# Stupid quantum question

1. Apr 11, 2010

### LocationX

Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?

2. Apr 11, 2010

### ansgar

1) detectors smear the energy

2) energy lines in atoms are smeared as well "time-energy uncertainty relation"

3. Apr 11, 2010

### genneth

Also:

3) The energy spectrum of a macroscopic body is not simply the atomic spectra

4) Putting two and three together, let's assume we have 100 atoms, each with 4 reasonably accessible energy levels. That gives 4^100 states. Generously, these will live within 100 eV of each other. That mean the mean energy spacing is about 6*10^-59 eV, or 10^-77 joules. Using the energy-time uncertainty relation, to resolve this energy difference, you would need to make an observation for about 3*10^35 years. The current probably age of the universe is 1.4*10^10 years. There are many more than 100 atoms in a typical macroscopic sample, and a lot more than 4 levels per atom.

4. Apr 11, 2010

### LocationX

You have a macroscopic piece of solid carbon. Each carbon atom has the same energy levels thus the emission lines should correspond to the same discrete energy levels. Would we still see a continuous spectra?

When we say 100 atoms, does this include any atoms? What about mercury gas? Or Hydrogen gas? When we heat up mercury gas or hydrogen gas and look at it through a diffraction grating we see the emission spectra for that specific element. Thus, is it necessary to mandate that the blackbody must be multi-elemental in order to get a continuum?

Last edited: Apr 11, 2010
5. Apr 12, 2010

### ansgar

no solids have energy BANDS not energy lines.

you know how a black body works so that suggestion is not correct ;)

6. Apr 12, 2010

### LocationX

Where do energy bands come from? And also what about the second part of that question?

and the sun is a "solid".. yet it behaves like a blackbody. It would be nice to have some concrete answers because this flaky terminology is just confusing

7. Apr 12, 2010

### ansgar

you should go for google or something first, then come back with more questions later - more specific questions. Me or anyone else can not give you a full blown lesson in solid state physics,

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

8. Apr 13, 2010

### Sillyboy

I think the harmonious_oscillator model can explain this phenomenon well. And frequecy of oscillation is not the same!

9. Apr 13, 2010

### LocationX

Why does the detector not smear the energy for hydrogen gas when it is heated? We see the emission lines very clearly thought a diffraction grating.

3. Is a bunch of hydrogen not a macroscopic body? By mandating that it has to be macroscopic, the we shouldnt see spectra lines for hydrogen gas when heated in a tube.

4. So let me get this straight. We have 100 hydrogen atoms. However, if we blast the heck out of these atoms, they will jump into higher and higher energy states. In your example, you're saying that the energy spacing is within 100eV of each other. However, for H gas, the energy spacing is much less for hydrogen, just several eV. If you blast it too much with thermal energy, it will ionize thus making plasma (?). So here's the thing, even though hydrogen gas has much lower energy spacing than 100eV, the energy spectra does not smear to the detector (the human eye). We see discrete energy levels corresponding to the emission spectra of hydrogen. Why does this not spread and a blackbody would spread the energy spectra? I really feel like I'm missing the big picture.

The wiki article says:

"It is due to the diffraction of the quantum mechanical electron waves in the periodic crystal lattice."

Now, do they mean the electrons bound to the atoms? The free electrons? The electrons in the valance band? Conduction band? doped electrons?

Assuming these energy bands represent the electrons bound to the lattice, then why are they bands? As genneth says, " let's assume we have 100 atoms, each with 4 reasonably accessible energy levels. That gives 4^100 states." Then i can understand there will be a smear of different energy states that make it look like a continuous energy distribution. However, in realize, these are actually very packed discrete energy levels. Is this the right picture?

Then why are there bands? What causes the regions where no electrons can have that energy level? Wikipedia says "the diffraction of the quantum mechanical electron waves". Why would bound electrons diffract off of anything? Unless this was talking about an electron traveling inside the lattice, which makes it even confusing as to why that would effect the various energy levels of the electrons bound to the lattice.

And why doesnt a blackbody have energy bands if solids have energy bands?

A harmonic oscillator describes the bound electron to the nucleolus right? However, how would a simple model like that exhibit band structure?

10. Apr 13, 2010

### Sillyboy

No, The black body radiation should be explained by the harmonious oscillation of the atoms.
Please refer to Planck's explaination, you should think it is so, not the eclctronic motion.

11. Apr 13, 2010

### jrosen13

Thats right, it is not transitions between atomic levels that leads to blackbody radiation, but instead if we have a collection of harmonic oscillators at some temperature and we ask how the different states will be populated. Each state corresponds to a different frequency of vibration, hence a different wavelength of light. The reason we do not observe the discreet changes in frequency is that we are talking about atoms, not electrons, and they are many orders of magnitude heavier, so the level spacing is so small that we can consider it to be continuous.

12. Apr 14, 2010

### LocationX

And the harmonic oscillators represent the atoms in a lattice? For example heating steal will cause it to glow various colors depending on the temperature. Then this glow is due to the phonon vibrations?

"Each state corresponds to a different frequency of vibration, hence a different wavelength of light." What would cause photons to be emitted from the blackbody? Unlike the absorption and radiation from electrons going up and down energy states in an atom, what do these "frequency of vibrations" do to emit photons? Are they constantly wanting to go back to the ground state thus the phonons emit photons?

If a blackbody was heated to exactly the same temperature, completely uniformly, then wouldnt we see ONE frequency for photon emission from the blackbody?

" The reason we do not observe the discreet changes in frequency is that we are talking about atoms, not electrons, and they are many orders of magnitude heavier, so the level spacing is so small that we can consider it to be continuous." Level spacing of what? Levels between the phonon energy levels? The level spacing of the discrete harmonic oscillator energies?

Why do the changes in energy levels of the electron in orbit (from excitation) contribute the black body radiation? And finally, do blackbodies have energy bands?

13. Apr 14, 2010

### ansgar

"Why does the detector not smear the energy for hydrogen gas when it is heated? We see the emission lines very clearly thought a diffraction grating."

"very clearly" is not equal to "infinite thin"... they are smeared, but not much. This is the issue of building good detectors, compare how good a germanium detector is to resolve gamma rays compared with a scintillator...

14. Apr 14, 2010

### f95toli

Real atoms do NOT have "discrete" energy levels. Real energy levels -be it in atoms or in solids etc.- always have a finite width. Even if you take the simplest case of a single atom you will find that the energy levels are "smeared" due to the coupling to the vacuum (the eigen- energies you get from solving the SE corresponds to the centres of Lorentzian distributions).

Note also that the usual mathematical relation between bandwidth and lifetime applies, excited states with a finite lifetimes (=all real states) will always correspond to levels with finite width.

15. Apr 14, 2010

### jrosen13

First off, good question. To be honest I am not really sure if it is easy to understand, but depending on the emissivity of the material there should be some connection between vibrations and photons, that is to say that thermal energy in vibrations can be radiated away.

This is definitely not correct, since at some finite temperature we have a thermal population of the levels according the the Bose-Einstein distribution which is similar to the Boltzmann distribution at high temperatures.

16. Apr 14, 2010

### LocationX

So a Sodium atom does not have discrete energy levels? Even if NIST has measured these very energy levels to precise eV INCLUDING spin orbit coupling and hyperfine splitting? What you're saying doesnt make sense. How would you even include the coupling to vacuum in the Hamiltonian? Are you talking about coupling to virtual particles?

Again, my original question as not been answered. What makes spectra likes for hydrogen gas when heated, versus increasing the temperature for steal and seeing a spectra of photon energies.

17. Apr 14, 2010

### LocationX

Who said I had a bunch of bosons? We keep a piece of steal at a temperature that makes it glow red, thus we have just confined it to emit a certain wavelength. What's wrong with that?

Again, we heat to a very specific temperature T=X>>0K. Boltzmann distribution tells us the number of states possible at T=X. So am I correct to say that blackbodies obey the Boltzmann distribution at high temperature and thus we have lots and lots of energy states that could be occupied? But that is only for ONE temperature. If it could occupy so many different states with just one temperature, then why does a piece of metal glow red at a specific temperature and not have a huge spectrum of various photon energies at T=X?

18. Apr 14, 2010

### LocationX

Ok, sorry. My question is actually related to why does hydrogen gas produce a discrete spectra and blackbodies produce a continuous spectra. I'm looking for a physical, microscopic reasoning regarding this. As in what happens inside atoms of blackbodies that allows this continuous emission of photons? I understand single atoms and their various discrete excitation levels before ionization, but I do not understand the microscopic behavior of blackbodies.

19. Apr 14, 2010

### jrosen13

The harmonic oscillator origin of the blackbody distribution is certainly not a microscopic model, it is just a fortunate thing that the problem reduces to such simple a model. The microscopic physics is not universal either, as for instance the emissivity of different materials varies and has complicated dependence on the material. The entire discussion hinges on an understanding of what blackbody radiation is, and essentially it is a thermodynamic distribution for a large collection of independently quantized systems which becomes a continuous distribution when taking the thermodynamic limit of a very very very large number of oscillators.

20. Apr 15, 2010

### f95toli

I guess it depends on what you mean by "discrete". If you mean "single valued" then no, they do not (but if you only mean well-separated in energy, then yes they are discrete) . The values you find tabulated by NIST and others refers to the energy of the centres of the "bands" that form the levels.
Now, in levels that are weakly coupled to the vacuum (or other levels, remember that ANY form of interaction that results in a finite transition probability means a finite lifetime, which in turn means a "broadening" of the level) the width can be very narrow (and in atoms they never overlap) but it is never truly zero; you can only get zero width if you do NOT take the coupling to the vacuum (or more generally the environment) into account.

One nice thing about experiments on "artificial atoms" (e.g. qubits) where you can adjust the coupling to the environment is that you can actually study this process experimentally; if you use spectroscopy to map out the width of the a levels you can see how the width increases (Q decreases) as you gradually "turn on" the coupling.

If you want an "easy" explanation for how this is handled mathematically (i.e. coupling to the continuum of states that form the vacuum, and o you do not need virtual photons) I would recommend
Cohen-Tannoudji's book on Atom-Photon interactions; although the physics will be described in any book that deals with open quantum systems (this is very much standard textbook stuff).

21. Apr 15, 2010

### jrosen13

I think I did.

Nothing, depending on what you mean by certain.

It is clear that your understanding of a distribution is not very good. The reason it is called a blackbody distribution is that it is a probability distribution for all the wavelengths that will be emitted. This is basic stuff, can you see IR? Would you be able to see the part of the spectrum emitted in IR or UV? Not with your eyes. So your "gut" feeling for red meaning one wavelength is just simply not correct. Its a lot of different wavelengths. As it gets gotter, does it go like the rainbow ROYGBIV? Not exactly. The sun is a blackbody, does it emit at just one frequency? No. Its a distribution! When it gets really hot, it will just look white, i.e. "white hot", because it has a lot of different wavelengths.

22. Apr 15, 2010

### Dickfore

Because blackbodies are not individual atoms.

23. Apr 16, 2010

### LocationX

and neither is hydrogen gas... but we see discrete spectra

24. Apr 16, 2010

### LocationX

Once again, my fundamental question has not been answered. It's is obvious that I do not understand blackbody radiation. I know what it describes, but those are only words without substance. Saying a blackbody is a probability distribution for all wave lengths is the same as saying BE distribution is a distribution for bosons. It means nothing. Let me explain again why I clearly do not understand blackbodies.

We take hydrogen gas. We thermally excite hydrogen gas. There are lots of atoms, that all excite and then drop back down thus producing the discrete spectra that we see. The explanation for ONLY discrete spectra lines are from the electrons having discrete energy levels in their various orbitals.

Now lets think about a blackbody. A blackbody is also made of atoms. Lets for example make all the atoms identical, thus its all the same element. We excite these atoms and they should in principle also emit discrete energy lies. But why do they not? What is physically happening in the atoms that causes them to produce a continuous spectra? That's the other thing. Why doesnt the electrons in the atoms ionize in a blackbody?

25. Apr 16, 2010

### Dickfore

Hydrogen gas is a collection of particles that interact negligibly with each other. That is why the spectrum emitted by a single structural unit of the hydrogen gas is the same as the one emitted by a collection of them. I meant to say blackbodies are not a collection of non-interacting particles.