Quantum mechanics/The hydrogen atom

Spherical harmonics are still useful in the presence of V(r)
In the past Lesson we wrote the Hamiltonian for a system of two particles with a potential energy that depends only on their distance:

$$\left \{ -\frac{\hbar^2}{2\mu} \left ( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \hat{\Lambda} (\theta, \phi) \right ) + V(r) \right \} \psi (r, \theta, \phi) = E \psi (r, \theta, \phi)$$ (Eq. 1)

We have described the solution in the particular case where r is constant (r = R):

$$\left \{ -\frac{\hbar^2}{2\mu R^2} \hat{\Lambda} (\theta, \phi) \right \} Y_{lm} (\theta, \phi) = E_l Y_{lm} (\theta, \phi)$$ (Eq. 2)

We can hope that the solution in the general case has the following structure:

$$\psi (r, \theta, \phi) = R(r) Y_{lm} (\theta, \phi)$$ (Eq. 3)

i.e. it is a product of a function of only the coordinate r times the spherical harmonics, eigenfunctions of the rigid rotor.

We are searching for the solution of:

$$\left \{ -\frac{\hbar^2}{2\mu} \left ( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right ) - \frac{\hbar^2}{2\mu r^2} \hat{\Lambda} (\theta, \phi) + V(r) \right \} R(r) Y_{lm} (\theta, \phi) = E R(r) Y_{lm} (\theta, \phi)$$ (Eq. 4)

which can be rewritten as (see Exercises to see why):

$$\left \{ -\frac{\hbar^2}{2\mu} \left ( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right ) + \frac{\hbar^2 l(l+1)}{2\mu r^2} + V(r) \right \} R(r) = E R(r)$$ (Eq. 5)

We have obtained an equation for R(r) which does not contain (θ, φ). So it is true that the solution for a generic potential V(r) can be expressed as the product of an angular part (the known spherical harmonic) and a radial part (which depends on the quantum number l) and is the solution of Equation 5.

Eigenvalues and eigenfunctions of the hydrogen atom
The potential energy between an electron and a proton is

$$V(r) = -\frac{e^2}{4 \pi \varepsilon_0} \frac{1}{r}$$ (Eq. 6)

So the radial equation to be solved is

$$\left \{ -\frac{\hbar^2}{2\mu} \left ( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right ) + \frac{\hbar^2 l(l+1)}{2\mu r^2} - \frac{e^2}{4 \pi \varepsilon_0} \frac{1}{r} \right \} R(r) = E R(r)$$ (Eq. 7)

The eigenvalues (given without proof) are:

$$E_n = -\frac{\mu e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2 n^2}$$ (Eq. 8)

And they depend only on a new quantum number n named the principal quantum number which can take the values 1,2,3,... . However the total eigenfunction also contains the angular part as proposed in Equation 3 which will simply be the spherical harmonics.

The radial part R(r) must depend on the quantum number l, because it is a solution of Equation 7 which contains l. The total eigenfunctions have the following structure (note the indexes/quantum numbers):

$$\Psi_{nlm} (r, \theta, \phi) = R_{nl} (r) Y_{lm} (\theta, \phi)$$ (Eq. 9)

For each n the allowed values of the angular momentum quantum number l are l = 0, ..., n-1 and, for each l, the allowed values of the magnetic quantum numbers are m = -l, -l+1, ... , l.

It is often said that l determines the shape of the orbital and m its orientation. The orbitals are also called s, p, d, f, g for l = 0, 1, 2, 3, 4 respectively. So, when an orbital is denoted, for example, as 3d it is meant n=3 and l=2.

The hydrogen atom radial function
The functions Rnl(r) are tabulated. You need to remember just their general structure: Rnl(r) = (normalization) × (polynomial in r ÷ a) × (decaying exponential ~ exp(-r ÷ a) ).

A few examples are given below: where a ≈ 0.52 Å.
 * 1s: Rundefined(r) = 2a-3/2exp(-r/a)
 * 2s: Rundefined(r) = (2a)-3/2(2 - r/a)exp(-r/2a)
 * 2p: Rundefined(r) = 3-1/2(2a)-3/2(r/a)exp(-r/2a)

These are easy to plot considering that the number of nodes is n-l and that they tend to be more extended for larger values of n.

Multiple integrals involving polar coordinates
When we have to integrate a function of x,y,z over all space, we write a triple integral in this way:

$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f(x,y,z) \,dx\,dy\,dz$$

When we use polar coordinates in 3D the element of volume dxdydz must be substituted by r2sinθdrdθdφ. To integrate a function over all space one has to write:

$$\int_0^{\infty} \int_0^{\pi} \int_0^{2\pi} g(r, \theta, \phi) r^2 \sin\theta \,dr\,d\theta\,d\phi$$ (Eq. 11)

Remembering that for all space, 0 < r < ∞ ; 0 < θ < π ; 0 < φ < 2π. If the angular and radial part are separable, the triple integral can be solved separately:

$$\int_0^{\infty} \int_0^{\pi} \int_0^{2\pi} A(r) B(\theta, \phi) r^2 \sin\theta \,dr\,d\theta\,d\phi = \left [ \int_0^{\pi} \int_0^{2\pi} B(\theta, \phi) \sin\theta \,d\theta\,d\phi \right ] \left [ \int_0^{\infty} A(r) r^2 \,dr \right ]$$

To practice with these integrals, we will check that the 1s wavefunction of the hydrogen atom is normalized.

$$\Psi_{\mbox{1s}} = \Psi_{n=1,l=0,m=0} = \left[ R_{n=1,l=0}(r) \right] \left[ Y_{l=0,m=0}(\theta, \phi) \right] = \left[ 2a^{-3/2} \exp(-r/a) \right] \left[ (4\pi)^{-1/2} \right]$$

$$|\Psi_{\mbox{1s}}|^2 = \left [ (4a)^{-3} \exp(-2r/a) \right ] \left [ (4\pi)^{-1} \right ]$$

The wavefunction is normalized if the integral of |Ψ1s|² = 1. In this particular case

$$\left [ \int_0^\pi \int_0^{2\pi} \left [ (4\pi)^{-1} \right ] \sin\theta \,d\theta\,d\phi \right ] \left [ \int_0^\infty \left [ 4a^{-3} \exp(-2r/a) \right ] r^2 \,dr \right ] = 1$$

It is easy to show that both terms in the square parentheses are equal to 1.

$$\int_0^\pi \int_0^{2\pi} \sin\theta \,d\theta\,d\phi = (4\pi)^{-1} \left ( \int_0^\infty \sin\theta \,d\theta \right ) \left ( \int_0^{2\pi} \,d\phi \right ) = (4\pi)^{-1} (2)(2\pi) = 1$$

$$\left [ (4a)^{-3} \right ] \int_0^\infty \exp(-2r/a) r^2 \,dr = \left [ (4a)^{-3} \right ] a^3 \int_0^\infty \exp(-2r/a)(r/a)^2 \,d(r/a) = 4\int_0^\infty x^2 e^{-2x} \,dx = 1$$

[In the last integral we substitute r/a for x]

Radial distribution function
The probability density of finding an electron at distance r is called the radial distribution function and it is given by Pnl(r) = r2 |Rnl(r)|2 (Eq. 12)

The reason for the extra factor r2 can be seen immediately if we express the probability of finding an electron at any angle θ or φ and at a distance between R1 and R2:

$$\left [ \int_0^\pi \int_0^{2\pi} |Y_{lm} (\theta, \phi)|^2 \sin\theta \,d\theta\,d\phi \right ] \left [ \int_{R_1}^{R_2} |R_{nl} (r)|^2 r^2 \,dr \right ] = \int_{R_1}^{R_2} |R_{nl} (r)|^2 r^2 \,dr$$

[The first term is one because spherical harmonics are normalized]


 * 1) Explain why it is possible to write Equation 5 from Equation 4.
 * 2) Evaluate all the constants in Equation 8 showing that the energy levels of the hydrogen atom are $$E_n = -\frac{13.6}{n^2}$$ (where the energy is expressed in electronvolts)
 * 3) What is the ionization energy of the hydrogen atom?
 * 4) Plot the radial wavefunction and radial distribution function for the H orbitals 1s, 2s, 2p. Indicate if there are nodal planes.
 * 5) What is the distance where it is most likely to find an electron in the ground state of the hydrogen atom?
 * 6) Show that the radial equation for the H atom, the He1+ ion, and the Li2+ ion can be written as $$\left \{ -\frac{\hbar^2}{2\mu} \left ( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right ) + \frac{\hbar^2 l(l+1)}{2\mu r^2} - \frac{Ze^2}{4\pi\varepsilon_0} \frac{1}{r} \right \} R(r) = ER(r)$$ where Z=1,2,3 respectively. Atoms with only one electron are hydrogen-like atoms.
 * 7) How can you write the energy and the wavefunction for all hydrogen-like atoms with any value of Z?
 * 8) Using your answer to Question 7, calculate the ionization potential of He1+. Find the distance where it is most likely to find the electron in He1+.

Next: Lesson 8 - Operators and Measurements