# Runge-Lenz vector and closed orbits

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.2.1.

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The energy levels of hydrogen, when calculated from the Coulomb potential alone (ignoring various perturbations) depend only on the principal quantum number ${n}$ according to

$\displaystyle E=-\frac{1}{n^{2}}\frac{\mu e^{4}}{2\hbar^{2}} \ \ \ \ \ (1)$

The quantization arises entirely from the requirement that the radial function remain finite for large ${r}$, and makes no mention of the angular quantum numbers ${l}$ and ${m}$. Thus each energy level (each value of ${n}$) has a degeneracy of ${n^{2}}$, with ${2l+1}$ degenerate states for each ${l}$. Each symmetry is associated with the conservation of some dynamical quantity, with the degeneracy in ${m}$ due to conservation of angular momentum.

Shankar points out that, in classical mechanics, any potential with a ${\frac{1}{r}}$ dependence conserves the Runge-Lenz vector, defined for the hydregen atom potential as

$\displaystyle \mathbf{n}=\frac{\mathbf{p}\times\boldsymbol{\ell}}{\mu}-\frac{e^{2}}{r}\mathbf{r} \ \ \ \ \ (2)$

where I’ve used ${\mu}$ for the electron mass to avoid confusion with the ${L_{z}}$ quantum number ${m}$.

Although it doesn’t make sense to talk about the orbit of the electron in quantum mechanics, classically we can see that the conservation of ${\mathbf{n}}$ implies that the orbit is closed. We can see this as follows.

First, using

$\displaystyle \boldsymbol{\ell}=\mathbf{r}\times\mathbf{p} \ \ \ \ \ (3)$

we have

 $\displaystyle \mathbf{n}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu}\mathbf{p}\times\left(\mathbf{r}\times\mathbf{p}\right)-\frac{e^{2}}{r}\mathbf{r}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mu}\mathbf{r}\times\left(\mathbf{p}\cdot\mathbf{p}\right)-\mathbf{p}\left(\mathbf{r}\cdot\mathbf{p}\right)-\frac{e^{2}}{r}\mathbf{r}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{p^{2}}{\mu}-\frac{e^{2}}{r}\right)\mathbf{r}-\mathbf{p}\left(\mathbf{r}\cdot\mathbf{p}\right) \ \ \ \ \ (6)$

In the second line, we used the vector identity

$\displaystyle \mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right)=\mathbf{B}\left(\mathbf{A}\cdot\mathbf{C}\right)-\mathbf{C}\left(\mathbf{A}\cdot\mathbf{B}\right) \ \ \ \ \ (7)$

Since we’re dealing with a bound state, ${r}$ must always remain finite, so it must have a maximum value. At this point ${\frac{dr}{dt}=0}$, which means that there is no radial motion, which in turn means that all motion at that point must be perpendicular to ${\mathbf{r}}$. In other words

$\displaystyle \mathbf{p}\cdot\mathbf{r}_{max}=0 \ \ \ \ \ (8)$

Also, from conservation of energy, we have

$\displaystyle E=\frac{p^{2}}{2\mu}-\frac{e^{2}}{r} \ \ \ \ \ (9)$

so at ${r_{max}}$ we have

$\displaystyle p^{2}=2\mu\left(E+\frac{e^{2}}{r_{max}}\right) \ \ \ \ \ (10)$

Plugging these into 6, we get

 $\displaystyle \mathbf{n}$ $\displaystyle =$ $\displaystyle \left(2E+\frac{2e^{2}}{r_{max}}-\frac{e^{2}}{r_{max}}\right)\mathbf{r}_{max}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(2E+\frac{e^{2}}{r_{max}}\right)\mathbf{r}_{max} \ \ \ \ \ (12)$

Exactly the same argument applies to the case where ${r}$ is a minimum: again ${\frac{dr}{dt}=0}$ so ${\mathbf{r}\cdot\mathbf{p}=0}$ and we end up with

$\displaystyle \mathbf{n}=\left(2E+\frac{e^{2}}{r_{min}}\right)\mathbf{r}_{min} \ \ \ \ \ (13)$

If ${\mathbf{n}}$ is conserved (constant), then it must be parallel or anti-parallel to both ${\mathbf{r}_{max}}$ and ${\mathbf{r}_{min}}$, and the latter two vectors must therefore always have the same direction. In other words, the particle reaches its maximum (and minimum) distance always at the same point in its orbit, meaning that the orbit is closed.

In a general (elliptical) orbit, ${r_{max}>r_{min}}$ so ${\frac{e^{2}}{r_{max}}<\frac{e^{2}}{r_{min}}}$. Since ${E<0}$ for a bound orbit, we therefore must have

 $\displaystyle 2E+\frac{e^{2}}{r_{max}}$ $\displaystyle <$ $\displaystyle 0\ \ \ \ \ (14)$ $\displaystyle 2E+\frac{e^{2}}{r_{min}}$ $\displaystyle >$ $\displaystyle 0 \ \ \ \ \ (15)$

This in turn implies that ${\mathbf{n}}$ is anti-parallel to ${\mathbf{r}_{max}}$ and parallel to ${\mathbf{r}_{min}}$.

For a circular orbit, both ${r}$ and ${p}$ are constant, so both the kinetic and potential energies are also constant. From the virial theorem, we know that, for ${V\propto\frac{1}{r}}$

$\displaystyle \left\langle T\right\rangle =-\frac{1}{2}\left\langle V\right\rangle \ \ \ \ \ (16)$

Thus

 $\displaystyle E$ $\displaystyle =$ $\displaystyle T+V\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V}{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{2r} \ \ \ \ \ (19)$

Thus from 12, we see that ${\mathbf{n}=0}$ for a circular orbit.

# Hydrogen atom – radial function at large r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.4.

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When solving the 3-d Schrödinger equation for a spherically symmetric potential, the radial function has the asymptotic form for large ${r}$, and for the energy ${E<0}$:

$\displaystyle U_{El}\left(r\right)\underset{r\rightarrow\infty}{\longrightarrow}Ar^{\pm me^{2}/\kappa\hbar^{2}}e^{-\kappa r} \ \ \ \ \ (1)$

where

$\displaystyle \kappa\equiv\sqrt{\frac{2m\left|E\right|}{\hbar^{2}}} \ \ \ \ \ (2)$

For the hydrogen atom, the function ${U_{El}}$ is obtained from a series solution of the differential equation with the result

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (3)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa r \ \ \ \ \ (6)$

To keep the wave function finite at large ${r}$, we require the series to terminate, which leads to the quantized energy levels, given by

$\displaystyle E_{n}=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (7)$

The series in 4 terminates at a value of ${k=n-l-1}$, so the function ${v_{El}}$ is a polynomial in ${\rho}$, and thus in ${r}$, of degree ${n}$. Since the actual radial function is

$\displaystyle R_{nl}=\frac{U_{El}}{r} \ \ \ \ \ (8)$

we have that ${R_{nl}}$ is a polynomial of degree ${n-1}$ in ${r}$ multiplied by the exponential ${e^{-\rho}=e^{-\kappa r}}$. That is, for large ${r}$

$\displaystyle R_{nl}\sim r^{n-1}e^{-\kappa r} \ \ \ \ \ (9)$

To show that this is consistent with 1, we use 7 and 2.

 $\displaystyle n$ $\displaystyle =$ $\displaystyle \sqrt{\frac{me^{4}}{2\hbar^{2}\left|E_{n}\right|}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{me^{4}}{2\hbar^{2}}}\sqrt{\frac{2m}{\hbar^{2}}}\frac{1}{\kappa}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{\kappa\hbar^{2}} \ \ \ \ \ (12)$

Comparing this with 1, we see that

$\displaystyle r^{n}=r^{me^{2}/\kappa\hbar^{2}} \ \ \ \ \ (13)$

so the condition is satisfied.

# Hydrogen atom – a sample wave function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.3.

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The wave function for the hydrogen atom can be obtained by a series solution of the differential equation, leading to the result (which I’ve rewritten in Shankar’s notation, although my original post used Griffiths’s notation):

$\displaystyle \psi_{nlm}\left(r,\theta,\phi\right)=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

Here, we have

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (2)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (3)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r \ \ \ \ \ (4)$

The energy levels of the hydrogen atom are

$\displaystyle E=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (5)$

where ${n=1,2,3,\ldots}$. The coefficients ${C_{k}}$ in 3 are given by a recursion relation

 $\displaystyle C_{k+1}$ $\displaystyle =$ $\displaystyle \frac{-e^{2}\lambda+2\left(k+l+1\right)}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}\ \ \ \ \ (6)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \sqrt{-\frac{2m}{\hbar^{2}E}} \ \ \ \ \ (7)$

Combining ${\lambda}$ and ${E}$, the formula becomes, for a given ${n}$

$\displaystyle C_{k+1}=\frac{2\left(k+l+1\right)-2n}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}$

The coefficient ${C_{0}}$ which starts everything off is determined by normalization.

As an example, we can find the wave function ${\psi_{210}}$. In this case ${n=2}$ and ${l=1}$ so the first term in the recursion, with ${k=0}$ gives ${k+l+1=2}$ and ${C_{1}=0}$. The full wave function is then

 $\displaystyle \psi_{210}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\rho^{2}e^{-\rho}C_{0}Y_{1}^{0} \ \ \ \ \ (8)$

To evaluate ${\rho}$ we use the energy for ${n=2}$:

$\displaystyle E_{2}=-\frac{me^{4}}{8\hbar^{2}} \ \ \ \ \ (9)$

This gives

$\displaystyle \rho=\sqrt{\frac{2m^{2}e^{4}}{8\hbar^{4}}}r=\frac{me^{2}}{2\hbar^{2}}r=\frac{r}{2a_{0}} \ \ \ \ \ (10)$

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

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (11)$

Plugging everything into 8, using ${Y_{1}^{0}=\sqrt{\frac{3}{4\pi}}\cos\theta}$, we have

$\displaystyle \psi_{210}=\sqrt{\frac{3}{4\pi}}\frac{C_{0}}{4a_{0}^{2}}re^{-r/2a_{0}}\cos\theta \ \ \ \ \ (12)$

Normalizing gives the condition

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{210}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi=1 \ \ \ \ \ (13)$

Working out the integral (using software or tables) gives

 $\displaystyle \frac{3}{2}a_{0}C_{0}^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{3a_{0}}} \ \ \ \ \ (15)$

So the final wave function is

$\displaystyle \psi_{210}=\frac{1}{\sqrt{32\pi a_{0}^{3}}}\frac{r}{a_{0}}e^{-r/2a_{0}}\cos\theta \ \ \ \ \ (16)$

which agrees with Shankar’s equation 13.1.27.

# 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.