# 16: Linear Momentum and Electronic Spectroscopy

Recap of Lecture 15

Last lecture continued the discussion of the 3D rigid rotor. We discussed the three aspect of the solutions to this system: The wavefunctions (the spherical harmonics), the energies (and degeneracies) and the TWO quantum numbers ($$J$$ and $$m_J$$) and their ranges. We discussed that the components of the angular momentum operator are subject to the Heisenberg uncertainty principle and cannot be know to infinite precision simultaneously, however the magnitude of angular momentum and any component can be. This results in the vectoral representation of angular momentum

This is one example of several cyclic permutations of the fundamental commutation relations satisfied by the components of an orbital angular momentum:

$[L_x, L_y] = {\rm i}\,\hbar\, L_z \label{6.3.17a}$

$[L_y, L_z] = {\rm i}\,\hbar\, L_x \label{6.3.17b}$

$[L_z, L_x] = {\rm i}\,\hbar\, L_y \label{6.3.17c}$

Therefore, two orthogonal components of angular momentum (for example $$L_x$$ and $$L_y$$) are complementary and cannot be simultaneously known or measured, except in special cases such as $$\displaystyle L_{x}=L_{y}=L_{z}=0$$.

We can introduce a new operator $$\hat{L}^2$$:

$\hat{L}^2 = L_x^{\,2}+L_y^{\,2}+L_z^{\,2} \label{6.3.5}$

That is the magnitude of the Angular momentum squared.

It is possible to simultaneously measure or specify $$L^2$$ and any one component of $$L$$; for example, $$L^2$$ and $$L_z$$. This is often useful, and the values are characterized by ($$J$$) and ($$m_J$$). In this case the quantum state of the system is a simultaneous eigenstate of the operators $$L^2$$ and $$L_z$$, but not of $$L_x$$ or $$L_y$$.

Illustration of the vector model of orbital angular momentum. Image used with permission (Public domain; Maschen).

Since the angular momenta are quantum operators, they cannot be drawn as vectors like in classical mechanics. Nevertheless, it is common to depict them heuristically in this way. Depicted above is a set of states with quantum numbers $${\displaystyle J =2}$$, and $${\displaystyle m_{J}=-2,-1,0,1,2}$$ for the five cones from bottom to top.

Since $${\displaystyle |L|={\sqrt {L^{2}}}=\hbar {\sqrt {6}}}$$, the vectors are all shown with length $${\displaystyle \hbar {\sqrt {6}}}$$. The rings represent the fact that $${\displaystyle L_{z}}$$ is known with certainty, but $${\displaystyle L_{x}}$$ and $${\displaystyle L_{y}}$$ are unknown; therefore every classical vector with the appropriate length and z-component is drawn, forming a cone.

The expected value of the angular momentum for a given ensemble of systems in the quantum state characterized by $${\displaystyle \J}$$ and $${\displaystyle m_{J }}$$ could be somewhere on this cone while it cannot be defined for a single system (since the components of $${\displaystyle L}$$ do not commute with each other).

## Spherical Harmonics in Rigid Rotors

The wavefunction of a rigid rotor can be separte (via the separation of variables approach) into product of two functions:

$\color{red} | Y(\theta, \phi) \rangle = \Theta(\theta) \cdot \Phi(\phi) \rangle$

The $$\Phi$$ function is found to have quantum number $$m$$. $$\Phi_m (\phi) = A_m e^{im\phi}$$, where $$A_m$$ is the normalization constant and $$m = 0, \pm1, \pm2 ... \pm\infty$$. The $$\Theta$$ function was solved and is known as Legendre polynomials, which have quantum numbers $$m$$ and $$\ell$$. When $$\Theta$$ and $$\Phi$$ are multiplied together, the product is known as spherical harmonics with labeling $$Y_{J}^{m} (\theta, \phi)$$.

$$m_J$$
$$J$$
$$\Theta ^{m_J}_J (\theta)$$
$$\Phi (\varphi)$$
$$Y^{m_J}_J (\theta , \varphi)$$
0
0
$$\dfrac {1}{\sqrt {2}}$$
$$\dfrac {1}{\sqrt {2 \pi}}$$
$$\dfrac {1}{\sqrt {4 \pi}}$$
0
1
$$\sqrt {\dfrac {3}{2}}\cos \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}$$
$$\sqrt {\dfrac {3}{4 \pi}}\cos \theta$$
1
1
$$\sqrt {\dfrac {3}{4}}\sin \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$$
$$\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{i \varphi}$$
-1
1
$$\sqrt {\dfrac {3}{4}}\sin \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$$
$$\sqrt {\dfrac {3}{8 \pi}}\sin \theta e^{-i \varphi}$$
0
2
$$\sqrt {\dfrac {5}{8}}(3\cos ^2 \theta - 1)$$
$$\dfrac {1}{\sqrt {2 \pi}}$$
$$\sqrt {\dfrac {5}{16\pi}}(3\cos ^2 \theta - 1)$$
1
2
$$\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}e^{i \varphi}$$
$$\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{i\varphi}$$
-1
2
$$\sqrt {\dfrac {15}{4}} \sin \theta \cos \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}e^{-i\varphi}$$
$$\sqrt {\dfrac {15}{8\pi}} \sin \theta \cos \theta e^{-i\varphi}$$
2
2
$$\sqrt {\dfrac {15}{16}} \sin ^2 \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}e^{2i\varphi}$$
$$\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{2i\varphi}$$
-2
2
$$\sqrt {\dfrac {15}{16}} \sin ^2 \theta$$
$$\dfrac {1}{\sqrt {2 \pi}}e^{2i\varphi}$$
$$\sqrt {\dfrac {15}{32\pi}} \sin ^2 \theta e^{-2i\varphi}$$

The above figure show the spherical harmonics $$Y_J^M$$, which are solutions of the angular Schrödinger equation of a 3D rigid rotor.

## Microwave Spectroscopy Probes Rotations

I will skip over this topic in lecture, but you have a workseet on it in your discussion and I expect you to master this concept. Not that hard actually: once you get the eigenstate for a system you can couple in a spectroscopy, in this microwave spectroscopy.

The permanent electric dipole moments of polar molecules can couple to the electric field of electromagnetic radiation. This coupling induces transitions between the rotational states of the molecules. The energies that are associated with these transitions are detected in the far infrared and microwave regions of the spectrum. For example, the microwave spectrum for carbon monoxide spans a frequency range of 100 to 1200 GHz, which corresponds to 3 - 40 $$cm^{-1}$$.

The selection rules for the rotational transitions are derived from the transition moment integral by using the spherical harmonic functions and the appropriate dipole moment operator, $$\hat {\mu}$$.

$\mu _T = \int Y_{J_f}^{m_f*} \hat {\mu} Y_{J_i}^{m_i} \sin \theta \,d \theta \,d \varphi \label {5.9.1}$

Evaluating the transition moment integral involves a bit of mathematical effort. This evaluation reveals that the transition moment depends on the square of the dipole moment of the molecule, $$\mu ^2$$ and the rotational quantum number, $$J$$, of the initial state in the transition,

$\mu _T = \mu ^2 \dfrac {J + 1}{2J + 1} \label {5.9.2}$

and that the selection rules for rotational transitions are

$\color{red} \Delta J = \pm 1 \label {5.9.3}$

$\color{red} \Delta m_J = 0, \pm 1 \label {5.9.4}$

A photon is absorbed for $$\Delta J = +1$$ and a photon is emitted for $$\Delta J = -1$$.

The energies of the rotational levels are given by

$E = J(J + 1) \dfrac {\hbar ^2}{2I}$

and each energy level has a degeneracy of $$2J+1$$ due to the different $$m_J$$ values.

Each energy level of a rigid rotor has a degeneracy of $$2J+l$$ due to the different $$m_J$$ values.

The transition energies for absorption of radiation are given by

$\Delta E_{states} = E_f - E_i = E_{photon} = h \nu = hc \bar {\nu} \label {5.9.5}$

$h \nu =hc \bar {\nu} = J_f (J_f +1) \dfrac {\hbar ^2}{2I} - J_i (J_i +1) \dfrac {\hbar ^2}{2I} \label {5.9.6}$

Since microwave spectroscopists use frequency, and infrared spectroscopists use wavenumber units when describing rotational spectra and energy levels, both $$\nu$$ and $$\bar {\nu}$$ are included in Equation $$\ref{5.9.6}$$, and $$J_i$$ and $$J_f$$ are the rotational quantum numbers of the initial (lower) and final (upper) levels involved in the absorption transition. When we add in the constraints imposed by the selection rules, $$J_f$$ is replaced by $$J_i + 1$$, because the selection rule requires $$J_f – J_i = 1$$ for absorption. The equation for absorption transitions then can be written in terms of the quantum number $$J_i$$ of the initial level alone.

$h \nu = hc \bar {\nu} = 2 (J_i + 1) \dfrac {\hbar ^2}{2I} \label {5.9.7}$

Divide Equation $$\ref{5.9.7}$$ by $$h$$ to obtain the frequency of the allowed transitions,

$\nu = 2B (J_i + 1) \label {5.9.8}$

where $$B$$, the rotational constant for the molecule, is defined as

$B = \dfrac {\hbar ^2}{2I} \label {5.9.9}$

The rotation spectrum of $$^{12}C^{16}O$$ at 40 K.

This microwave spectrum can be decomposed into eigenstates thusly

Energy levels and line positions calculated in the rigid rotor approximation. Image used with permission from Wikipedia

Example $$\PageIndex{1}$$: Carbon Monoxide

Calculate the bondlength of carbon monoxide if the $$J = 0$$ to $$J = 1$$ transition for $$^{12}C^{16}O$$ in its microwave spectrum is $$1.153 \times 10^{5} MHz$$.

Solution:

Assuming that $$^{12}C^{16}O$$ can be treated as a ridged rotor

$\nu = 2B(J + 1)$

with $$J = 1, 2, 3, . . .$$ and $$B = \dfrac{h}{8\pi^2l}$$

For the $$J = 0$$ to $$J = 1$$ transition,

$\dfrac {1}{2}\nu = B = \dfrac{h}{8\pi^2l}$

$\dfrac {1}{2}1.153 \times 10^{11} s^{-1} = B = \dfrac{6.626 \times 10^{-34} J.s}{8\pi^2\mu r^2}$

We can find $$\mu$$ and use the relationship $$r^2 = \dfrac{1}{\mu}$$ to find $$r$$.

$\mu = \dfrac{(12.00)(15.99)}{27.99} 1.661 \times 10^{-27} kg = 1.139 \times 10^{-26} kg$

$r^2 = \dfrac{6.626 \times 10^{-34} J.s}{4pi^2 1.139 \times 10^{-26} kg 1.153 \times 10^{11} s^-1}$

= $$1.13 \times 10^{-10} m$$ = 113 pm

Thus the bondlength in carbon monoxide is 113 pm.

## The Eigenvalue Problem for the Hydrogen Atom

Step 1: Define the potential for the problem

For the Hydrogen atom, the potential energy is given by the Coulombic potential, which is

$\color{red}V(r) = -\dfrac {e^2}{4\pi \epsilon_0 r}$

Step 2: Define the Schrödinger Equation for the problem

With every quantum eigenvalue problem, we define the Hamiltonian as such:

$\hat {H} = T + V$

The potential is defined above and the Kinetic energy is given by

$T = -\dfrac {\hbar^2}{2m_e} \bigtriangledown^2$

The Hamiltonian for the Hydrogen atom becomes

$\hat {H} = -\dfrac {\hbar^2}{2m_e}\bigtriangledown^2 - \dfrac {e^2}{4\pi \epsilon_0 r}\label {1}$

and since the potential has no time-dependence, we can se the time independent Schrödinger Equation

$\hat {H} | \psi (x,y,z) \rangle = E | \psi (x,y,z) \rangle$

Step 3: Solve the Schrödinger Equation for the problem

The potential has a spherical symmetry (i..e, depends only on $$r$$ and not typically in terms of $$x$$, $$y$$ and $$z$$), so switching to spherical coordinates is useful. The new eigenvalue problem is

$\hat {H}\psi(r,\theta,\phi)$

$= -\dfrac {\hbar^2}{2m_e} \left [\dfrac {1}{r^2} \dfrac {d}{d r} \left(r^2 \dfrac {d \psi(r,\theta,\phi)}{d r}\right) + \dfrac {1}{r^2 \sin(\theta)} \dfrac {d}{d \theta} \left(\sin(\theta) \dfrac {d \psi(r,\theta,\phi)}{d \theta}\right) + \dfrac {1}{r^2 \sin^2(\theta)} \dfrac {d^2 \psi(r,\theta,\phi)}{d \phi^2} \right] - \dfrac {e^2}{4\pi\epsilon_0 r} \psi(r,\theta,\phi)$

$= E\psi (r,\theta,\phi) \label{2}$

Multiplying equation $${2}$$ by $$2m_e r^2$$ and moving $$E$$ to the left side gives

$\hbar^2 \left(\dfrac {d}{d r} r^2 \dfrac {d \psi(r,\theta,\phi) }{d r}\right) - \hbar^2 \left[\dfrac {1}{\sin (\theta)} \left(\dfrac {d}{d \theta} \sin (\theta) \dfrac {d \psi(r,\theta,\phi) }{d \theta}\right) + \dfrac {1}{\sin^2 (\theta)} \dfrac {d^2 \psi(r,\theta,\phi) }{d \phi^2} \right] - 2m_e r^2 \left [\dfrac {e^2}{4\pi\epsilon_0 r} + E \right] \psi (r,\theta,\phi) = 0 \label {3}$

Although Equation $$\ref{3}$$ is a complex equation, it can be simplified by using three formulas,

$\hat{L}^2 = - \hbar^2 \left [\dfrac {1}{\sin (\theta)} (\dfrac {d}{d \theta} \sin (\theta) \dfrac {d \psi}{d \theta}) + \dfrac {1}{\sin^2 (\theta)} \dfrac {d^2 \psi}{d \phi^2}\right] \label{ 4a}$

$\color{red} \psi(r,\theta,\phi) = R(r)Y_{\ell}^{m}(\theta,\phi) \label{4b}$

where

$\hat {L}^2 Y_{\ell}^{m} = \hbar^2 \ell(\ell + 1) Y_{\ell}^{m} \label{4c}$

Using these three equations for Equation $$\ref{3}$$ gives

$-\dfrac {\hbar^2}{2m_e r^2} \dfrac {d}{d r} (r^2 \dfrac {d R(r)}{d r}) + \left[\dfrac {\hbar^2 \ell(\ell+1)}{2m_e r^2} - \dfrac{e^2}{4\pi\epsilon_0 r} - E \right] R(r) = 0 \label {5}$

A solution for $$R(r)$$ can be found with a quantum number $$n$$, and then $$E_n$$ is solved as

$\color{red} E_n = -\dfrac {m_e e^4}{8\epsilon_0^2 h^2 n^2}\label{6}$

with $$n=1,2,3 ...\infty$$

Step 4: Do something with the Eigenstates and associated energies

Let's talk about the solutions first.

## Ranges of the Three Quantum Numbers

In solving these types of differential equations, there are limits on $$\ell$$ and $$m_{\ell}$$, but not $$n$$.

• $$n = 1, 2, 3, ...\infty$$
• $$\ell = 0, 1, 2, ... (n-1)$$
• $$m_{\ell} = 0, \pm 1, \pm 2, ... \pm \ell$$

## Orthonormality

The normalization condition for the hydrogen atomic wavefunction is given by

$\int_0^{\infty} r^2 dr \int_0^{\pi} \sin(\theta) d\theta \int_0^{2\pi} \psi^*_{n \ell\ m} \psi_{n \ell\ m} d\phi = d_{nn'} d_{\ell\ell'} d_{mm'}\label{7a}$

Each $$n$$, $$\ell$$, $$m$$ value correspond to a specific orbital, which can be represented by a wavefunction. For example, $$\psi_{100}$$ corresponds to a wavefuntion with $$n=1, \ell=0, m_{\ell} = 0$$. This is the 1s orbital. Thus $$\psi_{100}$$ can be referred to as $$\psi_{1s}$$. With the same reasoning, $$\psi_{210}$$ refers to $$\psi_{2p_z}$$

$\color{red} | \psi(r,\theta,\phi) \rangle = |R(r) \rangle |Y_{\ell}^{m}(\theta,\phi) \rangle \label{7b}$

## Angular Part

From our work on the rigid rotor, we know that the eigenfunctions of the angular momentum operator are the Spherical Harmonic functions (Table M4), $$Y (\theta ,\phi )$$,

Spherical Harmonics 5 as commonly displayed, sorted by increasing energies and aligned for symmetry.

The Spherical Harmonic functions provide information about where the electron is around the proton, and the radial function R(r) describes how far the electron is away from the proton. The $$R(r)$$ functions that solve the radial differential equation, are products of the associated Laguerre polynomials times the exponential factor, multiplied by a normalization factor $$(N_{n,l})$$ and $$\left (\dfrac {r}{a_0} \right)^l$$.

$\color{red} R (r) = \underbrace{N_{n,l} \left ( \dfrac {r}{a_0} \right ) ^l}_{\text{Normalization}} \times \overbrace{L_{n,l} (r)}^{\text{Laguerre Poly}} \times \underbrace{e^{-\frac {r}{n {a_0}}}}_{\text{exponential}} \label {6.1.17}$

The decreasing exponential term overpowers the increasing polynomial term so that the overall wavefunction exhibits the desired approach to zero at large values of $$r$$. The first six radial functions are provided in Table below. Note that the functions in the table exhibit a dependence on $$Z$$, the atomic number of the nucleus.

Radial wave functions for wavefunctions of the first three shells. the Bohr radius $$a_o$$ is approximately 52.9 pm (half an angstrom).
$$n$$
$$l$$
$$R(r)$$
1
0
$$2 \left( \dfrac{Z}{a_o} \right)^{3/2} e^{ -Zr/a_o}$$
2
0
$$\dfrac{1}{2\sqrt{2}} \left( \dfrac{Z}{a_o} \right)^{3/2}\left [2-\dfrac{Zr}{a_o} \right]e^{ -Zr/2a_o}$$
2
1
$$\dfrac{1}{2\sqrt{6}} \left( \dfrac{Z}{a_o} \right)^{3/2}\left [\dfrac{Zr}{a_o} \right] e^{ -Zr/2a_o}$$
3
0
$$\dfrac{2}{81 \sqrt{3}} \left( \dfrac{Z}{a_o} \right)^{3/2}\left [27 -18 \dfrac{Zr}{a_o} +2 \left( \dfrac{Zr}{a_o}\right)^2 \right] e^{ -Zr/3a_o}$$
3
1
$$\dfrac{4}{81 \sqrt{6}} \left( \dfrac{Z}{a_o} \right)^{3/2}\left [6 \dfrac{Zr}{a_o} - \left( \dfrac{Zr}{a_o}\right)^2 \right] e^{ -Zr/3a_o}$$
3
2
$$\dfrac{4}{81 \sqrt{30}} \left( \dfrac{Z}{a_o} \right)^{3/2} \left ( \dfrac{Zr}{a_o} \right)^2 e^{ -Zr/3a_o}$$

Java simulation of particles in boxes :https://phet.colorado.edu/en/simulation/bound-states

## Energy

The motion of the electron in the hydrogen atom is not free. The electron is bound to the atom by the attractive force of the nucleus and consequently quantum mechanics predicts that the total energy of the electron is quantized. The expression for the energy is:

$E_n =\dfrac{-2 \pi^2 m e^4Z^2}{n^2h^2} \label{17.1}$

with $$n = 1,2,3,4...$$

where $$m$$ is the mass of the electron, $$e$$ is the magnitude of the electronic charge, $$n$$ is a quantum number, $$h$$ is Planck's constant and $$Z$$ is the atomic number (the number of positive charges in the nucleus). Equation $$\ref{17.1}$$ applies to any one-electron atom or ion. For example, He+ is a one-electron system for which Z = 2. We can again construct an energy level diagram listing the allowed energy values (Figure $$\PageIndex{1}$$).

These are obtained by substituting all possible values of n into Equation $$\ref{17.1}$$. As in our previous example, we shall represent all the constants which appear in the expression for $$E_n$$ by teh Rydberg constant $$R$$ and we shall set $$Z = 1$$, i.e., consider only the hydrogen atom.

$E_n = \dfrac{-R}{n^2} \label{17.2}$

with $$n = 1,2,3,4...$$

## Laguerre polynomials

The first six Laguerre polynomials. Image used with permission from Wikipedia.

The constraint that $$n$$ be greater than or equal to $$l +1$$ also turns out to quantize the energy, producing the same quantized expression for hydrogen atom energy levels that was obtained from the Bohr model of the hydrogen atom discussed in Chapter 2.

$E_n = - \dfrac {m_e e^4}{8 \epsilon ^2_0 h^2 n^2}$

Comparison to Bohr Theory

It is interesting to compare the results obtained by solving the Schrödinger equation with Bohr’s model of the hydrogen atom. There are several ways in which the Schrödinger model and Bohr model differ.

1. First, and perhaps most strikingly, the Schrödinger model does not produce well-defined orbits for the electron. The wavefunctions only give us the probability for the electron to be at various directions and distances from the proton.
2. Second, the quantization of angular momentum is different from that proposed by Bohr. Bohr proposed that the angular momentum is quantized in integer units of $$\hbar$$, while the Schrödinger model leads to an angular momentum of $$\sqrt{(l (l +1) \hbar ^2}$$.
3. Third, the quantum numbers appear naturally during solution of the Schrödinger equation while Bohr had to postulate the existence of quantized energy states. Although more complex, the Schrödinger model leads to a better correspondence between theory and experiment over a range of applications that was not possible for the Bohr model.

Methods for separately examining the radial portions of atomic orbitals provide useful information about the distribution of charge density within the orbitals.

Figure 6.2.2: Radial function, R(r), for the 1s, 2s, and 2p orbitals.

We could also represent the distribution of negative charge in the hydrogen atom in the manner used previously for the electron confined to move on a plane (Figure $$\PageIndex{1}$$), by displaying the charge density in a plane by means of a contour map. Imagine a plane through the atom including the nucleus. The density is calculated at every point in this plane. All points having the same value for the electron density in this plane are joined by a contour line (Figure $$\PageIndex{3}$$). Since the electron density depends only on r, the distance from the nucleus, and not on the direction in space, the contours will be circular. A contour map is useful as it indicates the "shape" of the density distribution.

$$n$$

$$\ell$$

$$m$$ Eigenstates
Hydrogen-like atomic wavefunctions for $$n$$ values $$1,2,3$$: $$Z$$ is the atomic number of the nucleus, and $$\rho = \dfrac {Zr}{a_0}$$, where $$a_0$$ is the Bohr radius and r is the radial variable.
$$n=1$$ $$\ell=0$$ $$m=0$$ $$\psi_{100} = \dfrac {1}{\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} e^{-\rho}$$
$$n=2$$ $$\ell=0$$ $$m=0$$ $$\psi_{200} = \dfrac {1}{\sqrt {32\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (2-\rho)e^{\dfrac {-\rho}{2}}$$
$$\ell=1$$ $$m=0$$ $$\psi_{210} = \dfrac {1}{\sqrt {32\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho e^{-\rho/2} \cos(\theta)$$
$$\ell=1$$ $$m=\pm 1$$ $$\psi_{21\pm\ 1} = \dfrac {1}{\sqrt {64\pi}} \left(\dfrac {Z}{a_0}\right)^{\dfrac {3}{2}} \rho e^{-\rho/2} \sin(\theta) e^{\pm\ i\phi}$$
$$n=3$$ $$\ell=0$$ $$m=0$$ $$\psi_{300} = \dfrac {1}{81\sqrt {3\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (27-18\rho +2\rho^2)e^{-\rho/3}$$
$$\ell=1$$ $$m=0$$ $$\psi_{310} = \dfrac{1}{81} \sqrt {\dfrac {2}{\pi}} \left(\dfrac {Z}{a_0}\right)^{\dfrac {3}{2}} (6r - \rho^2)e^{-\rho/3} \cos(\theta)$$
$$\ell=1$$ $$m=\pm 1$$ $$\psi_{31\pm\ 1} = \dfrac {1}{81\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (6\rho - \rho^2)e^{-r/3} \sin(\theta)e^{\pm\ i \phi}$$
$$\ell=2$$ $$m=0$$ $$\psi_{320} = \dfrac {1}{81\sqrt {6\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3}(3cos^2(\theta) -1)$$
$$\ell=2$$ $$m=\pm 1$$ $$\psi_{32\pm\ 1} = \dfrac {1}{81\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3} \sin(\theta)\cos(\theta)e^{\pm\ i \phi}$$
$$\ell=2$$ $$m=\pm 2$$

$$\psi_{32\pm\ 2} = \dfrac {1}{162\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3}{\sin}^2(\theta)e^{\pm\ 2i\phi}$$