The Schrödinger equation for the hydrogen atom and hydrogen-like cations

So what does it look like? The Schrödinger equation for single-electron Coulomb systems in spherical coordinates is

$$-{\hbar^2 \over 2m_e r^2} \left[r{\partial^2 \over \partial ... ...er 4\pi\epsilon_0 r}\psi(r,\theta,\phi) = E\psi(r,\theta,\phi)$$

This type of equation is an example of a partial differential equation, which is no simple task to solve. However, solving it gives both the allowed values of the angular momentum discussed above and the allowed energies $E_n$, which agree with the simpler Bohr model. Thus, we do not need to assume anything except the validity of the Schrödinger equation, and the allowed values of energy and angular momentum, together with the corresponding wave functions, all emerge from the solution.

Obviously, we are not going to go through the solution of the Schrödinger equation, but we can understand something about its mechanics and the solutions from a few simple considerations. Remember that the Schrödinger equation is set up starting from the classical energy, which we said takes the form

$${p_r^2 \over 2m_e} + {L^2 \over 2m_e r^2} - {Ze^2 \over 4\pi\epsilon_0 r} = E$$

which we can write as

$$E = \varepsilon_r + {L^2 \over 2m_e r^2}$$

where

$$\varepsilon_r = {p_r^2 \over 2m_e} - {Ze^2 \over 4\pi\epsilon_0 r}$$

The term $L^2 /(2m_e r^2)$ is actually dependent only on $\theta$ and $\phi$, so it is purely angular. Given the separability of the energy into radial and angular terms, the wave function can be decomposed into a product of the form

$$\psi(r,\theta,\phi) = R(r)Y(\theta,\phi)$$

Solution of the angular part for the function $Y(\theta,\phi)$ yields the allowed values of the angular momentum $L^2$ and the $z$-component $L_z$. The functions $Y(\theta,\phi)$ are then characterized by the integers $l$ and $m$, and are denoted $Y_{lm}(\theta,\phi)$. They are known as spherical harmonics. Here we present just a few of them for a few values of $l$.

for $l=0$, there is just one value of $m$, $m=0$, and, therefore, one spherical harmonic, which turns out to be a simple constant:

$$Y_{00}(\theta,\phi) = {1 \over \sqrt{4\pi}}$$

For $l=1$, there are three values of $m$, $m=-1,0,1$, and, therefore, three functions $Y_{1m}(\theta,\phi)$. These are given by

$$Y_{11}(\theta,\phi) = -\left({3 \over 8\pi}\right)^{1/2} \sin\theta e^{i\phi}$$

$$Y_{1-1}(\theta,\phi) = \left({3 \over 8\pi}\right)^{1/2} \sin\theta e^{-i\phi}$$

$$Y_{10}(\theta,\phi) = \left({3 \over 4\pi}\right)^{1/2} \cos\theta$$

Remember that

$$e^{i\phi} = \cos\phi + i\sin\phi$$

$$e^{-i\phi} = \cos\phi - i\sin\phi$$

For $l=2$, there are five values of $m$, $m=-2,-1,0,1,2$, and, therefore, five spherical harmonics, given by

$$Y_{22}(\theta,\phi) = \left({15 \over 32\pi}\right)^{1/2}\sin^2\theta e^{2i\phi}$$

$$Y_{2-2}(\theta,\phi) = \left({15 \over 32\pi}\right)^{1/2}\sin^2\theta e^{-2i\phi}$$

$$Y_{21}(\theta,\phi) = -\left({15 \over 8\pi}\right)^{1/2}\sin\theta\cos\theta e^{i\phi}$$

$$Y_{2-1}(\theta,\phi) = \left({15 \over 8\pi}\right)^{1/2}\sin\theta\cos\theta e^{-i\phi}$$

$$Y_{20}(\theta,\phi) = \left({5 \over 16\pi}\right)^{1/2} \left(3\cos^2\theta - 1\right)$$

The remaining function $R(r)$ is characterized by the integers $n$ and $l$, as this function satisfies the radial part of the Schrödinger equation, also known as the radial Schrödinger equation:

$$-{\hbar^2 \over 2m_e} {1 \over r}{d^2 \over dr^2}\left(rR_{n... ...}(r) - {Ze^2 \over 4\pi \epsilon_0 r}R_{nl}(r) = E_n R_{nl}(r)$$

Note that, while the functions $Y_{lm}(\theta,\phi)$ are not particular to the potential $V(r)$, the radial functions $R_{nl}(r)$ are particular for the Coulomb potential. It is the solution of the radial Schrödinger equation that leads to the allowed energy levels. The boundary conditions that lead to the quantized energies are $rR_{nl}(0) = 0$ and $rR_{nl}(\infty) = 0$. The radial parts of the wave functions that emerge are given by (for the first few values of $n$ and $l$):

$$R_{10}(r) = 2\left({Z \over a_0}\right)^{3/2}e^{-Zr/a_0}$$

$$R_{20}(r) = {1 \over 2\sqrt{2}} \left({Z \over a_0}\right)^{3/2} \left(2-{Zr \over a_0}\right)e^{-Zr/2a_0}$$

$$R_{21}(r) = {1 \over 2\sqrt{6}} \left({Z \over a_0}\right)^{3/2}{Zr \over a_0}e^{-Zr/2a_0}$$

$$R_{30}(r) = {2 \over 81\sqrt{3}} \left({Z \over a_0}\right)^{3/2} \left[27-18{Zr \over a_0} + 2\left({Zr \over a_0}\right)^2\right]e^{-Zr/3a_0}$$

where $a_0$ is the Bohr radius

$$a_0 = {4\pi\epsilon_0\hbar^2 \over e^2 m_e} = 0.529177\times 10^{-10}\;{\rm m}$$

The full wave functions are then composed of products of the radial and angular parts as

$$\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)$$

At this points, several comments are in order. First, the integers $n,l,m$ that characterize each state are known as the quantum numbers of the system. Each of them corresponds to a quantity that is classically conserved. The number $n$ is known as the principal quantum number, the number $l$ is known as the angular quantum number, and the number $m$ is known as the magnetic quantum number.

As with any quantum system, the wave functions $\psi_{nlm}(r,\theta,\phi)$ give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. The probability of finding the electron in a small volume element $dV$ of space around the point ${\bf r}= (r,\theta,\phi)$ is

$$P({\rm electron\ in\ }dV{\rm about\ }r,\theta,\phi) = \vert\psi_{nlm}(r,\theta,\phi)\vert^2 dV$$

$$= R_{nl}^2(r)\vert Y_{lm}(\theta,\phi)\vert^2 dV$$

What is $dV$? In Cartesian coordinates, $dV$ is the volume of a small box of dimensions $dx$, $dy$, and $dz$ in the $x$, $y$, and $z$ directions. That is,

$$dV = dx dy dz$$

In spherical coordinates, the volume element $dV$ is a small element of a spherical volume and is given by

$$dV = r^2 \sin\theta dr d\theta d\phi$$

which is derivable from the transformation equations.

If we integrate $dV$ over a sphere of radius $R$, we should obtain the volume of the sphere:

$$V = \left[\int_0^R r^2 dr\right] \left[\int_0^{\pi}\sin^2\theta d\theta\right] \left[\int_0^{2\pi}d\phi\right]$$

$$= \left[\left.{r^3 \over 3}\right\vert _0^R\right] \left[\left.-\cos\theta\right\vert _0^{\pi}\right] \left[\left.\phi\right\vert _0^{2\pi}\right]$$

$$= {R^3 \over 3}\times 2\times 2\pi$$

$$= {4 \over 3}\pi R^3$$

which is the formula for the volume of a sphere of radius $R$.

Example: The electron in a hydrogen atom ($Z=1$) is in the state with quantum numbers $n=1$, $l=0$ and $m=0$. What is the probability that a measurement of the electron's position will yield a value $r\geq 2a_0$?

The wave function $\psi_{100}(r,\theta,\phi)$ is

$$\psi_{100}(r,\theta,\phi) = {2 \over a_0^{3/2}} \left({1 \over 4\pi}\right)^{1/2}e^{-r/a_0}$$

Therefore, the probability we seek is

$$P(r\geq 2a_0) = \int_0^{2\pi}\int_0^{\pi} \int_{2a_0}^{\infty} \vert\psi_{100}(r,\theta,\phi)\vert^2 r^2\sin\theta dr d\theta d\phi$$

$$= {4 \over a_0^3}{1 \over 4\pi} \left[\int_0^{2\pi}d\phi\right] \le... ...pi}\sin\theta d\theta\right] \left[\int_{2a_0}^{\infty}r^2 e^{-2r/a_0}dr\right]$$

$$= {1 \over a_0^3 \pi}\cdot 2\pi\cdot 2\int_{2a_0}^{\infty} r^2 e^{-2r/a_0}dr$$

Let $x=2r/a_0$. Then

$$P(r\geq 2a_0) = {1 \over 2}\int_4^{\infty} x^2 e^{-x}dx$$

After integrating by parts, we find

$$P(r\geq 2a_0) = 13e^{-4} \approx 0.24 = 24\%$$

which is relatively large given that this is at least two Bohr radii away from the nucleus!

The part of the probability involving the product

$$P_{nl}(r) = r^2 R_{nl}^2(r)$$

is known as the radial probability distribution function or simply the radial distribution function. $P_{nl}(r)dr$ is the probability that a measurement of the electron's position yields a value in a radial shell of thickness $dr$ and radius $r$ as shown in the figure below:

Figure: Illustration of a radial shell for the radial distribution with $n=1$, $l=0$.

What the radial probability distribution shows is that the electron cannot be sucked into the nucleus because $P_{nl}(0) = 0$. Hence, as we shrink the radial shell into the nucleus, the probability of finding the electron in that shell goes to 0.

Another point concerns the number of allowed states for each allowed energy. Remember that each wave function corresponds to a probability distribution in which the electron can be found for each energy. The more possible states there are, the more varied the electronic properties and behavior of the system will be.

For $n=1$, there is one energy $E_1$ and only one wave function $\psi_{100}$.

For $n=2$, there is one energy $E_2$ and four possible states, corresponding to the following allowable values of $l$ and $m$

$$l=0\;\;\;m=0$$

$$l=1\;\;\;m=-1,0,1$$

Thus, there are four wave functions $\psi_{200}$, $\psi_{21-1}$, $\psi_{210}$, and $\psi_{211}$. Whenever there is more than one wave function corresponding to a given energy level, then that energy level is said to be degenerate. In the above example, the $n=2$ energy level is fourfold degenerate.