A hydrogen atom consists of a nucleus, which is a single proton, and one electron. The electron receives the Coulomb force from the nucleus, and its potential energy depends only on the distance r from the nucleus: U(r) = -\dfrac{e^2}{4\pi\varepsilon_0 r}. It is lower closer to the nucleus and approaches 0 far away.
The nucleus is about 1836 times heavier than the electron, so here the nucleus is taken to be at rest. The wave function \psi of the electron is a function on three-dimensional space, and the Schrödinger equation is -\dfrac{\hbar^2}{2m_\mathrm{e}}\nabla^2\psi + U(r)\psi = E\psi. \nabla^2\psi is the sum of the second derivatives of \psi with respect to x, y and z.
In Bohr's model, the electron goes around a circle of a fixed radius, and the energy was E_n = -\dfrac{13.6\,\mathrm{eV}}{n^2}. This article checks that the same energies come out of the Schrödinger equation without assuming either circular orbits or a condition on angular momentum.
The curve in the figure is the potential energy U(r) drawn on both sides of the nucleus, and the horizontal lines are the energies found below. Starting from -13.6 eV for n = 1, the lines get closer together as n grows and approach 0.
At energies above 0, the electron can move infinitely far away from the nucleus. The energy is then not discrete and can take any value. The electron of a hydrogen atom is bound to the nucleus when its energy is negative.
Because the potential energy depends only on the distance r, the wave function can be solved for as a product of a function of distance and a function of angles, \psi = R(r)\,Y(\theta, \varphi). The angular part Y is the same as in the article on angular momentum: it describes a state in which the angular momentum has magnitude L^2 = l(l+1)\hbar^2 and z component L_z = m\hbar.
For the radial part, putting u(r) = rR(r) gives an equation of the same form as the one-dimensional Schrödinger equation: -\dfrac{\hbar^2}{2m_\mathrm{e}}\dfrac{d^2u}{dr^2} + \left(U(r) + \dfrac{l(l+1)\hbar^2}{2m_\mathrm{e}r^2}\right)u = Eu.
The second term in the parentheses acts to push an electron with angular momentum away from the nucleus, and is called the centrifugal term. It has the same form as the rotational part of the kinetic energy of a rotating body in classical mechanics, \dfrac{L^2}{2mr^2}. For l = 0 this term is absent, and the electron can come right up to the nucleus.
The figure draws the sum of the Coulomb potential energy (thin line) and the centrifugal term (dotted line) for l = 0, 1, 2. Distance is measured in units of the Bohr radius a_0 = 0.0529 nm. The depths of the wells differ so much that the vertical scale is changed for each l. The larger l, the higher the curve near the nucleus, and the well becomes shallower and moves farther out.
The bottom of the well is at r = 2a_0 with -6.80 eV for l = 1, and at r = 6a_0 with -2.27 eV for l = 2. There are no states with energy below the bottom of the well, so the larger l, the higher the energy of the lowest state. The horizontal lines in the figure are the energies found in the next scene: states with l = 1 start from n = 2, and states with l = 2 start from n = 3.
The radial equation can be solved for any energy E. Starting from u = 0 at the nucleus and solving outward, for most values of E the solution grows exponentially, in the positive or negative direction, far away. For such a solution, the probability of the electron being far from the nucleus becomes unboundedly large, and it cannot describe an electron bound to the nucleus.
A solution that approaches 0 far away is obtained only for discrete energies. The figure shows the case l = 0: for E = -13.6 eV and E = -3.40 eV the solution approaches 0 far away, while for values between them or outside them it diverges. This is the same as for the harmonic oscillator, where only the solutions going to 0 far away remained.
The solution for l = 0 and E = -13.6 eV is u = re^{-r/a_0}, and putting it into the equation confirms that it holds. a_0 = \dfrac{4\pi\varepsilon_0\hbar^2}{m_\mathrm{e}e^2} is the Bohr radius, and the energy is E_1 = -\dfrac{\hbar^2}{2m_\mathrm{e}a_0^2} = -13.6 eV.
In general, the energies at which a solution going to 0 far away exists are E_n = -\dfrac{13.6\,\mathrm{eV}}{n^2}, where n is an integer 1 or larger. n is called the principal quantum number. If n_r is the number of times the solution u crosses 0 along the way, then n = n_r + l + 1, so for l = 1 even n_r = 0 gives n = 2. This is why, in the previous scene, the states with l = 1 started from n = 2.
The same energies as in Bohr's model were obtained without assuming either circular orbits or a condition on angular momentum. In 1926 Schrödinger carried out this calculation in his first paper and explained the spectrum of the hydrogen atom.
A state of the electron in a hydrogen atom is fixed by three numbers: the principal quantum number n, the azimuthal quantum number l and the magnetic quantum number m. n is an integer 1 or larger, l is an integer from 0 to n - 1, and m is an integer from -l to l. States with l = 0, 1, 2, 3 are called s, p, d and f, and the state with n = 2 and l = 1 is written 2p, for example.
The energy depends only on n, not on l or m. The number of states belonging to one n is 2l + 1 summed from l = 0 to n - 1, which is n^2: 1, 4 and 9 states for n = 1, 2, 3. That states with different l have the same energy is special to a potential energy proportional to \dfrac{1}{r}.
The probability of finding the electron at a distance between r and r + dr is proportional to |u(r)|^2dr. The figure draws this distribution for each state, with the highest point scaled to the same height. For 1s it is largest at r = a_0, which agrees with the radius of the orbit in Bohr's model. The electron is, however, spread out around that distance, and the average distance is 1.5a_0.
States with small l, such as 2s and 3s, have distances along the way at which the probability is 0. These are called nodes, and the dots on the axis in the figure mark them. The number of nodes is n - l - 1, and the number of peaks is n - l.
For 1s, 2p and 3d, where l = n - 1, there is only one peak, and its position r = n^2a_0 agrees with the orbit radii a_0, 4a_0 and 9a_0 of Bohr's model. The dotted line in the figure marks this radius.
For the same n, states with smaller l also have probability right near the nucleus. This difference is why, in atoms with two or more electrons, 2s has a lower energy than 2p, and it comes up again in the article on the periodic table.
The energies are the same as in Bohr's model, but the electron is described in a completely different way. In Bohr's model, the electron goes around a circle of radius a_0, and its position and speed are always fixed. In the solution of the Schrödinger equation, the position of the electron is fixed only as a probability distribution, which spreads around the nucleus as on the right of the figure.
The dots on the right of the figure are placed, on a plane through the nucleus, in proportion to the probability density of 1s. The dotted circle has radius a_0.
The angular momentum of the lowest state also differs. In Bohr's model it is L = \hbar, but the 1s state has l = 0, so its angular momentum is 0. An electron in 1s is not going around the nucleus; it is spread out in the same way in every direction.
An electron in 1s is most likely to be found near r = a_0, but it can also be found right near the nucleus and much farther away. The probability of finding it at r < a_0 is 32%, and at r > 3a_0 it is 6%.
Bohr's model gave the correct energies of the hydrogen atom, but it could not calculate the energies of atoms with two or more electrons, such as helium, or the strengths of spectral lines. The Schrödinger equation can treat these with the same approach.
Looked at closely, a single line in the hydrogen spectrum is made of several lines slightly apart. Explaining this fine splitting requires the effects of relativity and the spin of the electron. Spin is taken up in the article on the Stern-Gerlach experiment.