LOG#151. Bohrlogy (I).

The Bohr model of the hydrogen atom is a cool and nice “toy model”. It serves as a prototype in many applications, even if it is not fully “quantum”. It does provide many applications. In this post, and the followings, we will elaborate some of the most unknown uses of this “atomic” model.

Firstly, take the free non-relativistic hamiltonian (energy) for a point particle and add it a potential energy term U(R):

(1) $\begin{equation*} H=E=T+U=T+U(R)=\dfrac{p^2}{2m}+U(R)\end{equation*}$

In order to apply the Bohr model and its quantization conditions in the (almost) most general condition, choose a homogeneous arbitrary potential energy

(2) $\begin{equation*} U(R)=kR^D\end{equation*}$

Homogeneity of the potential energy means, mathematically, that

(3) $\begin{equation*} U(aR)=a^DU(R)\end{equation*}$

The given “UR” potential energy comes from a central force:

(4) $\begin{equation*} F(R)=-\dfrac{dU}{dR}=-kDR^{D-1}\end{equation*}$

Now, we will study the problem of circular orbits and their quantization rule from the Bohr postulates:

i) Angular momentum (or phase space) is quantized: $L=n\hbar=mvR=pR$ , with $n\in \mathbb{Z}^+$

ii) The centripetal force matches the central force providing the given $U(R)$ , since

(5) $\begin{equation*} F_C=F(R)\leftrightarrow\dfrac{mv^2}{R}=kDR^{D-1}\end{equation*}$

(6) $\begin{equation*} mv^2=kDR^D\end{equation*}$

(7) $\begin{equation*} m^2v^2=p^2=mkDR^D\end{equation*}$

and then

(8) $\begin{equation*} \boxed{p^2=mkDR^D}\end{equation*}$

Using the conditions, we get

(9) $\begin{equation*} p^2R^2=n^2\hbar^2\end{equation*}$

(10) $\begin{equation*} mkDR^DR^2=n^2\hbar^2\end{equation*}$

Therefore, the radii of the circular orbits are quantized (space quantization!) as follows

(11) $\begin{equation*} \boxed{R_n(D)=\sqrt[D+2]{\dfrac{n^2\hbar^2}{mkD}}}\end{equation*}$

Momentum is related to radius via

(12) $\begin{equation*} p_n(D)=\dfrac{n\hbar}{R_n(D)}\end{equation*}$

and thus

(13) $\begin{equation*} \boxed{p_n(D)=n\hbar\sqrt[D+2]{\dfrac{mkD}{n^2\hbar^2}}=\sqrt[D+2]{mkD(n\hbar)^D}}\end{equation*}$

Now, we can calculate the energies (note that we could also use the virial theorem for the homogeneous potential energy, but since it is a theorem not usually known by some kind of readers, I will use the brute force approach):

(14) $\begin{equation*} E=T+U=\dfrac{p^2}{2m}+kR^D\end{equation*}$

Inserting the values of the quantized radii and momenta into the energy formula (hamiltonian), we obtain, after some algebra,

(15) $\begin{equation*} E_n(D)=k^{\frac{2}{D+2}}\left(\dfrac{n^2\hbar^2}{m}\right)^{\frac{D}{D+2}}\dfrac{D+2}{2D^{D/D+2}}\end{equation*}$

(16) $\begin{equation*} \boxed{E_n(D)=C(k,D)\left(\dfrac{n^2\hbar^2}{m}\right)^{D/D+2}}\end{equation*}$

where

(17) $\begin{equation*} \boxed{C(k,D)=k^{\frac{2}{D+2}}\dfrac{D+2}{2D^{D/D+2}}=\left(\dfrac{D+2}{2}\right)\dfrac{k^{2/D+2}}{D^{D/D+2}}}\end{equation*}$

Velocities and accelerations are also quantized in this model, and it is something that sometimes is not remarked:

(18) $\begin{equation*} \boxed{v_n(D)=\dfrac{p_n(D)}{m}=\sqrt[D+2]{\dfrac{kD}{m^{D+1}}(n\hbar)^D}=\sqrt[D+2]{\dfrac{kD}{m}\left(\dfrac{n\hbar}{m}\right)^D}}\end{equation*}$

(19) $\begin{equation*} \boxed{a_n(D)=\dfrac{v_n^2}{R_n}=\sqrt[D+2]{\left(\dfrac{kD}{m}\right)^3\left(\dfrac{n^2\hbar^2}{m^2}\right)^{D-1}}=\sqrt[D+2]{\left(\dfrac{kD}{m}\right)^3\left(\dfrac{n\hbar}{m}\right)^{2(D-1)}}}\end{equation*}$

Due to the fact that accelerations are quantized, forces and angular frequencies are also quantized. The quantized forces read

(20) $\begin{equation*} \boxed{F_n(D)=ma_n(D)=\sqrt[D+2]{(kD)^3\left(\dfrac{n^2\hbar^2}{m}\right)^{D-1}}}\end{equation*}$

(21) $\begin{equation*} m\omega_n^2R_n\rightarrow \omega_n^2(D)=\sqrt[D+2]{(kD)^4\dfrac{(n^2\hbar^2)^{D-2}}{m^{2D}}}\end{equation*}$

(22) $\begin{equation*} \omega_n(D)=\sqrt[D+2]{(kD)^2\dfrac{(n\hbar)^{D-2}}{m^D}}\end{equation*}$

The angular frequencies are related to the energies through a pure number:

(23) $\begin{equation*} \dfrac{E_n(D)}{\omega_n(D)}=\Omega_n(D)\end{equation*}$

such as

(24) $\begin{equation*} \boxed{\Omega_n(D)=\left(\dfrac{D+2}{2}\right)\left(\dfrac{n\hbar}{D}\right)=\left(\dfrac{D}{2}+1\right)\left(\dfrac{n\hbar}{D}\right)=\left(\dfrac{D+2}{2}\right)\dfrac{n\hbar}{D}=n\hbar\left(\dfrac{1}{2}+\dfrac{1}{D}\right)}\end{equation*}$

Finally, the periods are also (generally) quantized. Firstly, we will derive the generalized Kepler law for our homogenous potential and central power law for the the force, in the classical sense. Circular orbits satisfy

(25) $\begin{equation*} m\omega^2R=kDR^{D-1}\end{equation*}$

(26) $\begin{equation*} \omega=\dfrac{2\pi}{T}\end{equation*}$

and thus

(27) $\begin{equation*} m\dfrac{4\pi^2}{T^2}R=kDR^{D-1}\end{equation*}$

and hence

(28) $\begin{equation*} \boxed{T^2=\dfrac{4\pi^2m}{kD}R^{2-D}}\end{equation*}$

In the case of an inverse dth-force law (as those related to theories with extra dimensions), we identify

(29) $\begin{equation*} D=-1-d\end{equation*}$

so the previous generalized Kepler third law is rewritten

(30) $\begin{equation*} \boxed{T^2=-\dfrac{4\pi^2m}{k(1+d)}R^{3+d}}\end{equation*}$

and the case $d=0$ , with $k=-GMm$ , correspond to the usual newtonian case. Now, we proceed with the quantization procedure. It yields

(31) $\begin{equation*} \boxed{T_n^2(D)=\dfrac{4\pi^2m}{kD}\left(\dfrac{n^2\hbar^2}{mkD}\right)^{\frac{2-D}{2+D}}}\end{equation*}$

(32) $\begin{equation*} \boxed{T_n(D)=\sqrt{\dfrac{4\pi^2m}{kD}\left(\dfrac{n^2\hbar^2}{mkD}\right)^{\frac{2-D}{2+D}}}}\end{equation*}$

Remark (I): If $k>0$ , repulsive force, $D>0$ in order to provide positive periods. Indeed, there is subtle point here, since $k>0$ with negative $D$ provides attractive force. Otherwise, $k<0$ provides attractive force if $D>0$ and repulsive force if $D<0$ . Moreover, the potential energy $U(R)=kR^D$ is attractive if $k<0$ and repulsive if $k>0$ for any D.

Remark (II): There are 2 special cases for the potential energy. The case with D=-1, the so-called “keplerian” problem (or newtonian/coulombian case), and the case with D=+2, the so-called “harmonic oscillator” case. Other uncommon but known case of power-law potential is the so-called string potential, where D=+1 (not -1, like the keplerian case). The string potential is useful in the quantum bouncing ball problem and the quark confinement.

Remark (III): The case with D=-1-d correspond to a world with “d” spatial extra dimensions. The case with D=+2d corresponds to the case “hyperharmonic”, and the case D=2d-1 to the “hyperstring” (d-brane) potential (string or 1-brane, 2-brane, 3-brane,…). The list of potential energies and the corresponding forces would be

1st. Harmonic oscillator: $U(R)=k_2R^2$ , $F(R)=-2k_2R$

2nd. The hyperharmonic oscillator: $U(R)=k_{2d}R^{2d}$ , $F(R)=-2dk_{2d}R^{2d-1}$

3rd. The keplerian/newtonian/coulombian problem: $U(R)=k_{-1}R^{-1}$ , $F(R)=k_{-1}R^{-2}$

4th. The hyperkeplerian/hypernewtonian/hypercoulombian problem: $U(R)=k_{-1-d}R^{-1-d}$ , $F(R)=(1+d)k_{-1-d}R^{-2-d}$

5th. The string potential: $U(R)=k_1R$ , $F(R)=-k_1$

6th. The hyperstring potential: $U(R)=k_{2d-1}R^{2d-1}$ , $F(R)=-(2d-1)k_{2d-1}R^{2d-2}=-(2d-1)k_{2d-1}R^{2(d-1)}$

Example 1. The harmonic oscillator. We have D=2, and $U(R)=kR^2$ . Please, note that generally the normalization constant is not k but $k/2$ for the potential energy in the harmonic oscillator (D=2). Therefore, we can compute all the above quantities (radius, velocity, momentum, acceleration, force,…) plugging D=2 in the formulae, to provide

$U(R)=kR^2$ , $F(R)=-2kR$

a) $L=n\hbar$

b) $p_n^2=2mkR_n^2$

c) $p_n=\sqrt{2mkR_n}=\sqrt[4]{2mk}(n\hbar)^{1/2}$

d) $R_n=\sqrt[4]{1/2mk}(n\hbar)^{1/2}$

e) $v_n=\sqrt[4]{2k/m^3}(n\hbar)^{1/2}$

f) $a_n=\sqrt[4]{8k^3}{m^5}(n\hbar)^{1/2}$

g) $F_n=ma_n=\sqrt[4]{8k^3/m}(n\hbar)^{1/2}$

h) $E_n=\sqrt{2k/m}n\hbar$

i) $\omega_n=E_n/\hbar=\sqrt{2k}{m}n=\omega_0n$

j) $T_n^2=T^2=\dfrac{2\pi^2m}{k}$ . Curiously, the periods (or the time) are nor quantized for the harmonic oscillator, the period is “constant” somehow but you can consider multiples of it even then it is constant (not quantized) itself from the above.

Example 2. The keplerian hydrogen atom. We write $D=-1$ and $k=-k_Ce^2$ , where $k_C$ is the Coulomb constant and $e$ is the absolute value of the electric charge of a single electron/proton. We define $\alpha=k_Ce^2/\hbar c$ as the fine structure constant, the Bohr radius $a_B=\hbar^2/mk_Ce^2=\lambda_C/\alpha$ and $\lambda_C=\hbar/mc$ as the reduced Compton wavelength of the electron. Thus, we get the quantities