# The real hydrogen atom

## Problem

The structures of atoms are described by quantum mechanics. This is a complicated theory which developed over time following a great deal of experimental input. Now, it is used routinely to model the behaviour of electrons, atoms and molecules. At its heart lies the Schrodinger equation, which is an equation for an object called the wavefunction, written
as $\psi({\bf x}, t)$. The square of the wavefunction is a probability density function such that $|(\psi{\bf x}, t)|^2dV$ is the probability of finding the particle in a small ball of size $dV$ centred on ${\bf x}$ at time
$t$. Solving the Schrodinger equation allows chemists to determine the physical form of the electron orbitals of atoms.

The general Schrodinger equation is very complicated, but for the motion of an electron around a single hydrogen nucleus reduces to three relatively simple equations when we use spherical polar coordinates $(r, \theta, \phi)$ (in terms of the globe $\phi$ is the longitude and $\theta$ is the latitude of a point) to describe the location of the electron through three functions $R(r), P(\theta),
F(\phi)$

$$

\frac{1}{R}\frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8\pi^2\mu}{h^2}\left(Er^2+\alpha r\right)=l(l+1)\;,

$$

$$

\frac{\sin\theta}{P}\frac{d}{d\theta}\left[\sin\theta \frac{dP}{d\theta}\right] +l(l+1)\sin^2\theta =m^2_l\;,

$$

$$

\frac{1}{F}\frac{d^2F}{d\phi^2}=-m^2_l\;.

$$

These equations involve the effective mass $\mu$ of the electron in terms of the mass of the electron and proton $m_e$ and $m_p$ as $\mu = \frac{m_em_p}{m_e+m_p}$, $\alpha = \frac{e^2}{4\pi\epsilon_0}$, the energy $E$ of the particle and Planck's constant $h =6.626068\times 10^{-34}$ m$^2$ kg s$^{-1}$.

There are three quantum numbers associated with these equations:

The principal quantum number $n=1, 2, 3, \dots$

The orbital quantum numbers $l = 0, 1, \dots, n-1$.

The magnetic quantum number $m= -l, -l+1, \dots, l -1, l$.

The energy $E$ of the electron depends on the principal quantum number as

$$

E=-\frac{\mu e^4}{8 n^2h^2\epsilon^2_0}\;.

$$

Solving this tricky set of equations gives the wave function for the electron to be $\psi({\bf x}, t)=R(r)P(\theta)F(\phi)$, from which the likely positions of the electron can be determined.

The s, p, d and f shells correspond to solutions with $l=0,1,2$ and $3$ respectively.

The goal of the quantum chemist is to attempt to extract solutions from these equations.

Task 0: Familiarise yourself with the structure of these equations. For example, what does the lowest energy ($n=1$) set of equations look like?

Task 1: Show that if $(R_1, P_1, F_1)$ is a solution to the set of equations then $(aR_1, bP_1, cF_1)$ will also be a solution for any non-zero constants $a, b$ and $c$.

Task 2: Start simply: do constant values for either $R, P$ or $F$ satisfy the equations for particular choices of $n, l, m$?

Task 3: Solving the $P$ equation will clearly be likely to involve trigonometric functions. Do $P=\sin\theta, \cos\theta, \tan\theta$ solve the $P$ equation for particular choices of $l$ and $m$? How about other simple combinations of $\sin$ and $\cos$?

Task 4: The $R$ equation contains many constants. Group these together as $a_0 = \frac{\pi e^2\mu}{h^2\epsilon_0}$ and see how the equation simplifies a little.

Task 5: On physical grounds we would expect the chance of the electron being very far away from the nucleus to be very small, tending to zero as $r$ increases. Why would a function $R = f(r)e^{-ar}$ satisfy this condition for any polynomial $f(r)$ and positive constant $a$? Try the simplest case $R=e^{-ar}$. Does this result in a solution? What about the possibilities for $R= (a+br)e^{-cr}$? for
constants $a, b, c$?

In this problem, we have only looked briefly into attempting to partially solve the Schrodinger equation for the simplest atom, and have taken the Schrodinger equation on trust. Exposure to the ideas raised in this problem will both improve your skills with differential equations and also ease the way into the true theory underlying quantum mechanics and atoms.To derive and solve these equations more generally requires some heavyweight analytical tools of the kind met at university. More complicated versions must be solved using clever numerical integration schemes.

If you wish to find out more about these equations see, for example, http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html#c2

## Getting Started

This problem is tough and will require very careful algebraic manipulation.

To have any chance of solving such equations requires you to look carefully at their structure and trial functional forms which have a chance of cancelling in the required way.

You will need to know about differential equations to be able to solve this problem.

## Student Solutions

Task 0: The lowest energy set of equation corresponds to $ n = 0$. This implies that $l =0$ and $m_l = 0$ also.

Thus:

a) $\frac{1}{R} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

b) $\frac{sin\theta}{P} \frac{d}{d\theta} \left[sin\theta \frac{dP}{d\theta}\right] = 0$

c) $\frac{1}{F} \frac{d^2F}{d\phi^2} = 0$

Task 1: If $R_1$ is a solution, then for $aR_1$:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

Since $a$ is a constant, it can be removed outside of the differential:

$\frac{a}{aR_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

$\frac{1}{R_1} \frac{d}{dr}\left[r^2 \frac{dR_1}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = 0$

Therefore, since $R_1$ is a solution, so is $aR_1$ for any non-zero constant $a$.

If $P_1$ is a solution, then for $bP_1$:

$\frac{sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{d(bP_1)}{d\theta}\right] = 0$

Since $b$ is a constant, it can be removed outside of the differential:

$\frac{b\ sin\theta}{bP_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$

$\frac{sin\theta}{P_1} \frac{d}{d\theta} \left[sin\theta \frac{dP_1}{d\theta}\right] = 0$

Therefore, since $P_1$ is a solution, so is $bP_1$ for any non-zero constant $b$.

If $F_1$ is a solution, then for $cF_1$:

$\frac{1}{cF_1} \frac{d^2(cF_1)}{d\phi^2} = 0$

Since $c$ is a constant, it can be removed outside of the differential:

$\frac{c}{cF_1} \frac{d^2F_1}{d\phi^2} = 0$

$\frac{1}{F_1} \frac{d^2F_1}{d\phi^2} = 0$

Therefore, since $F_1$ is a solution, so is $cF_1$ for any non-zero constant $c$.

Task 2: If R is a constant, then $\frac{dR}{dr} = 0$

$\therefore \frac{1}{R} \frac{d}{dr}\left[0\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

$ -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

Since $r > 0$, and $l> 0$, this equation is not satisfied by a constant value of $R$.

If P is a constant, then $\frac{dP}{d\theta} = m_l^2$

$\therefore \frac{sin\theta}{P} \frac{d}{d\theta} \left[0\right] + l(l+1)sin^2\theta= m_l^2$

$l(l+1)sin^2\theta = m_l^2$

The equation is not satisfied since there is still a dependence on $\theta$, which means that the wavefunction would only be permitted for certain values of $\theta$ Therefore a constant value of $P$ is not permissible.

If F is a constant, then:

$m_l^2 = 0$

$m_l = 0$

Task 3:

a) Let $P = sin\theta$

$\therefore \frac{dP}{d\theta} = cos\theta$

$\rightarrow \frac{sin\theta}{sin\theta} \frac{d}{d\theta} \left[sin\theta cos\theta \right] + l(l+1)sin^2 \theta= m_l^2$

$cos(2\theta) + l(l+1)sin^2 \theta = m_l^2$

Expanding $cos (2\theta)$ as $cos^2\theta - sin^2\theta$, and simplifying gives:

$cos^2\theta + (l^2 +l -1)sin^2\theta = m_l^2$

To eliminate the trigonometric terms, we wish to use the identity $cos^2\theta + sin^2\theta = 1$

$\therefore l^2 + l -2 = 0$

$(l+2)(l-1) = 1$

$\mathbf{ l =1}$ since $l > 0$

$\mathbf{m_l = \pm 1}$

b) Now let $P = cos\theta$

$\therefore \frac{dP}{d\theta} = -sin\theta$

$\frac{sin\theta}{cos\theta}\frac{d}{d\theta}\left[-sin^2\theta\right] + l(l+1)sin^2\theta = m_l^2$

Differentiating, simplifying and collecting terms yields:

$(l^2 + l -2) sin^2\theta = m_l^2$

For this to be valid, $\mathbf{m_l^2 = 0}$, which means that either $sin^2 \theta$ or $(l^2 + l -2)$ must be equal to zero.

Clearly, for the wavefunction to exist:

$l^2+ 1 -2 = 0$

$\mathbf{l=1}$ since $l > 0$

c) Substituting in $P = tan\theta$ yields an equation still containing trigonometry. There is no way for this to be removed, and so the original trial solution is invalid.

If you haven't tried any other trigonometric functions yet, why not try $cos^2\theta$, $sin(2\theta)$ and $sin^2\theta$...?

Task 4:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[Er^2 +\alpha r] = l(l+1)$

Given that $a_0 = \frac{\pi e^2 \mu}{h^2 \epsilon_0}$, $\alpha = \frac{e^2}{4\pi \epsilon_0}$, $E = -\frac{\mu e^4}{8n^2h^2\epsilon_0^2}$:

$\frac{1}{aR_1} \frac{d}{dr}\left[r^2 \frac{d(aR_1)}{dr}\right] -\frac{8 \pi^2 \mu}{h^2}[-\frac{\mu e^4 r^2}{8n^2h^2\epsilon_0^2} +\frac{e^2r}{4\pi \epsilon_0}] = l(l+1)$

$\frac{1}{R}\frac{d}{dr} \left[r^2\frac{dR}{dr}\right] + (\frac{a_0r}{n})^2 -2a_0r = l(l+1)$