.

Last edited by a moderator: Oct 6, 2004
7. Oct 6, 2004

### altered-gravity

Ok, that made me think. Perhaps te answer is simply in the Rayleigh and Raman scattering... Thanks very much to both of you, I will need time to think about it.

8. Oct 6, 2004

### Claude Bile

Not quite, the Born-Oppenheimer approximation says that we are able to separate the electronic part of the wavefunction from the rovibrational (rotation/vibration) part. The total wavefunction is the electronic wavefunction times the rovibrational wavefunction.

Claude.

9. Oct 7, 2004

### altered-gravity

OOps! Of course, you´re right. Thanks for correcting me.

Ok, the process that I was looking for is the scattering. But the scattering is a concerted process in which the absorbed and the emitted photons are connected by a virtual state with zero lifetime, so I may think that all the non-resonant energy (photons that match the cross section) is directly scattered without increasing linear momentum of the molecules. But the hot electrons emitted in laser ablation for example, or in photoelectric effect do so, they are accelerated without scattering. What´s the difference?

The more I think the more confused I get.... Je Je

10. Oct 7, 2004

### Gonzolo

Light is not necessarily scattered when it hits a solid. In ablation or in the photoelectric effect, much of its energy is used to eject the electron from the solid, this is indeed absorption. And much of the extra energy goes into the electron's velocity.

I am not so sure the linear momentum of a gas is not increased at all. I can see the cross section to be very small, but not sure it is zero. If photon direction changes, so should a particle's.

Last edited by a moderator: Oct 7, 2004
11. Oct 7, 2004

### altered-gravity

12. May 15, 2010

### george simpso

I was tracking down your thread "free particle spectrum" and others on google, but found nothing specific . Has no one calculated/measured it yet?
I can imagine the experiment: a discharge tube, say a meter long with transparent electrodes maintaining a 1 amp current flow at a variable potential difference up to speeds of 90%c. There would be a monochrometer passing variable narrow band emf from nicrowave to gamma down the tube to a detector that tells you Absorbance at a wavelength verses voltage.
My imaginary results would have a blip at spin flipping wavelength, superposed on a continua. The area under the absorbance vs wavelength curve properly calculated would be near to one electron.
Why can't the experiment and or oscillator strength calculation be made? Is there a fundamental problem like an electron can not have a transition (or any other) dipole?
Or, incoherent photon absorption would force a major fraction of electrons to leave the beam ( sic, angular dependence), and mess up "concentration"?...Or what.
Surely, you collective physicist could have come up with some appropriate, pointed answers in the four years since this thread was started.

13. May 15, 2010

### alxm

The first post already answered that question: A free particle has a continuous spectrum. It has only kinetic energy.

14. May 16, 2010

### george simpso

Thank you,alxm.
But I was hoping for an answer a bit more detailed than was found in the existing threads.
Can you refer to a calculation- say one showing the transition dipole is not zero, or some experimental observation,like the electron population between the earth and sun diminishes the suns spectra somewhat?
george simpso

15. May 16, 2010

### Bob S

Free electrons cannot absorb photons because energy and momentum conservation cannot both be satisfied simultaneously. The incident photons are instead scattered by Thomson (low energy Compton) scattering off of the electrons. This is an important hot-plasma electron temperature diagnostic. See

http://en.wikipedia.org/wiki/Plasma_diagnostics#Thomson_scattering

http://en.wikipedia.org/wiki/Thomson_scattering

Bob S

16. May 18, 2010

### george simpso

Thank you, Bob S.
I appreciated the references to Thomson Scattering. I'll try to track down all the scattering processes.
I feel certain that scatter is the result of an energy absorption process for the electron and an exchange of energy &/or momentum by the photon.
george simpso

17. May 18, 2010

### Dickfore

imagine a photon with energy $E_{\gamma}$ incident on a "composite particle" with a rest mass $M$. For simplicity, we will observe the phenomenon in the CM - frame. In this frame, the momentum of the particle is opposite of the momentum of the photon $p_{\gamma} = E_{\gamma}/c^{2}$.

After the collision, the particle absorbs the photon and its mass increases to $M' = M + \Delta M$. Momentum is conserved and the particle is at rest. Total energy is conserved, so we may write:

$$\frac{\Delta M}{M} = \frac{E_{\gamma}}{M c^{2}} - 1 + \sqrt{1 + (\frac{E_{\gamma}}{M c^{2}})^{2}}$$

One can view this increase in rest mass as excitation of the intrinsic degrees of freedom of the particle by an energy $\Delta \epsilon = \Delta M \, c^{2}$. We see that this dependence is a continuous function, so the absorbtion spectrum of the photon depends only on the excitation spectrum of the particle. Notice that the translational degrees of freedom where not limited and, therefore, they are not quantized.

18. May 18, 2010

### Bob S

Err......no.
For simplicity, let's consider the total absorption of a photon by an electron in the lab frame, where the electron is initially at rest. For a photon of energy Eγ = pc and momentum p = Eγ/c , the final total energy of the recoiling electron would be

Ee= m0c2 + Eγ.

But the momentum of the recoiling electron would be p = Eγ/c,

leading to a total energy of

[Ee]2= [m0c2]2 + [pc]2= [m0c2]2 + [Eγ]2

These two equations are irreconcilably different, because energy and momentum cannot both be conserved..

Bob S

19. May 18, 2010

### Dickfore

Is the electron a composite particle?

20. May 18, 2010