Are Planck's black body oscillators QHOs?

[snoopies622]

TL;DR Summary

Does the solution to the quantum harmonic oscillator problem explain Planck's quantum hypothesis?

What is the connection between

(1) Planck's quantum hypothesis that the oscillators of a blackbody can possess only discrete quantities of energy E=nhv n=1,2,3..

(2) the fact that the energy eigenvalues of the quantum harmonic oscillator wavefunction are also separated by intervals of hv.

Does (2) explain (1)?

 

[Demystifier]

Yes.

 

[vanhees71]

The oscillators are the modes of the electromagnetic field in a cavity. The Planck Law is the limit for a very big volume, i.e., the thermodynamic limit. In this sense the Planck law is nothing else than the Bose-Einstein distribution for photons.

 

[snoopies622]

hmm these seem like two different answers. It's always seemed odd to me that discussions about blackbody radiation involve modes (standing waves) within the walls of a cavity, when any object will emit blackbody radiation, whether it's hollow or solid.

 

[vanhees71]

The pure Planck spectrum, i.e., an ideal-black-body spectrum is simply best realized as radiation from a cavity, i.e., it's a spectrum completely independent of any properties of the cavity material, i.e., it depends only on temparature as the only variable parameter. Everything else in the Planck formula are just the fundamental natural constants (speed of light $c$, Planck's (modified) quantum of action $\hbar$, and the Boltzmann constant $k_{\text{B}}$, all of which are just defined at fixed values to define the SI units).

Observing, e.g., a glowing piece of metal, is only approximately a black-body spectrum.

 

[snoopies622]

So is there a relationship between the Bose-Einstein statistics and the QHO solutions? They seem to come from quite different places, yet . . coincidence?

 

[PeterDonis]

coincidence?

Not at all. What happens if you have a collection of QHOs, all of the same frequency, and you exchange two of them?

 

[snoopies622]

Everything stays the same. Ah!

I just mean, finding the solutions to the Schrodinger equation for the QHO doesn't seem intrinsically Bose-y. As in, it doesn't include an assumption about the nature of photons.

 

[PeterDonis]

finding the solutions to the Schrodinger equation for the QHO doesn't seem intrinsically Bose-y. As in, it doesn't include an assumption about the nature of photons.

Photons are not the only possible bosons.

 

[snoopies622]

So there must be a deep connection between matter and energy that I don't understand. One can arrive at the Schrodinger equation by plugging the de Broglie relation lamdba=h/p into a generic three dimensional wave equation. Then using the Hamiltonian from the classic harmonic oscillator, one gets energy eigenfunctions such that the energy levels are each separated by hf, where the f can be derived from classical physics alone. And yet hf is also precisely the energy of a photon with frequency f, which I assume doesn't need a harmonic oscillator to be created.

 

[PeterDonis]

using the Hamiltonian from the classic harmonic oscillator, one gets energy eigenfunctions

You can't get energy eigenfunctions from a classical Hamiltonian, because a classical Hamiltonian is not an operator.

 

[PeterDonis]

hf is also precisely the energy of a photon with frequency f, which I assume doesn't need a harmonic oscillator to be created.

A free quantum field, such as the quantum electromagnetic field whose corresponding particles are photons, mathematically is an infinite collection of harmonic oscillators. That's why the whole apparatus of creation and annihilation operators, energy eigenfunctions, etc., works the same way for harmonic oscillators and quantum fields.

 

[snoopies622]

You can't get energy eigenfunctions from a classical Hamiltonian, because a classical Hamiltonian is not an operator.

Sorry, I didn't word that clearly. I mean: you start with the classical Hamiltonian and then substitute the quantum operators. The momentum operator can be arrived at using lambda=h/p as the only quantum assumption.

 

[snoopies622]

A free quantum field, such as the quantum electromagnetic field whose corresponding particles are photons, mathematically is an infinite collection of harmonic oscillators. That's why the whole apparatus of creation and annihilation operators, energy eigenfunctions, etc., works the same way for harmonic oscillators and quantum fields.

That's what has my attention. The mathematics of the QHO and quantum electromagnetic field are the same, but physically they seem like pretty different situations.

 

[PeterDonis]

The momentum operator can be arrived at using lambda=h/p as the only quantum assumption.

I'm not sure what you mean here.

 

[PeterDonis]

The mathematics of the QHO and quantum electromagnetic field are the same, but physically they seem like pretty different situations.

That occurs a lot in physics: the same math turns out to describe very different physical situations. It just seems to be something that happens a lot.

 

[snoopies622]

I'm not sure what you mean here.

Well, if you start with a wave equation like psi=psi_0 exp i(kx-wt), then take the derivative with respect to x , then substitute lambda=h/p and move the constants around a bit, you get negative ih-bar times d-psi/dx = p psi. Then one can take the x derivative a second time and use KE=p^2/2m and get the KE portion of the Schrodinger equation.

 

[snoopies622]

That occurs a lot in physics: the same math turns out to describe very different physical situations. It just seems to be something that happens a lot.

Indeed. That's why I wondered if it was a coincidence or if there was a deep physical connection between the two scenarios that I'm not aware of.

 

[PeterDonis]

if you start with a wave equation like psi=psi_0 exp i(kx-wt)

Then you are already making an assumption about what momentum eigenstates look like. And of course once you do that, you will be led to a specific form for the momentum operator. But how do you know those are momentum eigenstates to begin with?

 

[snoopies622]

Interesting. I guess because the de Broglie hypothesis relates a momentum to a wavelength, I started with a function with one wavelength, as it were, and assumed that it had one momentum. One might think of a sinusoidal wave as the simplest wave associated with a given wavelength.

 

[PeterDonis]

I guess because the de Broglie hypothesis relates a momentum to a wavelength, I started with a function with one wavelength, as it were, and assumed that it had one momentum.

The de Broglie hypothesis is that waves can be associated with matter as well as radiation. But that hypothesis, by itself, doesn't tell you what an eigenstate of momentum looks like.

The usual way of obtaining the momentum operator is by starting with the fact that it is the generator of spatial translations. Then, once you have derived the operator that way, you can look at what its eigenstates look like and observe that they are plane waves.

One might think of a sinusoidal wave as the simplest wave associated with a given wavelength.

A pure sinusoidal wave is the only wave that has a single definite wavelength. Any other wave form will be a mixture of multiple wavelengths.

 

[snoopies622]

The usual way of obtaining the momentum operator is by starting with the fact that it is the generator of spatial translations.

Now that sounds interesting! Please provide a link for details.

 

[PeterDonis]

Now that sounds interesting! Please provide a link for details.

Most QM textbooks discuss this. See, for example, Ballentine, Sections 3.8 and 4.1.

 

[snoopies622]

Thanks! By the way,

The de Broglie hypothesis is that waves can be associated with matter as well as radiation. But that hypothesis, by itself, doesn't tell you what an eigenstate of momentum looks like.

I never read the 1923 de Broglie paper, does it not include the relation wavelength=h/p ? I've always seen it presented that way.

 

[snoopies622]

. . because if we start with the assumptions that matter has wave properties with wavelength= h/p AND that a pure sinusoidal wave is the only wave that has a single definite wavelength, then it hardly seems a stretch to conclude that matter with a single momentum value would be associated with a sinusoidal wave with wavelength h/p.

 

[PeterDonis]

does it not include the relation wavelength=h/p ?

It does, by analogy with the Planck relation frequency = h / E. But de Broglie did not make this hypothesis as a claim about eigenstates. He made it as a claim about observables: basically, that it should be possible to show effects like diffraction and interference with matter particles like electrons (i.e., particles with nonzero rest mass) as well as radiation, and that the wavelengths to be inferred from the diffraction or interference patterns would be related to the particle's momentum. But none of that implies that the particle must be in an eigenstate of momentum, or tells us what those eigenstates look like, any more than correlating radiation energies with frequencies using Planck's law tells us that the radiation is in an energy eigenstate (indeed, it is very difficult to produce photons in an energy eigenstate, and the states seen in most diffraction or interference experiments are coherent states, which are not energy eigenstates), or tells us what such eigenstates look like.