[web] [lect]

This is a revision section.

For full details of setting up and solving the hydrogen Schrödinger equation,

see my Quantum notes.

For full details of setting up and solving the hydrogen Schrödinger equation,

see my Quantum notes.

A *hydrogen-like atom*

is an atom consisting of a nucleus and just one electron; the nucleus can be bigger than just a
single proton, though. H atoms, He^{+} ions, Li^{2+} ions *etc.* are all hydrogen-like atoms. This means
the system consists of only two particles (left). Therefore the positions of both masses relative to the centre of
gravity remain constant, and we can simplify the situation to that of the *reduced mass*, $\mu=\frac{mM}{m+M}$

, in a point
located at the centre of gravity. This is not possible in a system of more than two particles (right) because the
distances keep changing.

Using the reduced mass, the *Hamiltonian*

consists of a kinetic energy term, $-\frac{\hbar^2}{2\mu}\hat{\nabla}^2$

, and the
potential energy part derives from the *Coulomb potential*, $-\frac{Ze^2}{4\pi\epsilon_0r}$, where $Z$ acknowledges the
fact that the nucleus of a hydrogen-like atom may contain more than one positive charge.

With the del operator in spherical co-ordinates

, the *Schrödinger equation*,
$\hat{H}\psi=E\psi$ is, then

$$\underbrace{-\frac{\hbar^2}{2\mu}\underbrace{\left[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^2\sin\vartheta}\frac{\partial}{\partial\vartheta}\left(\sin\vartheta\frac{\partial\psi}{\partial\vartheta}\right)+\frac{1}{r^2\sin^2\vartheta}\frac{\partial^2\psi}{\partial\varphi^2}\right]}_{\mathrm{\hat{\nabla}^2\psi}\;\textrm{in spherical co-ordinates}}}_{\textrm{kinetic energy term}}\underbrace{-\frac{Ze^2}{4\pi\epsilon_0r}\psi}_{\textrm{potential energy term}}=E\psi$$

This can be solved by double separation of variables, and the resulting wave functions are

$$\psi(r,\vartheta,\varphi)=NR_{n,l}(r)P_l^m(\cos\vartheta){\rm e}^{{\rm i}m\varphi}$$

where $N$ is a normalisation constant ensuring that $\int\psi^{\ast}\psi{\rm d}V=1$ when integrating over all space,
$R_{n,l}$ is the *radial wave function*

,
$P_l^m$ is a *Legendre polynomial*

describing the dependence on the polar angle $\vartheta$, and
${\rm e}^{{\rm i}m\varphi}$ is the azimuth part

of the wave function.
The latter two are often listed together
as *spherical harmonics*, $Y_{l,m}$

. The precise functional dependence of $R_{n,l}(r)$ and
$Y_{l,m}(\varphi,\vartheta)$ on the *quantum numbers* $n,l,m$

and the coordinates $r,\varphi,\vartheta$
is known and can be looked up.

**Exercise.** The different states at the same $n$ are all degenerate in a hydrogen-like atom. Taken together, they add up to a spherically symmetric probability density. What is the total degeneracy (number of states with different $l,m$) for a given value of $n$?

n= |
1 | 2 | 3 | ... | |||||||||||

l= |
0 | 0 | 1 | 0 | 1 | 2 | 0...(n-1) | ||||||||

m= |
0 | 0 | -1 | 0 | +1 | 0 | -1 | 0 | +1 | -2 | -1 | 0 | +1 | +2 | -l...+l |

The quantum numbers distinguish each state uniquely: $n$ determines the radial spread of the wave function and
its energy, $l$ describes the shape of the wave function, and $m$ its orientation in space relative to some
external axis such as a magnetic field or a chemical bond.
The quantum numbers are not independent; the choice of $n$ limits the choice of $l$, which in
turn limits the choice of $m$. A fourth quantum number, $s$, does not follow directly from
solving the Schrödinger equation but is to do with spin

. The
possible combinations of quantum numbers are given in the table.

Note that the energy of a state (*i.e.* of a wave function) depends only on $n$ but not on the
other quantum numbers. This *degeneracy* is only strictly true for the hydrogen-like atom; any
approximate solutions for higher atoms cause a dependence of the energy eigenvalue of a state on all
quantum numbers.

**Exercise.** Given that the wavelength of the first line of the Balmer series of hydrogen is $\lambda_{23}=\mathrm{656\,nm}$, what is the wavelength of the first line of the Paschen series of hydrogen?

And what is the wavelength for the first Paschen line for ionised helium, $\mathrm{He^+}$?

[solution]

The energy of a state with quantum number $n$ is $$E_n=-\frac{\mu Z^2e^4}{8\epsilon_0^2h^2n^2}\qquad.$$
For *transitions*

between two states, the energy difference between the two states must either
be supplied (excitation) or emitted. Since the states have discrete energy levels, only fitting
*energy quanta* can be involved in the transition. The energy difference between two adjacent
states is
$$\Delta E_{n_1,n_2}=-\frac{\mu Z^2e^4}{8\epsilon_0^2h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\qquad;$$
thus spectral lines occur only at the corresponding energies (frequencies, wavelengths, wavenumbers,...).

This had been observed long before quantum mechanics - the first observations concern the Sun's spectrum.
The constants in front of the bracket are collectively known as the
*Rydberg constant*, $R_H$;
its value was determined experimentally before quantum physics provided the link to the fundamental
constants.

The different series of spectral lines discovered by Lyman, Balmer, and Paschen, correspond to the excitations from the $n$=1,2,3 states.

The wave function itself, a complex function with positive and negative values, doesn't tell us
much about the structure of the atom or any connectivity it may have with other atoms. The complex
square of the wave function represents the *probability density*

of finding the electron at a given
point in space when one looks (*i.e.* does an experiment). It does not say anything about where the
electron actually *is* at any moment, the solutions of the Schrödinger equation only
refer to which states are observable. The act of measuring throws the system into one of
the *eigenstates*

that are solutions of the Schrödinger equation.

A recurrent idea of statistics is that temporal averages are equivalent to ensemble averages, *i.e.*
look for a while at one object moving randomly and you will see the same set of orientations that you
would see when taking a snapshot of many identical objects moving randomly. In the same way, we can
interpret the probability density as a representation of the *electron density* in an atom, molecule, or
solid.

The radial solutions of the Schrödinger equation of the hydrogen atom, $R(r)$, are plotted
on the right. Each time the quantum number $n$ increases, an additional node is created. At
$n$=1, the radial function is all positive. Its maximum is at $r$=0, *i.e.* the point in
space with the highest probability density of finding the electron is actually *inside* the nucleus!
That is why the term *probability density* is used: As we move outward along the radius, the volume
of a shell of equal thickness is getting larger and larger, thereby spreading out the probability over
a larger volume.

Each time the quantum number $l$ is increased, one of the spherical nodes disappears again. It is replaced by a planar node that goes through the nucleus. Therefore, only $l$=0 electrons have a finite probability density at the nucleus.

The diagram on the right shows cross sections of the full wave function $\psi(r,\vartheta,\varphi)$ in the polar ($r$-$\vartheta$) plane. This representation highlights the transformation of spherical nodes into planar nodes as $l$ increases.

It is also apparent that the wave function is spreading out into space as $n$ increases, *i.e.*
that electrons with a small $n$ are, on balance, nearer the nucleus. Given that the energy eigenvalue
increases with $n$, that matches the semi-classical expectation that electrons have a lower energy if
they are deep down in the Coulomb potential.

One more observation: As $l$ increases, the additional planar nodes cause the wave function to become less and less symmetric. This is compensated by the increasing number of equivalent states having the same $n$ and $l$ but different $m$. (In the diagram, only $m$=0 states are shown.) All (2$l$+1) states with the same value of $n$ and $l$ together form a perfectly spherical distribution of probability density.

Note that since the probability density is the square of the wave function, it doesn't make any difference if the wave function is positive or negative.

l= |
0 | 1 | 2 | 3 | 4... |

letter: | s | p | d | f | g (then continue alphabetically) |

Instead of describing a state by listing the values of its quantum numbers, a common practice is to refer to them by main quantum number, $n$, followed by a letter representing the value of $l$ as shown in the table.

Thus, a 2p state is one with $n$=2 and $l$=1 (the $m$=-1,0,+1 cases are sometimes
distinguished as 2p_{x}, 2p_{y}, and 2p_{z}); a 3d state is one with $n$=3 and
$l$=2.

We now have accurate wave functions and their energies for hydrogen-like atoms. Solving the Schrödinger equation analytically is impossible for more complex systems. Instead, we can use the known system as a base and add complexity gradually, adjusting the wave functions and energies step by step. This perturbation approach also allows us to calculate the effect of interactions with the electron system, such as the interaction of radiation with matter on which spectroscopy is based.