Spherical harmonics are still useful in the presence of V(r)[edit  edit source]
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:
$\left\{{\frac {\hbar ^{2}}{2\mu }}\left({\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}\right){\frac {\hbar ^{2}}{2\mu r^{2}}}{\hat {\Lambda }}(\theta ,\phi )+V(r)\right\}R(r)Y_{lm}(\theta ,\phi )=ER(r)Y_{lm}(\theta ,\phi )$ (Eq. 4)
which can be rewritten as (see Exercises to see why):
$\left\{{\frac {\hbar ^{2}}{2\mu }}\left({\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}\right)+{\frac {\hbar ^{2}l(l+1)}{2\mu r^{2}}}+V(r)\right\}R(r)=ER(r)$ (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[edit  edit source]
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
$\left\{{\frac {\hbar ^{2}}{2\mu }}\left({\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}\right)+{\frac {\hbar ^{2}l(l+1)}{2\mu r^{2}}}{\frac {e^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}\right\}R(r)=ER(r)$ (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, ... , n1 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[edit  edit source]
The functions R_{nl}(r) are tabulated. You need to remember just their general structure: R_{nl}(r) = (normalization) × (polynomial in r ÷ a) × (decaying exponential ~ exp(r ÷ a) ).
A few examples are given below:
 1s: R_{n=1,l=0}(r) = 2a^{3/2}exp(r/a)
 2s: R_{n=2,l=0}(r) = (2a)^{3/2}(2  r/a)exp(r/2a)
 2p: R_{n=2,l=1}(r) = 3^{1/2}(2a)^{3/2}(r/a)exp(r/2a)
where a ≈ 0.52 Å.
These are easy to plot considering that the number of nodes is nl and that they tend to be more extended for larger values of n.
Multiple integrals involving polar coordinates[edit  edit source]
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 r^{2}sinθdrdθdφ. 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 Ψ_{1s}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[edit  edit source]
The probability density of finding an electron at distance r is called the radial distribution function and it is given by P_{nl}(r) = r^{2} R_{nl}(r)^{2} (Eq. 12)
The reason for the extra factor r^{2} can be seen immediately if we express the probability of finding an electron at any angle θ or φ and at a distance between R_{1} and R_{2}:
$\left[\int _{0}^{\pi }\int _{0}^{2\pi }Y_{lm}(\theta ,\phi )^{2}\sin \theta \,d\theta \,d\phi \right]\left[\int _{R_{1}}^{R_{2}}R_{nl}(r)^{2}r^{2}\,dr\right]=\int _{R_{1}}^{R_{2}}R_{nl}(r)^{2}r^{2}\,dr$
[The first term is one because spherical harmonics are normalized]
Exercise
 Explain why it is possible to write Equation 5 from Equation 4.
 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)
 What is the ionization energy of the hydrogen atom?
 Plot the radial wavefunction and radial distribution function for the H orbitals 1s, 2s, 2p. Indicate if there are nodal planes.
 What is the distance where it is most likely to find an electron in the ground state of the hydrogen atom?
 Show that the radial equation for the H atom, the He^{1+} ion, and the Li^{2+} 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 hydrogenlike atoms.
 How can you write the energy and the wavefunction for all hydrogenlike atoms with any value of Z?
 Using your answer to Question 7, calculate the ionization potential of He^{1+}. Find the distance where it is most likely to find the electron in He^{1+}.
Solutions

 In this Section it was found ${\frac {\hbar ^{2}}{2mr^{2}}}\Lambda (\theta ,\phi )Y_{ml}(\theta ,\phi )={\frac {\hbar ^{2}}{2mr^{2}}}l(l+1)Y_{ml}(\theta ,\phi )$ $\left\{{\frac {\hbar ^{2}}{2\mu R^{2}}}\Lambda (\theta ,\phi )\right\}\psi (\theta ,\phi )=E\psi (\theta ,\phi )$
 2
 The ground state energy of the hydrogen atom is 13.6 eV. The energy with an electron and a proton at infinite distance is zero. The ionization energy is 13.6 eV.
 Only 2s has a nodal plane for r=2a.
 The general probability is $P_{nl}(r)=r^{2}R_{nl}(r)^{2}$. For the 1s orbital it is $P_{\mbox{1s}}(r)=Ar^{2}\exp(2r/a)$. A is the irrelevant normalization costant. The maximum is found by setting the derivative of P_{1s}(r) = 0. We get $2r\exp(2r/a)2{\frac {2r}{a}}\exp(2r/a)=0\Rightarrow r=a$.
 ${\frac {Ze^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ is the interaction energy between a nucleus with charge +Ze and an electron with charge e.
 In the hydrogenlike atom you have the term ${\frac {Ze^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ instead of ${\frac {e^{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ as seen in the hydrogen atom. The solution of the hydrogen atom is valid for all hydrogenlike atoms if you substitute e^{2} with Ze^{2} in the eigenvalues and the radial functions. The modified eigenvalues are therefore $E_{n}={\frac {\mu Z^{2}e^{4}}{32\pi ^{2}\varepsilon _{0}^{2}\hbar ^{2}n^{2}}}$ and the modified eigenfunctions have a / Z instead of a because $a={\frac {4\mu \varepsilon _{0}\hbar ^{2}}{\mu e^{2}}}$ and $a/Z={\frac {4\mu \varepsilon _{0}\hbar ^{2}}{\mu Ze^{2}}}$
 Ionization Potential = 13.6 x Z^{2} = 13.6 x 4 eV.


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