5

Is it possible to recover the old Bohr-Sommerfeld model from the QM description of the atom by turning off some parameters?

Can we use Ehrenfest's theorem (or some other scheme) to reduce the QM model to the Bohr-Sommerfeld model? If not, why not? The issue seems to be significant because it might shed some deep conceptual issues.

Qmechanic
  • 201,751
QSA
  • 194
  • As you are asking by Sommerfeld model too, an interesting reading is L. C. Biedenharn ''The “Sommerfeld Puzzle” revisited and resolved'' – arivero Oct 20 '13 at 02:09

2 Answers2

4

No, the Bohr-Sommerfeld model is conceptually a classical toy model (which has only been fudged to imply some selected quantization features similar to quantum mechanics) so it is inequivalent to the proper quantum mechanical description or any approximation of it. The agreement of the Bohr-Sommerfeld model with the right quantum mechanical results is a coincidence, a special feature of the hydrogen atom.

The only limit in which quantum mechanics reduces to the "Bohr-Sommerfeld physics" is the limit of long distances and high momenta in which $\Delta p\cdot \Delta x\gg \hbar$ and in which both quantum mechanics and the Bohr-Sommerfeld model reduce to classical physics without any Bohr-like quantization restrictions. But this limit is clearly not relevant for the description of low-lying states of the hydrogen atom.

Well, some proper interpretations of the Bohr quantization rules also emerge in the semiclassical (next to leading) WKB approximation of quantum mechanics. But one must be careful about the interpretation and various shifts and subtleties. For example, $\int p\,dq$ over the phase space contours is a multiple of $2\pi\hbar$. In Bohr's old picture, this statement applied to allowed closed trajectories of particles. In quantum mechanics, however, it applies to boundaries of phase space regions corresponding to $N$ microstates. The interpretations are slightly different.

In QM, there are typically no closed trajectories as the initial localized wave packets spread, and in Feynman's approach, one sums over all classical trajectories, whether they obey classical equations of motion and Bohr's quantization conditions or not.

See also

Bohr Model of the Hydrogen Atom - Energy Levels of the Hydrogen Atom

All the criticism of my answer below is completely invalid.

Community
  • 1
Luboš Motl
  • 179,018
  • 1
    -1: This is incorrect, and this has been pointed out to you many times: Bohr Sommerfeld agrees with QM in the correspondence limit $\hbar\rightarrow 0$ (not the same as $\Delta P \Delta X$ large, but that's just a blooper, not the real mistake) both to zeroeth order (as you say) and also to first order in $\hbar$ as you falsely imply is not true. Stop saying this. – Ron Maimon Nov 10 '12 at 14:11
  • @RonMaimon and lubos, maybe you're both right? Isn't the Bohr-Sommerfield model to QM what CED is to QED? – John McAndrew Nov 10 '12 at 18:06
  • @JohnMcVirgo: Not exactly, Lubos feels that the ad-hoc nature of Bohr-Sommerfeld (and he's right about this) makes it not a true theory, but it is better than he gives it credit for, as it correctly gives the density of states and the level-spacing from classical dynamics, and this is something people are no longer taught. – Ron Maimon Nov 10 '12 at 18:29
  • No, Ron, the Bohr model is just a wrong classical model, a wrong modification of classical physics that isn't equivalent in any way to QM and that served as a great motivation to anticipate and look for a completely new framework for physics since early 1910s - and people had to wait for 15 more years. All the quantum numbers and degeneracies are wrong. For example, QM predicts $n^2$ degeneracy for principal number $n$ or $2n^2$ with the spin. None of these things is reproduced by the old Bohr model that really confuses and incorrectly identifies $n$ and $m$ and $l$, and so on. – Luboš Motl Nov 10 '12 at 18:38
  • On the other hand, the orbital plane in the Bohr model is arbitrary and continuous. In proper QM, it is effectively quantized because the total degeneracy is finite including the rotations in space. All the things are just fake, the hydrogen success is a coincidence that is likely due to the special simplicity of the simplest atoms, and all the agreements go completely away for all other atoms. – Luboš Motl Nov 10 '12 at 18:39
  • @LubošMotl: Yes, we know, we know, I changed my -1 to a +1 given that you now say exactly correct stuff. The question I interpreted as "is there any approximation to QM where the leading order in hbar account is exactly correct, and the subleading order is negligible", or "when is WKB exact?" This question you didn't adress. – Ron Maimon Nov 10 '12 at 18:47
  • Thank you for your answers and comments. I do have a better grasp now, although I have heard of similar arguments before but not as clear. However, I still have some issue that I don’t understand, but I hope I am not being malicious (or stupid). What really bothers me is not the quantization of the energy per se, but the numeric equivalency. Now, the potential in both the Bohr-Sommerfeld model and proper QM model are both follow the 1/r law and relates it to charge, but I still have a hard time of seeing how they came to give the same result, even for the lowest energy level. – QSA Nov 11 '12 at 01:02
  • Of course, the use of Schrodinger equation for the classical potential (1/r) itself is a bit awkward. Nevertheless, my thinking is now that the connection has to be made three ways i.e. classical, QM and QFT since it predicts 1/r law(Zee says that is the triumph of 20th century physics ,even though does not predict charge). So, I guess I am asking, the 1/r law which seem to force both models to predict the same energy is derived using QFT but how does it reduce to classical other than JUST mathematical equivalency. I wonder if all these questions are connected, or I am not making any sense. – QSA Nov 11 '12 at 01:02
  • Dear QSA, the $1/r$ potential is, much like the harmonic oscillator, very special, so the preferred energies happen to give the same results in old-Bohr-quantized classical physics as well as quantum physics. This is also seen in the Coulomb scattering etc. The very value of the hydrogen binding energy is incidentally the most natural product of powers of the parameters so it's not shocking that it appears at several places. None of those agreements generalize beyond the $1/r$ and $r^2$ potentials. – Luboš Motl Nov 12 '12 at 17:40
3

Some version of Bohr-Sommerfeld quantization is exact for classically integrable systems (i.e., systems that have a fairly large symmetry group in a sense that can be made precise), and hence in particular for the hydrogen atom.

However, already the 3-body problem is nonintegrable, and even the semiclassical versions of Bohr-Sommerfeld (keeping the first nonlclassical corrections) lead to messy formulas, though involving some interesting mathematics (e.g., Gutzwiller's trace formula).

Numerically, Bohr's model is quite misleading beyond hydrogen.

Arnold Neumaier
  • 44,880
  • Thanks Arnold, But my question is very specific. I understand the algebra that makes both the energy in proper QM equal to Bohr atom energy {no spin}. I find this an astonishing coincidence, don't you think. Or am I missing something? – QSA Nov 13 '12 at 13:39
  • I mean there are at least three constants that got aligned properly arithmetically with the exact powers. – QSA Nov 13 '12 at 13:52
  • @QSA: The only coincidence is the fact that the 2-body problem is completely integrable and has more symmetries than an arbitrary rotationally invariant system. But there are many completely integrable systems, so this coincidence is more of the kind that a plane shape is a square or a circle rather than of a more general kind. - Much of physics is the exploitation of this kind of coincidences to get a handle on more generic systems by means of perturbation theory around these special cases. – Arnold Neumaier Nov 13 '12 at 15:50
  • Sorry Arnold, But I hope I don't stretch your patience. Ok, the Bohr Model is a 2-body, but are you considering Schrodinger equation (for hydrogen) as a 2-body problem? I mean we say the three methods of QM (matrix, Schrodinger, path integral) are mathematically equivalent, we even say the physics is the same. But in our case there is no apparent mathematical or physical intuition to reduce one to another. One is an algebraic manipulation and the other we solve a differential equation. – QSA Nov 13 '12 at 19:52
  • @QSA: The Schroedinger equation for hydrogen is indeed the quantum 2-body problem for the 1/r potential. This is why it is exactly solvable, and why one gets exact Bohr-Sommerfeld rules. - In the Heisenberg picture, the classical and the quantum case are very similar, and the Schroedinger picture is equivalent to the Heisenberg picture, with a nearly trivial physical interpretation of the equivalence. So I don't know what you are complaining about. – Arnold Neumaier Nov 14 '12 at 09:11
  • Thanks, your answer is clearer now. But my problem is that you and lubos seems to say that the energy (=e^4m/h^2) is coincidental but not surprising. My view is that it looks coincidental but very surprising. It is not like we had a classical problem(it never was) and then we found an Ehrenfest's theorem to make the correspondence. It is ok though I can ponder it better now. Thank you all. – QSA Nov 15 '12 at 00:52