1
$\begingroup$

I found this question here and the answers got me thinking.

Take a case where $\Psi(x,t)$ is the linear combination of two eigenvectors of a charged particle:

$\Psi(x,t)=c_1\psi_1e^{-iE_1t/\hbar}+c_2\psi_2e^{-iE_2t/\hbar}$.

At $t=0$ the wave function is:

$\Psi(x,0)=c_1\psi_1+c_2\psi_2$.

The probability density distribution is:

$P(x,t)=\Psi^*(x,t)\Psi(x,t)$.

$P(x,t)=|c_1\psi_1|^2+|c_2\psi_2|^2+c_1c_2^*\psi_1\psi_2^*e^{-i(E_1-E_2)t/\hbar}+c_1^*c_2\psi_1^*\psi_2e^{-i(E_1-E_2)t/\hbar}$.

So $P(x,t)$ is (obviously) not constant but contains elements that oscillate with angular frequency $\omega=(E_1-E_2)/\hbar$ and frequency $\nu=(E_1-E_2)/h$.

Does this explain Bohr's equation: $E_1-E_2=h\nu$?

$\endgroup$
3
$\begingroup$

I don't think your argument works, for a few reasons. All you've done is shown that interference terms arise with a frequency given by the difference between the energy levels. The problem is that there's no connection to a photon here, which is what I assume you mean by "Bohr's equation." You are correct that you're seeing kind of the same effect, but the reason is that you've assumed your conclusions. Let me explain:

As soon as you write the first equation, you're asserting that $h$ or $\hbar$ is the conversion between frequency and energy, otherwise you wouldn't be able to write $E t / \hbar$ and have it make sense, dimensionally. Therefore, when you take an energy difference, you still use $h$ to get $\nu$ from $E$. What you've called "Bohr's formula" is just the statement that the difference between two atomic energy levels is $h \nu$ for some $\nu$. I suspect the actual information you want to prove is that this is then the frequency of an absorbed or emitted photon, but that just follows from conservation of energy. The real leap is just knowing that there exists a constant, called $h$ or $\hbar$, that can convert between frequency and energy.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Right, the "Bohr's formula" is actually just a special case of the Planck-Einstein relation where the energy is just contained in one photon. You might want to argue about why there aren't, say, two photons needed, or something, but that's just a "we haven't observed it yet" type of thing where those terms could probably work their way into Lagrangians, but we haven't needed to add them yet. $\endgroup$ – CR Drost Sep 25 '15 at 13:44
1
$\begingroup$

As pointed out in zeldredge's answer, what is missing is the connection to the photon. However, I do think you can fix this problem. If you couple a system that exhibits natural oscillations at certain frequencies to an external force that oscillates at some frequency $\omega$, then you get a resonance when $\omega$ matches one of the natural frequencies. Conservation of energy then yields that in a quantum mechanical treatment of the force field, the quanta that appear must have energies of the form $\hbar\omega$, which is indeed the case for a harmonic oscillator.

So, what one would be assuming here is that the electromagnetic field can indeed be described in terms of harmonic oscillators.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.