- #1
michael879
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I'm trying to get a deeper understanding of why Planck's constant shows up in so many seemingly unrelated places, and what I'm finding is that in most of the cases I've considered I can reduce it to a relation between some "physical" (depending on interpretation) frequency and an energy. The obvious example is for photons where E=\hbar\omega. Most of the places the constant shows up can be traced back to this relation, with a few exceptions. The two I'm focusing on now are the Hydrogen atom and the electron.
Using a semi-classical approach and neglecting l > 0 states, you can derive various details about the orbit of the electron in the hydrogen atom. For example,
<br /> r_n = n^2\dfrac{\hbar^2}{me^2}\\<br /> E_n = \dfrac{e^2}{2a_n}\\<br /> v_n = \sqrt{\dfrac{2E_n}{m}} = \dfrac{e^2}{\hbar n}\\<br /> \omega_n = \dfrac{v_n}{r_n} = \dfrac{n\hbar}{ma_0} = \dfrac{me^4}{n^3\hbar^3}\\<br />
Which gives the result E_n = \frac{1}{2}n\hbar\omega_n. Now this makes a little sense, but it deviates significantly from the usual expression E=\hbar\omega. Where does that factor of 2 come from, and why is there suddenly a dependence on n??
The next case is just simply with any fermion. The usually quoted formula for the de broglie frequency of a particle used E=\hbar\omega. However, if you look into Zitterbewegung you will find that the actual angular frequency predicted for a fermion is
\omega=\dfrac{2mc^2}{\hbar}
so that E=\frac{1}{2}\hbar\omega. So does this factor of two come from the half-integer spin of the electron? Is there some general formula E=L\omega and L is just quantized into units of \hbar?? If so how does that explain the hydrogen atom where L=0 for all those states?
Related to this: I came across this site (which I won't link to because it's complete bogus) which made the claim that \hbar is just a conversion constant between angular frequencies and energies, similar to the speed of light (i.e. it has no real meaning, we're just using the wrong units). They made the claim that E=\hbar\omega always holds, which I just showed is wrong. However they did seem to have a good point, since Planck's constant acts and is treated very similarly to the speed of light, in that its frequently set to 1 and frequencies and energies are usually interchangeable by some factor of it. I suppose it could be like Boltzmann's constant, which although it isn't a simple conversion factor between temperature and energy it is still a completely redundant and meaningless artifact of our unit system (temperature is energy per degree of freedom).
*edit*
Also, I found that any classical cylindrically symmetric system will obey:
L\omega = E - \dfrac{m^2}{E}
Where E is the total energy and m is the energy of the system when L=0. So the quantum description of particles matches the classical prediction for a massless rotating system (e.g. photon quantization and electron Zitterbewegung). The hydrogen atom is still a big mystery to me though
Using a semi-classical approach and neglecting l > 0 states, you can derive various details about the orbit of the electron in the hydrogen atom. For example,
<br /> r_n = n^2\dfrac{\hbar^2}{me^2}\\<br /> E_n = \dfrac{e^2}{2a_n}\\<br /> v_n = \sqrt{\dfrac{2E_n}{m}} = \dfrac{e^2}{\hbar n}\\<br /> \omega_n = \dfrac{v_n}{r_n} = \dfrac{n\hbar}{ma_0} = \dfrac{me^4}{n^3\hbar^3}\\<br />
Which gives the result E_n = \frac{1}{2}n\hbar\omega_n. Now this makes a little sense, but it deviates significantly from the usual expression E=\hbar\omega. Where does that factor of 2 come from, and why is there suddenly a dependence on n??
The next case is just simply with any fermion. The usually quoted formula for the de broglie frequency of a particle used E=\hbar\omega. However, if you look into Zitterbewegung you will find that the actual angular frequency predicted for a fermion is
\omega=\dfrac{2mc^2}{\hbar}
so that E=\frac{1}{2}\hbar\omega. So does this factor of two come from the half-integer spin of the electron? Is there some general formula E=L\omega and L is just quantized into units of \hbar?? If so how does that explain the hydrogen atom where L=0 for all those states?
Related to this: I came across this site (which I won't link to because it's complete bogus) which made the claim that \hbar is just a conversion constant between angular frequencies and energies, similar to the speed of light (i.e. it has no real meaning, we're just using the wrong units). They made the claim that E=\hbar\omega always holds, which I just showed is wrong. However they did seem to have a good point, since Planck's constant acts and is treated very similarly to the speed of light, in that its frequently set to 1 and frequencies and energies are usually interchangeable by some factor of it. I suppose it could be like Boltzmann's constant, which although it isn't a simple conversion factor between temperature and energy it is still a completely redundant and meaningless artifact of our unit system (temperature is energy per degree of freedom).
*edit*
Also, I found that any classical cylindrically symmetric system will obey:
L\omega = E - \dfrac{m^2}{E}
Where E is the total energy and m is the energy of the system when L=0. So the quantum description of particles matches the classical prediction for a massless rotating system (e.g. photon quantization and electron Zitterbewegung). The hydrogen atom is still a big mystery to me though
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