5.3 Energy Eigenstates of Hydrogen

5.3.1 Summary of the Eigenstates of Hydrogen

Putting the radial and angular parts of the eigenfunctions together, the whole spatial part of the electron eigenstate is

	$\displaystyle u_{n\ell m}(r,\theta,\phi)$	$\displaystyle=R_{n\ell}(r)Y_{\ell}^{m}(\theta,\phi)$		(5.55)
		$\displaystyle=N_{n\ell}\sqrt{\frac{(2\ell+1)(\ell-m)!}{4\pi(\ell+m)!}}\sum_{p=% 0}^{n-\ell-1}c_{p}\left(\frac{r}{a_{0}}\right)^{p+\ell}e^{-\frac{r}{na_{0}}}e^% {im\phi}P_{\ell}^{m}(\cos\theta),$		(5.56)

where $P_{\ell}^{m}$ are the associated Legendre polynomials and the principal, orbital angular momentum, and magnetic quantum numbers take the following values:

$\displaystyle n$	$\displaystyle=1,2,3,\dots$	(5.57)
$\displaystyle\ell$	$\displaystyle=0,1,2,\dots,n-1$	(5.58)
$\displaystyle m$	$\displaystyle=-\ell,-\ell+1,\dots,-1,0,1,\dots,\ell-1,\ell.$	(5.59)

These eigenstates have energy eigenvalues given by

E_{n}=-\frac{\hbar^{2}}{2\mu a_{0}^{2}n^{2}}=-\frac{R_{H}}{n^{2}},

(5.60)

where the (reduced) Bohr radius is given by

a_{0}=\frac{4\pi\varepsilon_{0}\hbar^{2}}{\mu e^{2}}.

(5.61)

The ground state energy $E_{1}$ is given by

E_{1}=-R_{H}=-$13.6\text{\,}\mathrm{eV}$.

(5.62)

This is the ionisation energy of hydrogen, i.e. the energy required to completely remove the electron from the influence of the proton.

Since the energy only depends on the principal quantum number, and for $n>1$ we can have multiple values of $\ell$ and $m$ , we have a set of degenerate states for all energy levels above the ground state. For example, for the first excited state $n=2$ , we can have $\ell=0,1$ and $m=-1,0,1$ , which leads to four degenerate states

u_{200},\,u_{210},\,u_{211},\,\text{and }u_{21\,-1},

(5.63)

all with energy $E_{2}=-\frac{R_{H}}{4}$ .

For a given $\ell$ , there are $2\ell+1$ possible values of $m$ . Thus in can be shown that for a given $n$ , the total degeneracy is

\sum_{\ell=0}^{n-1}(2\ell+1)=n^{2}.

(5.64)

The greater the principal quantum number $n$ for a given $\ell$ , the greater the expectation value of radial distance $\langle r\rangle$ i.e. the more spread out the probability distribution is. For a given $n$ , $\ell$ determines the number of polar nodes, which are the nodes as you go from $\theta=0$ to $\theta=\pi$ .

In Dirac notation, the eigenstates are labelled

\lvert n\ell m\rangle,

(5.65)

and they have the orthonormality condition

\langle n\ell m\mathclose{}|\mathopen{}n^{\prime}\ell^{\prime}m^{\prime}% \rangle=\delta_{nn^{\prime}}\delta_{\ell\ell^{\prime}}\delta_{mm^{\prime}}.

(5.66)

5.3.2 Spectroscopic Notation

For historical reasons, scientists working in atomic and nuclear physics adopt a notation for the eigenstates of hydrogen based on different series of spectral lines that were observed before electron orbitals were fully understood. Early spectroscopists denoted the first four values of $\ell$ by the letters s, p, d, and f, standing for “sharp”, “principal”, “diffuse”, and “fundamental” respectively. These names do not mean anything in the quantum mechanical description, but the notation remains widespread.

In this notation the ground state is called the $1\text{s}$ orbital, the first excited states are called the $2\text{s}$ and the $2\text{p}$ orbitals, the second excited states would be the $3\text{s}$ , $3\text{p}$ , and $3\text{d}$ orbitals, and so on. After $\ell=4$ , the letters continue in alphabetical order from g for $\ell=5$ onwards, missing out the letter “j”.

Note that a p orbital stands for three eigenstates with $m=-1,0,1$ , a d orbital stands for five eigenstates with $m=-2,-1,0,1,2$ , etc.

5.3.3 Hydrogen-like Atoms and Ions

We have solved for the motion of a single electron in a hydrogen atom, but what about electrons in other atoms?

We can apply the same model that we have developed so far to ions with a single electron such as $\text{He}^{+}$ (a helium nucleus with a single electron), $\text{Li}^{2+}$ (a lithium nucleus with a single electron), $\text{Be}^{3+}$ (a beryllium nucleus with a single electron), etc. In this case, the charge of the nucleus is now $Z e$ , where $Z$ is the atomic number, or the number of protons in the nucleus, equal to $2$ for helium, $3$ for lithium, $4$ for beryllium, etc. The potential energy then increases in magnitude by a factor of $Z$ , and the energy levels shift down by a factor of $Z^{2}$ .

E_{n}=-\frac{Z^{2}R_{H}}{n^{2}}.

(5.67)

The electrons are more strongly bound. Note that for our description of hydrogen, the reduced mass would also change slightly.

What about atoms with more than one electron? The dynamics get complicated, since we have to consider interactions between the electrons. For this reason, studying multi-electron atoms in general is out of the scope of these notes. What we can begin to look at are the alkali metals (lithium, sodium, potassium, rubidium, etc.). These atoms are neutral, so they have $Z$ protons in the nucleus and $Z$ electrons in orbitals, but $Z-1$ of these electrons are tightly bound in closed shells. This means that the most weakly bound outermost electron (the valence electron) is in its own shell. It is not really possible to say anything about the form of the eigenstates of the valence electron using the framework we have created so far, but by considering two extreme scenarios we can make claims about its energy levels.

If the valence electron is at very large radii, much greater than the inner electrons, then the nuclear charge would be “screened” such that the valence electron would only notice an effective charge of $+e$ coming from the centre. This would mean that the energy levels would be roughly the same as for hydrogen. In the other extreme case, where the valence electron is at very small radii, inside the inner electron shells, then it would see the full unscreened nuclear charge $+Ze$ . Therefore the actual potential energy curve must lie somewhere between these two extremes.

Spectroscopy experiments also show that for alkali metals the degeneracy between the different values of $\ell$ is broken. For a given $n$ , the smaller values of $\ell$ have slightly lower energy. Why is this? If we assume that the shape of the radial probability density will be roughly the same for the valence electron as for the hydrogen atom electron, then recall that for a given $n$ the radius of the first maxima increases for increasing $\ell$ . This means that for smaller $\ell$ (for a given $n$ ), the electron is more likely to be found closer in to the nucleus and therefore further into the unscreened regime where the Coulomb force is stronger. This results in these states being slightly more tightly bound and therefore being slightly lower in energy.

Quantitatively, we can write the energy levels for the valence electron in an alkali metal as

E_{n\ell}=-\frac{R_{H}}{(n-\Delta(n,\ell))^{2}},

(5.68)

where $\Delta(n,\ell)$ is the quantum defect, an empirically determined quantity which denotes how far the energy levels are shifted down compared to hydrogen. Notice that we label the energy levels $E_{n\ell}$ now instead of just $E_{n}$ because the states with different $\ell$ are now no longer degenerate.

For example, the $3\text{s}$ orbital in sodium has $\Delta(3,0)=1.37$ , the $3\text{p}$ orbital has $\Delta(3,1)=0.88$ , and the $3\text{d}$ orbital has $\Delta(3,2)=0.01$ , meaning this state has almost the same energy as $E_{3}$ for hydrogen.