This Quantum World/Implications and applications/Atomic hydrogen

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[edit] Atomic hydrogen

While de Broglie's theory of 1923 featured circular electron waves, Schrödinger's "wave mechanics" of 1926 features standing waves in three dimensions. Finding them means finding the solutions of the time-independent Schrödinger equation

$E\psi(\mathbf{r})=-{\hbar^2\over2m}\left(\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)\psi(\mathbf{r})+V(\mathbf{r})\,\psi(\mathbf{r}).$

with $V(\mathbf{r})=-e^2/r,$ the potential energy of a classical electron at a distance $r=|\mathbf{r}|$ from the proton. (Only when we come to the relativistic theory will we be able to shed the last vestige of classical thinking.)

$E\psi(\mathbf{r})=-{\hbar^2\over2m}\left(\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)\psi(\mathbf{r})-\frac{e^2}{r}V(\mathbf{r})\,\psi(\mathbf{r}).$

In using this equation, we ignore (i) the influence of the electron on the proton, whose mass is some 1836 times larger than that of he electron, and (ii) the electron's spin. Since relativistic and spin effects on the measurable properties of atomic hydrogen are rather small, this non-relativistic approximation nevertheless gives excellent results.

For bound states the total energy $E$ is negative, and the Schrödinger equation has a discrete set of solutions. As it turns out, the "allowed" values of $E$ are precisely the values that Bohr obtained in 1913:

$E_n=-{1\over n^2}\,{\mu e^4\over2\hbar^2},\qquad n=1,2,3,\dots$

However, for each $n$ there are now $n 2$ linearly independent solutions. (If $\psi_{1},\dots,\psi_{k}$ are independent solutions, then none of them can be written as a linear combination $\sum a_i\psi_i$ of the others.)

Solutions with different $n$ correspond to different energies. What physical differences correspond to linearly independent solutions with the same $n$ ?

Using polar coordinates, one finds that all solutions for a particular value $E n$ are linear combinations of solutions that have the form

$\psi(r,\phi,\theta)=e^{(i/\hbar)\,l_z\,\phi}\psi(r,\theta).$

$l z$ turns out to be another quantized variable, for $e^{(i/\hbar)\,l_z\,\phi}=e^{(i/\hbar)\,l_z\,(\phi\pm2\pi)}$ implies that $l_z=m\hbar$ with $m=0,\pm1,\pm2,\dots$ In addition, $| m |$ has an upper bound, as we shall see in a moment.

Just as the factorization of $\psi(t,\mathbf{r})$ into $e^{-(i/\hbar)\,E\,t}\,\psi(\mathbf{r})$ made it possible to obtain a $t$ -independent Schrödinger equation, so the factorization of $ψ(r,φ,θ)$ into $e^{(i/\hbar)\,l_z\,\phi}\,\psi(r,\theta)$ makes it possible to obtain a $φ$ -independent Schrödinger equation. This contains another real parameter $Λ,$ over and above $m,$ whose "allowed" values are given by $l(l+1)\hbar^2,$ with $l$ an integer satisfying $0\leq l\leq n-1.$ The range of possible values for $m$ is bounded by the inequality $|m|\leq l.$ The possible values of the principal quantum number $n,$ the angular momentum quantum number $l,$ and the so-called magnetic quantum number $m$ thus are:

$n = 1$	$l = 0$	$m = 0$
$n = 2$	$l = 0$	$m = 0$
	$l = 1$	$m=0,\pm1$
$n = 3$	$l = 0$	$m = 0$
	$l = 1$	$m=0,\pm1$
	$l = 2$	$m=0,\pm1,\pm2$
$n = 4$	$\dots$	$\dots$

Each possible set of quantum numbers $n, l, m$ defines a unique wave function $\psi_{nlm}(t,\mathbf{r}),$ and together these make up a complete set of bound-state solutions ( $E < 0$ ) of the Schrödinger equation with $V(\mathbf{r})=-e^2/r.$ The following images give an idea of the position probability distributions of the first three $l = 0$ states (not to scale). Below them are the probability densities plotted against $r .$ Observe that these states have $n - 1$ nodes, all of which are spherical, that is, surfaces of constant $r .$ (The nodes of a wave in three dimensions are two-dimensional surfaces. The nodes of a "probability wave" are the surfaces at which the sign of $ψ$ changes and, consequently, the probability density $| ψ | 2$ vanishes.)

Take another look at these images:


$ρ 2 p 0$	$ρ 3 p 0$	$ρ 3 d 0$	$ρ 4 p 0$

$ρ 4 d 0$	$ρ 4 f 0$	$ρ 5 d 0$	$ρ 5 f 0$

The letters s,p,d,f stand for l=0,1,2,3, respectively. (Before the quantum-mechanical origin of atomic spectral lines was understood, a distinction was made between "sharp," "principal," "diffuse," and "fundamental" lines. These terms were subsequently found to correspond to the first four values that $l$ can take. From $l = 3$ onward the labels follows the alphabet: f,g,h...) Observe that these states display both spherical and conical nodes, the latter being surfaces of constant $θ.$ (The "conical" node with $θ = 0$ is a horizontal plane.) These states, too, have a total of $n - 1$ nodes, $l$ of which are conical.

Because the "waviness" in $φ$ is contained in a phase factor $e i m φ,$ it does not show up in representations of $| ψ | 2 .$ To make it visible, it is customary to replace $e i m φ$ by its real part $cos(m φ),$ as in the following images, which do not represent probability distributions.

$ρ 4 f 1$	$ρ 4 f 3$	$ρ 5 f 1$	$ρ 5 f 2$
$ρ 5 f 3$	$ρ 5 g 1$	$ρ 5 g 2$	$ρ 5 g 3$

The total number of nodes is again $n - 1,$ the total number of non-spherical nodes is again $l,$ but now there are $m$ plane nodes containing the $z$ axis and $l - m$ conical nodes.

What is so special about the $z$ axis? Absolutely nothing, for the wave functions $ψ' n l m,$ which are defined with respect to a different axis, make up another complete set of bound-state solutions. This means that every wave function $ψ' n l m$ can be written as a linear combination of the functions $ψ n l m,$ and vice versa.

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This Quantum World/Implications and applications/Atomic hydrogen

From Wikibooks, the open-content textbooks collection

[edit] Atomic hydrogen

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