# Uncertainty principle and an estimate of the ground state energy of hydrogen

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 9, Exercise 9.4.3.

The uncertainty principle can be used to get an estimate of the ground state energy in some systems. In his section 9.4, Shankar shows how this is done for the hydrogen atom, treating the system as a proper 3-d system.

A somewhat simpler analysis can be done by treating the hydrogen atom as a one-dimensional system. The Hamiltonian is

$\displaystyle H=\frac{P^{2}}{2m}-\frac{e^{2}}{\left(R^{2}\right)^{1/2}} \ \ \ \ \ (1)$

where ${m}$ and ${e}$ are the mass and charge of the electron. The operators ${P}$ and ${R}$ stand for the 3-d momentum and position:

 $\displaystyle P^{2}$ $\displaystyle =$ $\displaystyle P_{x}^{2}+P_{y}^{2}+P_{z}^{2}\ \ \ \ \ (2)$ $\displaystyle R^{2}$ $\displaystyle =$ $\displaystyle X^{2}+Y^{2}+Z^{2} \ \ \ \ \ (3)$

If we ignore the expansions of ${P^{2}}$ and ${R^{2}}$ and treat the Hamiltonian as a function of the operators ${P}$ and ${R}$ on their own, we can use the uncertainty principle to get a bound on the ground state energy. By analogy with one-dimensional position and momentum, we assume that the uncertainties are related by

$\displaystyle \Delta P\cdot\Delta R\ge\frac{\hbar}{2} \ \ \ \ \ (4)$

By using coordinates such that the hydrogen atom is centred at the origin, and from the spherical symmetry of the ground state, we have

 $\displaystyle \left(\Delta P\right)^{2}$ $\displaystyle =$ $\displaystyle \left\langle P^{2}\right\rangle -\left\langle P\right\rangle ^{2}=\left\langle P^{2}\right\rangle \ \ \ \ \ (5)$ $\displaystyle \left(\Delta R\right)^{2}$ $\displaystyle =$ $\displaystyle \left\langle R^{2}\right\rangle -\left\langle R\right\rangle ^{2}=\left\langle R^{2}\right\rangle \ \ \ \ \ (6)$

We can then write 1 as

 $\displaystyle \left\langle H\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\left\langle P^{2}\right\rangle }{2m}-e^{2}\left\langle \frac{1}{\left(R^{2}\right)^{1/2}}\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle \simeq$ $\displaystyle \frac{\left\langle P^{2}\right\rangle }{2m}-\frac{e^{2}}{\left\langle \sqrt{\left\langle R^{2}\right\rangle }\right\rangle } \ \ \ \ \ (8)$

where in the last line we used an argument similar to that considered earlier, in which we showed that, for a one-dimensional system,

$\displaystyle \left\langle \frac{1}{X^{2}}\right\rangle \simeq\frac{1}{\left\langle X^{2}\right\rangle } \ \ \ \ \ (9)$

where the ${\simeq}$ sign means ‘same order of magnitude’. We can now write the mean of the Hamiltonian in terms of the uncertainties:

 $\displaystyle \left\langle H\right\rangle$ $\displaystyle \simeq$ $\displaystyle \frac{\left(\Delta P\right)^{2}}{2m}-\frac{e^{2}}{\Delta R}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \gtrsim$ $\displaystyle \frac{\hbar^{2}}{8m\left(\Delta R\right)^{2}}-\frac{e^{2}}{\Delta R} \ \ \ \ \ (11)$

We can now minimize ${\left\langle H\right\rangle }$:

 $\displaystyle \frac{\partial\left\langle H\right\rangle }{\partial\left(\Delta R\right)}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{4m\left(\Delta R\right)^{3}}+\frac{e^{2}}{\left(\Delta R\right)^{2}}=0\ \ \ \ \ (12)$ $\displaystyle \Delta R$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4me^{2}} \ \ \ \ \ (13)$

This gives an estimate for the ground state energy of

$\displaystyle \left\langle H\right\rangle _{g.s.}\simeq-\frac{2me^{4}}{\hbar^{2}} \ \ \ \ \ (14)$

The actual value is

$\displaystyle E_{0}=-\frac{me^{4}}{2\hbar^{2}} \ \ \ \ \ (15)$

so our estimate is too large (in magnitude) by a factor of 4. For comparison, the estimate worked out by Shankar for the 3-d case is

$\displaystyle \left\langle H\right\rangle \gtrsim-\frac{2me^{4}}{9\hbar^{2}} \ \ \ \ \ (16)$

This estimate is too small by around a factor of 2.

# Uncertainties in the harmonic oscillator and hydrogen atom

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 9, Exercises 9.4.1 – 9.4.2.

Here we’ll look at a couple of calculations relevant to the application of the uncertainty principle to the hydrogen atom. When calculating uncertainties, we need to find the average values of various quantities. First, we’ll look at an average in the case of the harmonic oscillator.

$\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$

where ${H_{n}}$ is the ${n}$th Hermite polynomial. For ${n=1,}$ we have

$\displaystyle H_{1}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)=2\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (2)$

so

$\displaystyle \psi_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}\left(\frac{m\omega}{\hbar}\right)^{3/4}x\;e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)$

For this state, we can calculate the average

 $\displaystyle \left\langle \frac{1}{X^{2}}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)\frac{1}{x^{2}}dx\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\sqrt{\frac{\pi\hbar}{m\omega}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m\omega}{\hbar} \ \ \ \ \ (7)$

where we evaluated the Gaussian integral in the second line.

We can compare this to ${1/\left\langle X^{2}\right\rangle }$ as follows:

 $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)x^{2}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}x^{4}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\frac{3\sqrt{\pi}}{4}\left(\frac{\hbar}{m\omega}\right)^{5/2}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}\frac{\hbar}{m\omega}\ \ \ \ \ (11)$ $\displaystyle \frac{1}{\left\langle X^{2}\right\rangle }$ $\displaystyle =$ $\displaystyle \frac{2}{3}\frac{m\omega}{\hbar} \ \ \ \ \ (12)$

Thus ${\left\langle \frac{1}{X^{2}}\right\rangle }$ and ${\frac{1}{\left\langle X^{2}\right\rangle }}$ have the same order of magnitude, although they are not equal.

In three dimensions, we consider the ground state of hydrogen

$\displaystyle \psi_{100}\left(r\right)=\frac{1}{\sqrt{\pi}a_{0}^{3/2}}e^{-r/a_{0}} \ \ \ \ \ (13)$

where ${a_{0}}$ is the Bohr radius

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (14)$

with ${m}$ and ${e}$ being the mass and charge of the electron. The wave function is normalized as we can see by doing the integral (in 3 dimensions):

 $\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{2}dr \ \ \ \ \ (15)$

We can use the formula (given in Shankar’s Appendix 2)

$\displaystyle \int_{0}^{\infty}e^{-r/\alpha}r^{n}dr=\frac{n!}{\alpha^{n+1}} \ \ \ \ \ (16)$

We get

$\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}=\frac{4\pi}{\pi a_{0}^{3}}\frac{2!}{2^{3}}a_{0}^{3}=1 \ \ \ \ \ (17)$

as required.

For a spherically symmetric wave function centred at ${r=0}$,

$\displaystyle \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (18)$

with identical relations for ${Y}$ and ${Z}$. Since

 $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle x^{2}+y^{2}+z^{2}\ \ \ \ \ (19)$ $\displaystyle \left\langle r^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle x^{2}\right\rangle +\left\langle y^{2}\right\rangle +\left\langle z^{2}\right\rangle =3\left\langle X^{2}\right\rangle \ \ \ \ \ (20)$ $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\left\langle r^{2}\right\rangle \ \ \ \ \ (21)$

Thus

 $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\int\psi_{100}^{2}(r)r^{2}d^{3}\mathbf{r}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{3\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{4}dr\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{3a_{0}^{3}}\frac{4!}{2^{5}}a_{0}^{5}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{0}^{2}\ \ \ \ \ (25)$ $\displaystyle \Delta X$ $\displaystyle =$ $\displaystyle a_{0}=\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (26)$

We can also find

 $\displaystyle \left\langle \frac{1}{r}\right\rangle$ $\displaystyle =$ $\displaystyle \int\psi_{100}^{2}(r)\frac{1}{r}d^{3}\mathbf{r}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r\;dr\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{a_{0}^{3}}\frac{a_{0}^{2}}{4}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{a_{0}}\ \ \ \ \ (30)$ $\displaystyle \left\langle r\right\rangle$ $\displaystyle =$ $\displaystyle \int\psi_{100}^{2}(r)r\;d^{3}\mathbf{r}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{3}dr\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{a_{0}^{3}}\frac{6a_{0}^{4}}{16}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}a_{0} \ \ \ \ \ (34)$

Thus both ${\left\langle \frac{1}{r}\right\rangle }$ and ${\frac{1}{\left\langle r\right\rangle }}$ are of the same order of magnitude as ${1/a_{0}=me^{2}/\hbar^{2}}$.

# Helmholtz energy of a hydrogen atom

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.20.

The Helmholtz energy is defined as

$\displaystyle F\equiv U-TS \ \ \ \ \ (1)$

We can apply this to a hydrogen atom, at least in an approximate form. The ground state of hydrogen has quantum numbers ${n=1}$, ${\ell=m=0}$ and is non-degenerate. The first excited state is four-fold degenerate, as its quantum numbers are ${n=2}$ with ${\ell=m=0}$ or ${\ell=1}$ and ${m=\pm1,0}$. We can therefore say that the entropy of the ${n=2}$ state is

$\displaystyle S=k\ln4 \ \ \ \ \ (2)$

with ${k=8.62\times10^{-5}\mbox{ eV K}^{-1}}$.

If we take the energy of the ground state to be zero, then the first excited state has ${U=10.2\mbox{ eV}}$. The Helmholtz energy is therefore zero at a temperature of

$\displaystyle T_{0}=\frac{U}{S}=\frac{10.2}{\left(8.62\times10^{-5}\right)\ln4}=8.54\times10^{4}\mbox{ K} \ \ \ \ \ (3)$

For ${T>T_{0}}$, ${F<0}$ so the excited state is actually the preferred state, and a hydrogen atom in the ground state would spontaneously make the transition to the excited state. However, ${T_{0}}$ is so large that virtually all hydrogen atoms would be ionized at that temperature, as we saw earlier when discussing the Saha equation for stellar atmospheres. We saw there that the fraction of hydrogen atoms that are ionized is essentially 1.0 for temperatures above around 12000 K.

# Hydrogen spectrum: Lyman, Balmer and Paschen series

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problems 5.10 – 5.11.

$\displaystyle E_{n}=-\left(\frac{e^{2}}{4\pi\epsilon_{0}}\right)^{2}\frac{\mu}{2n^{2}\hbar^{2}}=-\frac{13.6\mbox{ eV}}{n^{2}} \ \ \ \ \ (1)$

When the electron jumps between these levels it absorbs (when jumping to a higher level) or emits (lower) a photon with an energy ${h\nu}$ equal to the difference between the energy levels. This gives rise to a characteristic spectrum which can be observed in stars.

For example, when an electron makes the transition from ${n=3}$ to ${n=1}$, it can do so directly, or via the pair of transitions ${3\rightarrow2\rightarrow1}$. In the former case

$\displaystyle E_{3\rightarrow1}=13.6\mbox{ eV}\left(1-\frac{1}{3^{2}}\right)=12.09\mbox{ eV} \ \ \ \ \ (2)$

which corresponds to a wavelength of

$\displaystyle \lambda_{3\rightarrow1}=\frac{c}{\nu}=\frac{hc}{E}=\frac{1240\mbox{ eV nm}}{12.09\mbox{ eV}}=102.6\mbox{ nm} \ \ \ \ \ (3)$

This is in the ultraviolet.

The pair of transitions gives

 $\displaystyle E_{3\rightarrow2}$ $\displaystyle =$ $\displaystyle 13.6\mbox{ eV}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.89\mbox{ eV}\ \ \ \ \ (4)$ $\displaystyle \lambda_{3\rightarrow2}$ $\displaystyle =$ $\displaystyle \frac{1240}{1.89}=656.47\mbox{ nm (visible)}\ \ \ \ \ (5)$ $\displaystyle E_{2\rightarrow1}$ $\displaystyle =$ $\displaystyle 13.6\mbox{ eV}\left(1-\frac{1}{2^{2}}\right)=10.2\mbox{ eV}\ \ \ \ \ (6)$ $\displaystyle \lambda_{2\rightarrow1}$ $\displaystyle =$ $\displaystyle \frac{1240}{10.2}=121.6\mbox{ nm (ultraviolet)} \ \ \ \ \ (7)$

Transitions to and from ${n=1}$ give the Lyman series (all in the ultraviolet); to and from ${n=2}$ (from higher energy levels) give the Balmer series (visible) and to and from ${n=3}$ (from higher energy levels) give the Paschen series (infrared). For all series, the most energetic photons come from transitions from essentially infinite ${n}$ to the base level. For Lyman, for example,

$\displaystyle E_{n\rightarrow1}=13.6\mbox{ eV}\left(1-\frac{1}{n^{2}}\right) \ \ \ \ \ (8)$

so the series limit occurs as ${n\rightarrow\infty}$ and is 13.6 eV. This corresponds to a wavelength of

$\displaystyle \lambda_{1}=\frac{1240}{13.6}=91.18\mbox{ nm (ultraviolet)} \ \ \ \ \ (9)$

For Balmer, the series limit is

 $\displaystyle E_{n\rightarrow2}$ $\displaystyle =$ $\displaystyle 13.6\mbox{ eV}\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)\rightarrow3.4\mbox{ eV}\ \ \ \ \ (10)$ $\displaystyle \lambda_{2}$ $\displaystyle =$ $\displaystyle \frac{1240}{3.4}=364.7\mbox{ nm (slightly ultraviolet)} \ \ \ \ \ (11)$

For Paschen,

 $\displaystyle E_{n\rightarrow3}$ $\displaystyle =$ $\displaystyle 13.6\mbox{ eV}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)\rightarrow1.51\mbox{ eV}\ \ \ \ \ (12)$ $\displaystyle \lambda_{3}$ $\displaystyle =$ $\displaystyle \frac{1240}{1.51}=820.6\mbox{ nm (slightly infrared)} \ \ \ \ \ (13)$

# Gravitational hydrogen atom

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problem 5.9.

To see that gravity is much weaker than the electrostatic force, we can repeat Bohr’s semi-classical derivation of the hydrogen energy levels, replacing the Coulomb force with the Newtonian gravitational force. We can do this with the following replacement:

$\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\rightarrow Gm_{e}m_{p} \ \ \ \ \ (1)$

giving energy levels of

 $\displaystyle E$ $\displaystyle =$ $\displaystyle -\left(Gm_{e}m_{p}\right)^{2}\frac{m_{e}}{2n^{2}\hbar^{2}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{4.23\times10^{-97}\mbox{ J}}{n^{2}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2.64\times10^{-78}\mbox{ eV}}{n^{2}} \ \ \ \ \ (4)$

compared to the actual energy levels of hydrogen:

$\displaystyle E=-\frac{13.6\mbox{ eV}}{n^{2}} \ \ \ \ \ (5)$

 $\displaystyle r_{n}$ $\displaystyle =$ $\displaystyle \frac{n^{2}\hbar^{2}}{Gm_{e}^{2}m_{p}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1.2\times10^{29}\mbox{ m}\right)n^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1.2\times10^{38}\mbox{ nm}\right)n^{2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(8\times10^{17}\mbox{ AU}\right)n^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1.27\times10^{13}\mbox{ ly}\right)n^{2} \ \ \ \ \ (10)$

Thus the ground state radius of gravitational hydrogen is many times larger than the visible universe, compared with the electrostatic Bohr radius of ${5.29177\times10^{-11}\mbox{ m}}$.

# Energy levels of hydrogen: Bohr’s semi-classical derivation

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problem 5.8.

We’ve seen how to derive the Bohr formula for the energy levels in hydrogen by solving the Schrödinger equation, but this isn’t the way Bohr originally derived the formula; in fact, he couldn’t have done it that way since he arrived at the formula before Schrödinger came up with his equation.

Although Bohr’s derivation isn’t correct, as it relies a lot on classical physics, it’s interesting to see how he did it. His idea was to take the electron as a classical particle in a circular orbit around the proton, and equate the Coulomb force of attraction with the centripetal force required to keep the electron in its orbit. In the centre of mass frame, this gives

$\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}r^{2}}=\frac{\mu v^{2}}{r} \ \ \ \ \ (1)$

where ${e}$ is the proton charge, ${\mu=m_{p}m_{e}/\left(m_{p}+m_{e}\right)}$ is the reduced mass, ${v}$ is the velocity of the electron in its orbit and ${r}$ is the separation of the electron and proton. The kinetic energy is therefore

$\displaystyle K=\frac{\mu v^{2}}{2}=\frac{e^{2}}{8\pi\epsilon_{0}r}=-\frac{U}{2} \ \ \ \ \ (2)$

where ${U}$ is the potential energy (this agrees with the virial theorem, where the binding force is electrostatic rather than gravitational). Therefore the total energy of the hydrogen atom is

$\displaystyle E=K+U=\frac{U}{2}=-\frac{e^{2}}{8\pi\epsilon_{0}r} \ \ \ \ \ (3)$

This is entirely classical physics, but it is at this point that Bohr introduced the quantum assumption. He proposed that the angular momentum of the system is quantized in units of Planck’s constant, so that the only allowed values are

$\displaystyle L=\mu rv=n\hbar \ \ \ \ \ (4)$

for some positive integer ${n}$. From 1, this gives

 $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}$ $\displaystyle =$ $\displaystyle \frac{L^{2}}{\mu r}=\frac{n^{2}\hbar^{2}}{\mu r}\ \ \ \ \ (5)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{8\pi\epsilon_{0}r}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\frac{e^{2}}{4\pi\epsilon_{0}}\right)^{2}\frac{\mu}{2n^{2}\hbar^{2}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{13.6\mbox{ eV}}{n^{2}} \ \ \ \ \ (8)$

As our quantum mechanical derivation assumed a stationary proton (and since ${m_{e}\ll m_{p}}$), ${\mu\approx m_{e}}$ and we get the Bohr formula for the energy levels of hydrogen. Although this derivation is obviously a lot simpler than the series method we had to use to solve the Schrödinger equation, it’s also obviously not correct, as it assumes that the electron is a solid particle in a fixed orbit about the proton. However, it seems a bit too much of a coincidence that the semi-classical derivation gives the right answer.

The derivation is easily generalized to an electron orbiting a nucleus containing ${Z}$ protons, since we just replace ${e^{2}}$ by ${Ze^{2}}$ to get

 $\displaystyle E_{Z}$ $\displaystyle =$ $\displaystyle -\left(\frac{Ze^{2}}{4\pi\epsilon_{0}}\right)^{2}\frac{\mu}{2n^{2}\hbar^{2}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -13.6\mbox{ eV}\frac{Z^{2}}{n^{2}} \ \ \ \ \ (10)$

The radius of the orbit in the ${n}$th quantum state (assuming the classical model) is, from 5

$\displaystyle r_{n}=\frac{4\pi\epsilon_{0}n^{2}\hbar^{2}}{\mu Ze^{2}} \ \ \ \ \ (11)$

With ${Z=n=1}$, this gives the Bohr radius:

$\displaystyle a=\frac{4\pi\epsilon_{0}\hbar^{2}}{\mu e^{2}}=5.29177\times10^{-11}\mbox{ m} \ \ \ \ \ (12)$

For ionized helium, ${Z=2}$ and the ground state radius is

$\displaystyle r=\frac{a}{2}=2.645885\times10^{-11}\mbox{ m} \ \ \ \ \ (13)$

The ground state energy is

$\displaystyle E=4\left(-13.6\mbox{ eV}\right)=-54.4\mbox{ eV} \ \ \ \ \ (14)$

For doubly ionized lithium, ${Z=3}$ so

 $\displaystyle r$ $\displaystyle =$ $\displaystyle \frac{a}{3}=1.763923\times10^{-11}\mbox{ m}\ \ \ \ \ (15)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle 9\left(-13.6\mbox{ eV}\right)=-122.4\mbox{ eV} \ \ \ \ \ (16)$

# Integral form of the Schrödinger equation: ground state of hydrogen

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.9.

We can check that the ground state of the hydrogen atom satisfies the integral form of the Schrödinger equation:

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)-\frac{m}{2\pi\hbar^{2}}\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (1)$

For the ground state of hydrogen

 $\displaystyle \psi_{100}\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}\ \ \ \ \ (2)$ $\displaystyle a$ $\displaystyle \equiv$ $\displaystyle \frac{4\pi\epsilon_{0}\hbar^{2}}{me^{2}}\ \ \ \ \ (3)$ $\displaystyle V\left(r\right)$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{4\pi\epsilon_{0}r}=-\frac{\hbar^{2}}{mar}\ \ \ \ \ (4)$ $\displaystyle E_{1}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2ma^{2}}\ \ \ \ \ (5)$ $\displaystyle k$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2mE_{1}}}{\hbar}=\frac{i}{a} \ \ \ \ \ (6)$

In this case, we’re not considering scattering, so the incident plane wave (the free particle) is not present, so ${\psi_{0}=0}$ in 1, and the integral equation becomes

 $\displaystyle \psi\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle -\frac{m}{2\pi\hbar^{2}}\left(-\frac{\hbar^{2}}{ma}\right)\frac{1}{\sqrt{\pi a^{3}}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{e^{-\left|\mathbf{r}-\mathbf{r}_{0}\right|/a}e^{-r_{0}/a}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|r_{0}}\sin\theta r_{0}^{2}d\phi d\theta dr_{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi^{3/2}a^{5/2}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{e^{-\left|\mathbf{r}-\mathbf{r}_{0}\right|/a}e^{-r_{0}/a}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}\sin\theta r_{0}d\phi d\theta dr_{0}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\pi a^{5}}}\int_{0}^{\infty}\int_{0}^{\pi}\frac{e^{-\left|\mathbf{r}-\mathbf{r}_{0}\right|/a}e^{-r_{0}/a}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}\sin\theta r_{0}d\theta dr_{0} \ \ \ \ \ (9)$

where we’ve taken the ${z}$ axis to be parallel to ${\mathbf{r}}$, since for the purposes of the integral, ${\mathbf{r}}$ is constant.

We have

$\displaystyle \left|\mathbf{r}-\mathbf{r}_{0}\right|=\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta} \ \ \ \ \ (10)$

so the integral becomes

$\displaystyle \frac{1}{\sqrt{\pi a^{5}}}\int_{0}^{\infty}\int_{0}^{\pi}\frac{e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}e^{-r_{0}/a}}{\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}}\sin\theta r_{0}d\theta dr_{0}=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a} \ \ \ \ \ (11)$

where we did the integral using Maple. This is just the original wave function 2 so the integral equation works out.

If you want to do the integral by hand, we do the ${\theta}$ integral first since, despite its appearance, it’s actually quite simple:

$\displaystyle \int_{0}^{\pi}\frac{e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}e^{-r_{0}/a}}{\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}}\sin\theta r_{0}d\theta=-\frac{ae^{-r_{0}/a}}{r}\left.e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}\right|_{0}^{\pi} \ \ \ \ \ (12)$

The value of the integral depends on whether ${r or ${r>r_{0}}$:

$\displaystyle -\frac{ae^{-r_{0}/a}}{r}\left.e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}\right|_{0}^{\pi}=\begin{cases} \frac{a}{r}\left(e^{2r/a}-1\right)e^{-\left(2r_{0}+r\right)/a} & rr_{0} \end{cases} \ \ \ \ \ (13)$

Using these results, we can split the integral over ${r_{0}}$ into two parts (0 to ${r}$ and ${r}$ to ${\infty}$). It is just a simple integral over exponential functions so the answer comes out fairly easily.

# Stark effect in hydrogen for n = 1 and n = 2

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.36.

The Zeeman effect occurs when an atom is placed in an external magnetic field, resulting in the interaction between field and the magnetic dipole moments of the atom causing splitting of the energy levels. The electrical analogue of the Zeeman effect, when an atom is placed in an external electric field, is called the Stark effect. We can use perturbation theory to analyze the effect on the energy levels of the electron.

The perturbation hamiltonian is, assuming the electric field points in the ${z}$ direction:

$\displaystyle H_{S}^{\prime}=eE_{ext}z=eE_{ext}r\cos\theta \ \ \ \ \ (1)$

To use perturbation theory, we’ll need the wave functions for unperturbed hydrogen, which are given in Griffiths as equation 4.89. For the ground state ${n=1}$, we have

$\displaystyle \left|100\right\rangle =\frac{2}{a^{3/2}}\frac{1}{\sqrt{4\pi}}e^{-r/a} \ \ \ \ \ (2)$

Since the ground state is non-degenerate, we can use non-degenerate perturbation theory:

 $\displaystyle E_{100,1}$ $\displaystyle =$ $\displaystyle \left\langle 100\right|H_{S}^{\prime}\left|100\right\rangle \ \ \ \ \ (3)$

Rather than writing out the integral, we observe that ${\left\langle 100\right|H_{S}^{\prime}\left|100\right\rangle }$ contains the integral of ${\cos\theta\sin\theta=\frac{1}{2}\sin2\theta}$ over ${\theta=0,..,\pi}$ which is zero, so ${E_{100,1}=0}$.

To analyze ${n=2}$, we need the four wave functions:

 $\displaystyle \left|200\right\rangle$ $\displaystyle =$ $\displaystyle R_{20}\left(r\right)Y_{0}^{0}\left(\theta,\phi\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}a^{3/2}}\left(1-\frac{r}{2a}\right)\frac{1}{\sqrt{4\pi}}e^{-r/2a}\ \ \ \ \ (5)$ $\displaystyle \left|211\right\rangle$ $\displaystyle =$ $\displaystyle R_{21}\left(r\right)Y_{1}^{1}\left(\theta,\phi\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\frac{3}{8\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}re^{-r/2a}\sin\theta e^{i\phi}\ \ \ \ \ (7)$ $\displaystyle \left|210\right\rangle$ $\displaystyle =$ $\displaystyle R_{21}\left(r\right)Y_{1}^{0}\left(\theta,\phi\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{3}{4\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}re^{-r/2a}\cos\theta\ \ \ \ \ (9)$ $\displaystyle \left|21-1\right\rangle$ $\displaystyle =$ $\displaystyle R_{21}\left(r\right)Y_{1}^{-1}\left(\theta,\phi\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{3}{8\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}re^{-r/2a}\sin\theta e^{-i\phi} \ \ \ \ \ (11)$

Since all four of these states have the same unperturbed energy, we need to use degenerate perturbation theory, so we’ll need to find the matrix ${W}$ with elements

$\displaystyle W_{a,b}=\left\langle a\right|H_{S}^{\prime}\left|b\right\rangle \ \ \ \ \ (12)$

where ${a}$ and ${b}$ represent one of the four states above.

First, we’ll look at the ${\theta}$ integrals. All matrix elements involve integrals of the form (remember that ${H_{S}^{\prime}}$ always contributes a ${\cos\theta}$ and the spherical volume element always contributes a ${\sin\theta}$):

$\displaystyle I_{nm}=\int_{0}^{\pi}\sin^{n}\theta\cos^{m}\theta d\theta \ \ \ \ \ (13)$

For the possible values of ${n}$ and ${m}$ in this problem, the only non-zero integrals of this form are

 $\displaystyle I_{12}$ $\displaystyle =$ $\displaystyle \frac{2}{3}\ \ \ \ \ (14)$ $\displaystyle I_{22}$ $\displaystyle =$ $\displaystyle \frac{\pi}{8} \ \ \ \ \ (15)$

${I_{12}}$ arises in ${\left\langle 200\right|H_{S}^{\prime}\left|210\right\rangle }$ (and its transpose) and ${I_{22}}$ arises in ${\left\langle 211\right|H_{S}^{\prime}\left|210\right\rangle }$ and ${\left\langle 210\right|H_{S}^{\prime}\left|21-1\right\rangle }$ (and their transposes). Thus these are the only possible non-zero entries in ${W}$. However, ${\left\langle 211\right|H_{S}^{\prime}\left|210\right\rangle }$ and ${\left\langle 210\right|H_{S}^{\prime}\left|21-1\right\rangle }$ involve integrating ${e^{\pm i\phi}}$ over ${\phi=0..2\pi}$ which gives zero. Thus the only non-zero matrix elements are ${\left\langle 200\right|H_{S}^{\prime}\left|210\right\rangle }$ (and its transpose). This gives

 $\displaystyle \left\langle 200\right|H_{S}^{\prime}\left|210\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}a^{3/2}}\left(\frac{3}{4\pi}\right)^{1/2}\frac{1}{\sqrt{24}a^{5/2}}\frac{1}{\sqrt{4\pi}}eE_{ext}\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\left(1-\frac{r}{2a}\right)re^{-r/a}\cos^{2}\theta r^{2}\sin\theta d\phi d\theta dr\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -3aeE_{ext} \ \ \ \ \ (17)$

(The integral can be done with software, or by hand using integration by parts.) The matrix ${W}$ is therefore

$\displaystyle W=\left[\begin{array}{cccc} 0 & 0 & -3aeE_{ext} & 0\\ 0 & 0 & 0 & 0\\ -3aeE_{ext} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (18)$

The eigenvalues are 0, 0, ${\pm3aeE_{ext}}$ so the ${n=2}$ state splits into 3 states, one with energy ${E_{2,0}}$ (degeneracy 2) and two with energies ${E_{2,0}\pm3aeE_{ext}}$ (each with degeneracy 1). The eigenvectors are ${\left[0,1,0,0\right]}$ and ${\left[0,0,0,1\right]}$ for eigenvalue 0, ${\left[-1,0,1,0\right]}$ for ${3aeE_{ext}}$ and ${\left[1,0,1,0\right]}$ for ${-3aeE_{ext}}$. Thus the ‘good’ states are

$\displaystyle \left|211\right\rangle ,\left|21-1\right\rangle ,\frac{1}{\sqrt{2}}\left(-\left|200\right\rangle +\left|210\right\rangle \right),\frac{1}{\sqrt{2}}\left(\left|200\right\rangle +\left|210\right\rangle \right) \ \ \ \ \ (19)$

The electric dipole moment of hydrogen is (treating the proton and electron as point charges):

 $\displaystyle \mathbf{p}$ $\displaystyle =$ $\displaystyle -e\mathbf{r}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -er\left(\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}}\right) \ \ \ \ \ (21)$

We can work out the expectation value of ${\mathbf{p}}$ in each of the ‘good’ states by straightforward integration: ${\left\langle \mathbf{p}\right\rangle =\left\langle a\right|\mathbf{p}\left|a\right\rangle }$ where ${a}$ stands for one of the ‘good’ states. Note that if ${a=\left|211\right\rangle }$ or ${a=\left|21-1\right\rangle }$, then ${\left\langle a\right|\mathbf{p}\left|a\right\rangle }$ has only a ${z}$ component that is non-zero, since the complex exponentials in ${\phi}$ cancel out and the integral of ${\sin\phi}$ or ${\cos\phi}$ in the ${x}$ or ${y}$ components is zero. Similarly, if ${a=\frac{1}{\sqrt{2}}\left(-\left|200\right\rangle +\left|210\right\rangle \right)}$ or ${a=\frac{1}{\sqrt{2}}\left(\left|200\right\rangle +\left|210\right\rangle \right)}$, the ${x}$ and ${y}$ components are again zero, since these wave functions are independent of ${\phi}$ so the integral of ${\sin\phi}$ or ${\cos\phi}$ in the ${x}$ or ${y}$ components gives zero again. Therefore, ${\left\langle \mathbf{p}\right\rangle }$ is always in the ${z}$ direction, and can be calculated from

$\displaystyle \left\langle \mathbf{p}\right\rangle =-e\left\langle a\right|r\cos\theta\left|a\right\rangle \hat{\mathbf{z}} \ \ \ \ \ (22)$

Doing the integrals results in

$\displaystyle \left\langle \mathbf{p}\right\rangle =0,0,3a\hat{\mathbf{z}},-3a\hat{\mathbf{z}} \ \ \ \ \ (23)$

respectively.

# Radiative decay of the Bohr hydrogen atom

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 11, Post 14.

The instantaneous power radiated by an accelerating point charge is given by the Larmor formula (valid for a charge moving at a speed ${v\ll c}$):

$\displaystyle P=\frac{\mu_{0}q^{2}a^{2}}{6\pi c} \ \ \ \ \ (1)$

One historic application of this formula was to Bohr’s early model of the hydrogen atom as an electron in a classical circular orbit around the proton, with the centripetal force provided by the Coulomb attraction. Since a particle moving in a circle is accelerating, it will radiate away energy, so its orbit should eventually decay until the electron crashes into the proton. This classical instability of atoms was one motivation behind the introduction of quantum theory, but that’s another story. Here, we’ll investigate how long it would take a Bohr hydrogen atom to decay.

First, we need to reassure ourselves that the electron is moving non-relativistically. From equating centripetal and Coulomb forces, we have

 $\displaystyle \frac{mv^{2}}{r}$ $\displaystyle =$ $\displaystyle \frac{q^{2}}{4\pi\epsilon_{0}r^{2}}\ \ \ \ \ (2)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{q}{\sqrt{4\pi\epsilon_{0}mr}}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle qc\sqrt{\frac{\mu_{0}}{4\pi mr}} \ \ \ \ \ (4)$

using ${c=1/\sqrt{\mu_{0}\epsilon_{0}}}$. Plugging in the numbers we get

$\displaystyle \frac{v}{c}=\frac{5.3\times10^{-8}}{\sqrt{r}} \ \ \ \ \ (5)$

The Bohr radius is ${a=5\times10^{-11}\mbox{ m}}$ so at that radius ${v/c=0.0075}$ so we’re safe here. As ${r}$ gets smaller, of course, ${v}$ will increase but since the dependence is on the square root, the rate of increase of ${v}$ is fairly small, so that even when ${r=a/100}$, ${v}$ has increased only to ${0.075c}$. So for most of its journey towards the proton, the electron is moving non-relativistically.

To work out how long it takes for the decay to occur, consider the energy radiated during a time ${dt}$, which is ${P\; dt}$. From conservation of energy, this must be equal to the amount of energy lost by the electron. The total energy of the electron is its kinetic plus potential energy, so

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{1}{2}mv^{2}-\frac{q^{2}}{4\pi\epsilon_{0}r}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q^{2}c^{2}\mu_{0}}{8\pi r}-\frac{q^{2}c^{2}\mu_{0}}{4\pi r}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q^{2}c^{2}\mu_{0}}{8\pi r} \ \ \ \ \ (8)$

Therefore, the energy lost is

$\displaystyle dE=-\frac{q^{2}c^{2}\mu_{0}}{8\pi r^{2}}dr \ \ \ \ \ (9)$

[We’ve taken ${dE}$ as negative, since the electron loses energy.] Putting these results together, we get

 $\displaystyle Pdt$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q^{2}a^{2}}{6\pi c}dt\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q^{2}c^{2}\mu_{0}}{8\pi r^{2}}dr\ \ \ \ \ (11)$ $\displaystyle dt$ $\displaystyle =$ $\displaystyle -\frac{3c^{3}}{4r^{2}a^{2}}dr \ \ \ \ \ (12)$

We need ${a}$ to solve this, but this is just the centripetal acceleration, so

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{v^{2}}{r}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q^{2}c^{2}\mu_{0}}{4\pi mr^{2}} \ \ \ \ \ (14)$

Therefore,

 $\displaystyle dt$ $\displaystyle =$ $\displaystyle -\frac{12\pi^{2}m^{2}}{q^{4}c\mu_{0}^{2}}r^{2}dr\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -3.159\times10^{20}r^{2}dr\ \ \ \ \ (16)$ $\displaystyle \int_{0}^{T}dt$ $\displaystyle =$ $\displaystyle -3.159\times10^{20}\int_{a}^{0}r^{2}dr\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\left(3.159\times10^{20}\right)\left(5\times10^{-11}\right)^{3}\ \ \ \ \ (18)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 1.31\times10^{-11}\mbox{ s} \ \ \ \ \ (19)$

Thus in classical electrodynamics, the hydrogen atom is so unstable that it would decay in a tiny fraction of a second. We should be grateful that quantum mechanics saves the universe.

# Spontaneous emission rates for hydrogen: general solution

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 9.22.

The spontaneous emission rate for a charge ${q}$ (such as an electron in an excited state in an atom) in a starting state ${\left|a\right\rangle }$ decaying to a state ${\left|b\right\rangle }$ is, from Einstein’s argument, and averaged over all propagation and polarization directions

$\displaystyle A=\frac{\omega_{0}^{3}\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{3\epsilon_{0}\pi\hbar c^{3}} \ \ \ \ \ (1)$

where ${\boldsymbol{\mathfrak{p}}}$ is the matrix element of the dipole moment

$\displaystyle \boldsymbol{\mathfrak{p}}=q\left\langle b\left|\mathbf{r}\right|a\right\rangle \ \ \ \ \ (2)$

For spherically symmetric potentials such as that for the hydrogen atom, we found that transitions are allowed from a state ${\left|n\ell m\right\rangle }$ only for ${\ell\rightarrow\ell\pm1}$ and ${m\rightarrow m,m\pm1}$. The hydrogen wave function in its most general form is

$\displaystyle \psi_{n\ell m}=R_{n}\left(r\right)Y_{\ell}^{m}\left(\theta,\phi\right) \ \ \ \ \ (3)$

where ${R_{n}}$ is the radial function and ${Y_{\ell}^{m}}$ is a spherical harmonic. To work out the transition rates we need to work out the matrix elements using these functions, subject to the selection rules given above. That is, we need the matrix elements for each of the three coordinates ${x}$, ${y}$ and ${z}$ between the starting state ${\left|n\ell m\right\rangle }$ and finishing state ${\left|n'\ell'm'\right\rangle }$. We won’t bother working out the ${R_{n}}$ part of these matrix elements; rather we’ll just leave this bit as an integral.

In order to get the total transition rate from a state ${\left|n\ell m\right\rangle }$ to a state with ${\ell'=\ell+1}$, we must work out the transition rates to all possible final values of ${m'}$ and add them up.

Fortunately, there are some formulas for the integrals of spherical harmonics that we will find very useful, so we’ll quote them here for reference.

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{0}^{0}\left(\theta\phi\right)Y_{\ell}^{-m}\left(\theta\phi\right)d\theta d\phi$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{4\pi}}\ \ \ \ \ (4)$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{1}^{0}\left(\theta\phi\right)Y_{\ell+1}^{-m}\left(\theta\phi\right)d\theta d\phi$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\sqrt{\frac{\left(\ell+m+1\right)\left(\ell-m+1\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}}\ \ \ \ \ (5)$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{1}^{1}\left(\theta\phi\right)Y_{\ell+1}^{-\left(m+1\right)}\left(\theta\phi\right)d\theta d\phi$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}}\sqrt{\frac{\left(\ell+m+1\right)\left(\ell+m+2\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}}\ \ \ \ \ (6)$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{1}^{1}\left(\theta\phi\right)Y_{\ell-1}^{-\left(m+1\right)}\left(\theta\phi\right)d\theta d\phi$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{3}{8\pi}}\sqrt{\frac{\left(\ell-m\right)\left(\ell-m-1\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}} \ \ \ \ \ (7)$

One other formula will come in handy:

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell_{1}}^{m_{1}}\left(\theta\phi\right)Y_{\ell_{2}}^{m_{2}}\left(\theta\phi\right)Y_{\ell_{3}}^{-m_{3}}\left(\theta\phi\right)d\theta d\phi=0\mbox{ if }m_{1}+m_{2}\ne m_{3} \ \ \ \ \ (8)$

To make use of these formulas, we note that the matrix elements we need to work out are ${\left\langle n'\ell'm'\left|x\right|n\ell m\right\rangle ,\left\langle n'\ell'm'\left|y\right|n\ell m\right\rangle }$ and ${\left\langle n'\ell'm'\left|z\right|n\ell m\right\rangle }$. In spherical coordinates, we have

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (9)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (10)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta \ \ \ \ \ (11)$

These can be written in terms of spherical harmonics:

 $\displaystyle x+iy$ $\displaystyle =$ $\displaystyle r\sin\theta e^{i\phi}=-r\sqrt{\frac{8\pi}{3}}Y_{1}^{1}\ \ \ \ \ (12)$ $\displaystyle x-iy$ $\displaystyle =$ $\displaystyle r\sqrt{\frac{8\pi}{3}}Y_{1}^{-1}\ \ \ \ \ (13)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\sqrt{\frac{4\pi}{3}}Y_{1}^{0} \ \ \ \ \ (14)$

For ${\ell'=\ell+1}$, the only non-zero matrix elements are (we can eliminate the zero elements using 8):

 $\displaystyle \left\langle n',\ell+1,m\left|z\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle I_{\ell+1}\left\langle \ell+1,m\left|\sqrt{\frac{4\pi}{3}}Y_{1}^{0}\right|\ell m\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I_{\ell+1}\sqrt{\frac{4\pi}{3}}\int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{1}^{0}\left(\theta\phi\right)Y_{\ell+1}^{-m}\left(\theta\phi\right)d\theta d\phi\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I_{\ell+1}\sqrt{\frac{\left(\ell+m+1\right)\left(\ell-m+1\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}}\ \ \ \ \ (17)$ $\displaystyle \left\langle n',\ell+1,m+1\left|x+iy\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -I_{\ell+1}\sqrt{\frac{8\pi}{3}}\int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{1}^{1}\left(\theta\phi\right)Y_{\ell+1}^{-\left(m+1\right)}\left(\theta\phi\right)d\theta d\phi\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I_{\ell+1}\sqrt{\frac{\left(\ell+m+1\right)\left(\ell+m+2\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}}\ \ \ \ \ (19)$ $\displaystyle \left\langle n',\ell+1,m-1\left|x-iy\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle n\ell m\left|x+iy\right|n',\ell+1,m-1\right\rangle ^*\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I_{\ell+1}\sqrt{\frac{8\pi}{3}}\left[\int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell+1}^{m-1}\left(\theta\phi\right)Y_{1}^{1}\left(\theta\phi\right)Y_{\ell}^{-m}\left(\theta\phi\right)d\theta d\phi\right]^*\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I_{\ell+1}\sqrt{\frac{\left(\ell+1-\left(m-1\right)\right)\left(\ell+1-\left(m-1\right)-1\right)}{\left(2\left(\ell+1\right)-1\right)\left(2\left(\ell+1\right)+1\right)}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I_{\ell+1}\sqrt{\frac{\left(\ell-m+2\right)\left(\ell-m+1\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}} \ \ \ \ \ (23)$

where

$\displaystyle I_{\ell+1}\equiv\int_{0}^{\infty}r^{3}R_{n\ell}\left(r\right)R_{n'\left(\ell+1\right)}\left(r\right)dr \ \ \ \ \ (24)$

To separate ${x}$ and ${y}$, we note that

$\displaystyle \left\langle n',\ell+1,m+1\left|x-iy\right|n\ell m\right\rangle =I_{\ell+1}\sqrt{\frac{8\pi}{3}}\int_{0}^{2\pi}\int_{0}^{\pi}Y_{\ell}^{m}\left(\theta\phi\right)Y_{1}^{-1}\left(\theta\phi\right)Y_{\ell+1}^{-\left(m+1\right)}\left(\theta\phi\right)d\theta d\phi=0 \ \ \ \ \ (25)$

because of 8, so adding and subtracting this from 19 we get

 $\displaystyle \left\langle n',\ell+1,m+1\left|x\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}I_{\ell+1}\sqrt{\frac{\left(\ell+m+1\right)\left(\ell+m+2\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}}\ \ \ \ \ (26)$ $\displaystyle \left\langle n',\ell+1,m+1\left|y\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle \frac{i}{2}I_{\ell+1}\sqrt{\frac{\left(\ell+m+1\right)\left(\ell+m+2\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}} \ \ \ \ \ (27)$

Similarly, from 23 we get

 $\displaystyle \left\langle n',\ell+1,m-1\left|x\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}I_{\ell+1}\sqrt{\frac{\left(\ell-m+2\right)\left(\ell-m+1\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}}\ \ \ \ \ (28)$ $\displaystyle \left\langle n',\ell+1,m-1\left|y\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{i}{2}I_{\ell+1}\sqrt{\frac{\left(\ell-m+2\right)\left(\ell-m+1\right)}{\left(2\ell+1\right)\left(2\ell+3\right)}} \ \ \ \ \ (29)$

Putting it all together, we get

 $\displaystyle \frac{\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{q^{2}}$ $\displaystyle =$ $\displaystyle \left|\left\langle n',\ell+1,m\left|z\right|n\ell m\right\rangle \right|^{2}+\left|\left\langle n',\ell+1,m+1\left|x\right|n\ell m\right\rangle \right|^{2}+\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left|\left\langle n',\ell+1,m+1\left|y\right|n\ell m\right\rangle \right|^{2}+\left|\left\langle n',\ell+1,m-1\left|x\right|n\ell m\right\rangle \right|^{2}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left|\left\langle n',\ell+1,m-1\left|y\right|n\ell m\right\rangle \right|^{2}\nonumber$ $\displaystyle \left|\boldsymbol{\mathfrak{p}}\right|^{2}$ $\displaystyle =$ $\displaystyle q^{2}I_{\ell+1}^{2}\frac{\left(\ell+m+1\right)\left(\ell-m+1\right)+\frac{1}{2}\left[\left(\ell+m+1\right)\left(\ell+m+2\right)+\left(\ell-m+2\right)\left(\ell-m+1\right)\right]}{\left(2\ell+1\right)\left(2\ell+3\right)}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q^{2}I_{\ell+1}^{2}\frac{\ell+1}{2\ell+1}\ \ \ \ \ (32)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{q^{2}I_{\ell+1}^{2}\omega_{0}^{3}}{3\epsilon_{0}\pi\hbar c^{3}}\frac{\ell+1}{2\ell+1} \ \ \ \ \ (33)$

For ${\ell'=\ell-1}$, we can do the same calculations.

 $\displaystyle \left\langle n',\ell-1,m\left|z\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle n',\ell,m\left|z\right|n,\ell-1,m\right\rangle ^*\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I_{\ell-1}\sqrt{\frac{\left(\ell+m\right)\left(\ell-m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}}\ \ \ \ \ (35)$ $\displaystyle \left\langle n',\ell-1,m-1\left|x-iy\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle n\ell m\left|x+iy\right|n',\ell-1,m-1\right\rangle ^*\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -I_{\ell-1}\sqrt{\frac{\left(\ell+m-1\right)\left(\ell+m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}}\ \ \ \ \ (37)$ $\displaystyle \left\langle n',\ell-1,m+1\left|x+iy\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -I_{\ell-1}\sqrt{\frac{\left(\ell-m-1\right)\left(\ell-m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}} \ \ \ \ \ (38)$

Separating ${x}$ and ${y}$ as before:

 $\displaystyle \left\langle n',\ell-1,m-1\left|x\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}I_{\ell-1}\sqrt{\frac{\left(\ell+m-1\right)\left(\ell+m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}}\ \ \ \ \ (39)$ $\displaystyle \left\langle n',\ell-1,m-1\left|y\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{i}{2}I_{\ell-1}\sqrt{\frac{\left(\ell+m-1\right)\left(\ell+m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}}\ \ \ \ \ (40)$ $\displaystyle \left\langle n',\ell-1,m+1\left|x\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}I_{\ell-1}\sqrt{\frac{\left(\ell-m-1\right)\left(\ell-m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}}\ \ \ \ \ (41)$ $\displaystyle \left\langle n',\ell-1,m+1\left|y\right|n\ell m\right\rangle$ $\displaystyle =$ $\displaystyle \frac{i}{2}I_{\ell-1}\sqrt{\frac{\left(\ell-m-1\right)\left(\ell-m\right)}{\left(2\ell-1\right)\left(2\ell+1\right)}} \ \ \ \ \ (42)$

Putting it together, we get

 $\displaystyle \left|\boldsymbol{\mathfrak{p}}\right|^{2}$ $\displaystyle =$ $\displaystyle q^{2}I_{\ell-1}^{2}\frac{\left(\ell+m\right)\left(\ell-m\right)+\frac{1}{2}\left[\left(\ell+m-1\right)\left(\ell+m\right)+\left(\ell-m-1\right)\left(\ell-m\right)\right]}{\left(2\ell-1\right)\left(2\ell+1\right)}\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q^{2}I_{\ell-1}^{2}\frac{\ell}{2\ell+1}\ \ \ \ \ (44)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{q^{2}I_{\ell-1}^{2}\omega_{0}^{3}}{3\epsilon_{0}\pi\hbar c^{3}}\frac{\ell}{2\ell+1} \ \ \ \ \ (45)$

This differs from the answer given in Griffiths in that he has ${2\ell-1}$ in the denominator rather than ${2\ell+1}$. I’m not sure if his answer is wrong or whether there’s something wrong in my calculation. As a check, the matrix elements above all make sense in that from 35 if a state starts ${m=\pm\ell}$ then ${m}$ cannot stay the same if ${\ell'=\ell-1}$, from 37, no transitions are allowed to ${m-1}$ if ${m=-\ell}$ or ${m=-\ell+1}$, and from 38, no transitions are allowed to ${m+1}$ if ${m=+\ell}$ or ${m=\ell-1}$. So my answer looks sensible, but I can’t guarantee it’s right.