Bohr model with all Quantum numbers for hydrogen atom

bohr model:

$$E_n=-\frac {\mathcal R}{n^2(1+\frac {m_e}{m_p})}$$

• can we developed bohr model with all Quantum Numbers of the Hydrogen Atom?

$R(r)$ Principal quantum number $$n=1,2,3,4,5,...,n$$

P(θ) Orbital quantum number $$\ell=n-1$$

F(φ) Magnetic quantum number

$$m_{\ell}=-\ell,\ell+1,0,\ell,\ell-1...$$

Spin quantum number $$m_s=+\frac {1}{2},-\frac {1}{2}$$

Q numbers

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There isn't any old Bohr model that would explain all properties of the Hydrogen atom; new quantum mechanics is needed at some point. –  Luboš Motl Aug 1 '12 at 9:16
@Luboš Motl i know there isn't but i don't know that is it possible? –  Salvador Aug 1 '12 at 9:19
By "isn't", I meant "it isn't possible". The $1/n^2$ scaling of the energy one could get from the old model rather naturally was a coincidence - this kind of a coincidence sometimes occurs when the equations are simple enough so that the solutions also have to be simple. However, the model really predicts that $l$ and $n$ and $m$ are all the same thing. The old model doesn't have enough room for the right $n^2$ or $2n^2$ degeneracy of the energy levels. Also, those kinds of old models completely fail for more complicated atoms. –  Luboš Motl Aug 1 '12 at 10:25
@LubošMotl: Your comments are false--- the result is not a coincidence, and the quantum numbers l and m were added by Sommerfeld before 1920, in the semiclassical Bohr picture. –  Ron Maimon Aug 2 '12 at 2:16

The Bohr model was extended by Arnold Sommerfeld. Sommerfeld was able to predict all the Hydrogen atom quantum numbers. This theory is referred to today by the name "Old quantum theory". Please see also, Hazhar Ghaderi's Bachelor thesis.

The basic idea of this theory is that closed classical trajectories (not necessarily circular) lead to discrete quantum numbers. The quantization is obtained through modified condition, now known by the name "Bohr-Sommerfeld condition"

The application of this theory becomes difficult for many electron atoms where exact classical solutions are not available.

This theory was abandoned after the discovery of the Schroedinger equation which gave precise predictions for many electron atoms spectra.

Now, we know that the old quantum theory is only an approximation to the correct quantum theory which is valid in general for large quantum numbers (First order in the WKB quantization).

However, for integrable systems, the Bohr-Sommerfeld is known to be exact or almost exact.

However, this is not the end of the story. In 1980 Sniatycki discovered a deep geometrical meaning of the Bohr-Sommerfeld condition within the theory of geometric quantization. Sniatycki's work is described briefly in Matthias Blau's lecture notes.

Heuristically Sniatycki's result states that there exists a coordinate system in which the wave functions are concentrated on subspaces of the phase space where the quantum numbers are fixed to their quantized values. These spaces are called "Bohr-Sommerfeld varieties".

Nowdays, This subject is under active research and was applied sussessfully to complicated quantization problems such as the quantization of the moduli space of flat connections.

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This is a good answer, but I was under the impression that there are classical integrable systems where the Bohr Sommerfeld answer drifts away from the correct answer at large quantum numbers by a more than constant amount, so that E(n) would be off by more than a constant "n" value at large n from the right answer. Is this false? for example, the particle in an $|x|$ potential. –  Ron Maimon Aug 2 '12 at 2:19

You can develop it for all but the spin quantum number, this was done by Sommerfeld.

Justifying the quantum condition

The basic law is the old quantization condition for integrable classical systems:

$$J = \oint p dq = nh = 2\pi n \hbar$$

This rule holds for each periodic motion separately, in cases where the motion is integrable and so multi-periodic. One should justify this rule, and there are two ways, Einstein's and Bohr's.

Bohr's justification is by considering that electromagnetic radiation is quantized, and that you should be able to couple the mechanical system to electromagnetism in arbitrary ways. Imagine that you couple the classical system to an electromagnetic field weakly, say by making the dipole moment be equal to the cosine of one of the angle variables. Then you get outgoing radiation with frequency equal to the orbital period (the outgoing waves are periodic because the coupling is weak, so the backreaction is negiligible). The angle variables are periodic with period $2\pi$, so the classical frequency of the orbit (in terms of orbit radians per second) is equal to the rate of change of the angle variable, which by Hamilton's equations is:

$${d\theta\over dt} = {\partial H\over \partial J} = f$$

Since the outgoing radiation comes in quanta of size hf, the energy carried away by the individual quanta causes losses to lower energy levels, in steps, and the condition for consistency is that the steps are of size $hf$. This means that the steps in energy are in increments given by the above. This clearly works to lowest order when J is integer spaced in unit of h, since:

$$H(J-h) = H(J) - h {\partial H \over \partial J} = H(J) - hf$$

So this justifies the quantum rule for each independent action variable.

Einstein's justification relies on the fact that the action J (the integral above) is adiabatic invariant. This means that if you change the Hamiltonian smoothly and slowly over time (the precise condition is that the Fourier transform of H shouldn't involve the frequencies which are comparable to any of the classical inverse periods), then the motion in the changing Hamiltonian keeps the J constant, so that the J value at the end is the same as the J value at the beginning.

But in quantum mechanics, if you smoothly change H, the energy level must be preserved, since at no point do you provide the frequency required for a transition. This means that whatever quantity you quantize, it should be adiabatic invariant. Planck's law for quantizing radiation tells you that the harmonic oscillator has $J=nh$, and the adiabatic hypothesis means that this rule should be general.

These rules are only precise to give the first-order level spacing, they do not take into account a possible additive constant.

Quantizing a central force--- the orbital and magnetic quantum number

Consider a particle described in polar coordiantes, $r$, $\theta$ and $\phi$, whose Lagrangian is

$$L = \int {1\over 2} (\dot{r}^2 + \dot{\theta}^2 + \cos^2(\theta) \dot{\phi}^2) - V(r)$$

The rotational motion has a conserved angular momentum, and the angular momentum conservation law gives you that $L_z$ is constant. The conjugate variable to $L_z$ is $\phi$, as you can see from the fact that $\phi$ doesn't appear in the action without a time-derivative, so it's conjugate momentum is conserved, and is nothing other than $L_z$.

Applying the quantum condition

$$\int L_z d\phi = 2\pi L_z = mh$$

gives you that the z-component of the angular momentum is quantized in integer multiples of $\hbar$, and the integer is "m" the magnetic quantum number. It is called this, because it gives the magnetic moment of the orbit, assuming no spin, and therefore the spinless Zeeman splitting.

This is not the entire angular motion, however. You need a second action and angle variable for the second variable $\theta$. $\theta$ won't do because it's not an angle. But

$$L^2$$

the length of the angular momentum is also conserved, and the conjugate variable to this quantity is the angle variable of the motion restricted to a plane. This variable I'll also call $\theta$, but it's another $\theta$, it's the angle between the central force origin and the particle when you rotate the orbit to be planar.

The quantum condition for this action-angle pair gives that:

$$2\pi L = l h$$

or that the total angular momentum is quantized in multiples of $\hbar$. From geometry, $m\le l$.

To solve the radial motion, you make an effective hamiltonian for r, by considering the Lagrangian for motion in r alone:

$$\int {1\over 2} \dot{r}^2 - {L^2\over 2r^2} + {1\over r}$$

The way to solve this is to explicitly find the momentum $p_r$ at every position $r$,

$$p_r = \sqrt{2E - {L^2\over r^2} + {1\over r} }$$

Then integrate $p_r$ between the classical turning points

$$2\int p_r dr = kh$$

Where k is a new quantum number. To do the integral, you take out a factor of ${1\over r^2}$ from inside the square root, complete the square (remembering that E is negative, and dimensionalize to get an integral of the form:

$$\int {\sqrt{ 1- u^2} \over u + C} du$$

which is elementary when you transform coordinates using $u=\cos(\theta)$. The result is that you get an infinite sequence of k orbits at any l, of energy

$$E \propto {1\over (k+l)^2}$$

identifying k+l as the principal quantum number n completes the derivation, and provides the restriction $l<n$, except you need to start k off at 1, not zero. This is the proper H spectrum, excluding spin.

Deficiencies

The semiclassical quantization procedure breaks the rotational symmetry. It has ambiguities at the boundaries, and it is generally hokey (does the H-atom ground state have l=0 or l=1? I did it so that it's l=0 above). But it's not arbitrary, and it is guaranteed to be correct to leading order. In the H-atom, it happens to be exact, up to the arbitrariness in the procedure.

In the old days, the spin was thought to be a breakdown in the quantization rule, requiring half-integer values sometimes, until Pauli and others understood the electron carried angular momentum itself.

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