Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The energy of a single planet in a gravitational potential is $$E=\frac{m\dot{r}^2}{2}+\frac{L^2}{2mr^2}-\frac{GMm}{r}$$ And the effective potential energy is defined as the last two terms. Note this satisfies $$F=-\nabla U$$ $$mr^2 \dot{\theta}-\frac{MmG}{r^2}=-\frac{d}{dr}(\frac{L^2}{2mr^2}-\frac{GMm}{r})$$ $$=\frac{L^2}{mr^3}-\frac{GMm}{r^2}$$

Similarly, in quantum mechanics I have encountered the energy of an electron in an electric potential as being $$E=\frac{\hbar^2}{2mr^2}-\frac{Zq_e^2}{4 \pi \epsilon_0r}$$

With the first term arising from energy due to confinement in a small volume (uncertainty principle).

I'm aware that this question is probably site-pollutingly basic, but:

  • I assume, by analogy, $\hbar$ is the electron's angular momentum (or integer multiples of $\hbar$: it's probably quantised). Does this form for $L_e$hold firm in post-1920/30's quantum mechanics, or is this result as spurious an idea as the Bohr atom model?
  • If I'm not incorrect, why aren't I? The energy-angular momentum equation is wholly classical, surely electrons behave very differently from this?
  • To what extent does the electron energy equation hold true, or is it again spurious?
  • Are there any other analogies exist between the equations that I've missed?
share|improve this question
1  
The orbital angular momentum of the electron in the H atom ground state is zero. –  John Rennie Mar 9 '13 at 11:21

1 Answer 1

up vote 1 down vote accepted

Your formula seems slightly odd to me as in QM one usually deals with operators and if for the planetary case we have the relation:

$$ E = \frac 1 2 m v^2 + U_{eff} = \frac 1 2 m v^2 + \frac{L^2}{2mr^2} + \frac{GMm}{R} $$

Then in Quantum Mechanics the Hamiltonian for the hydrogen like atom (only one electron) would look like:

$$ \hat H = \frac{\hat p^2}{2m} - \frac 1 {4\pi\varepsilon_0} \frac{Ze^2}{r} = - \frac{\hbar^2}{2m} \nabla^2 - \frac{Ze^2}{4\pi\varepsilon_0r} $$

Where then we use the Hamiltonian to solve the Schroedinger equation:

$$ \hat H \left| \Psi \right> = E \left| \Psi \right> $$

which is just a simple eigenvalue eigenfunction problem.

Now the thing is that we can recast the formula to a somewhat more understandable expression:

$$ \hat H = - \frac{\hbar^2}{2m} \frac 1 {r^2} \frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right) + \frac{\hat L^2}{2mr^2} - \frac{Ze^2}{4\pi\varepsilon_0r} $$

Where the operator $\hat L$ contains all the angular bits of the laplacian in spherical coordinates. Then, you can solve the Schroedinger equation to find the energy values.

As you see, your intuition was correct, that there is quite a lot of similarity between the two problems, but the actual formalism is quite different.

Note: $\hbar$ is a constant, so your premise that it is the angular momentum of the electron is not quite right, as it is just a prefactor. The total angular momentum is given by the formula of:

$$ L^2 = \hbar^2 \ell (\ell + 1) $$

Where $\ell$ is a non-negative integer value describing the angular momentum state of the system.

I hope that this helps you to get on the right track.

EDIT: I wasn't entirely correct with the first version of the additional note.

share|improve this answer
1  
To be extra clear: the difference between the formalisms is the difference between classical and quantum mechanics, not between the gravitational and electrostatic problem. There is a one to one correspondence between the two. It is a trivial excercise to calculate the quantised energy levels of the Earth in orbit around the sun (idealising to point particles, of course). –  Michael Brown Mar 9 '13 at 12:18
    
Yes, that's precsely what I meant about the difference in formalisms. Thanks. –  gns-ank Mar 9 '13 at 12:21
    
I did guess that it was probably integer multiples of $\hbar$ (although, yes, it's not true unless $l^2+l$ is a perfect square). –  Alyosha Mar 9 '13 at 12:37
    
I got the equation from Cheng and Warner's (excellent but) fairly non-rigorous introduction to QM, so your answer's much appreciated. Thank you! –  Alyosha Mar 9 '13 at 12:40
    
I see. Well I guess it is nice to have some connection with the classical mechanics in the begining. But once you get used to QM (yes, you need to get used to it) everything is clearer. Try and read some basic book on Quantum Mechanics, where you'd be able to find the postulates of Quantum Mechanics and the basic definition and interpretation of the Schroedinger equation. –  gns-ank Mar 9 '13 at 12:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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