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Section 7.3 The Hydrogen Atom

We are now ready to talk about the actual wavefunctions of the simplest atom: hydrogen. Specifically, we consider the wavefunctions for the electron that “orbits” around the single proton in the nucleus. We put the word “orbits” in quotes because the behavior of the electron is quite different from the behavior of a satellite orbiting around the Earth. Because of the small mass of an electron and the small size of an atom, quantum wave-like effects are quite significant.

Because of the Uncertainty Principle, we can't specify both the position and the velocity of the electron. An electron orbiting in an atom is typically smeared out into a probability wave around the atom, so there really isn't a “position” per se. And because of the fact that the motion of the electron (whatever motion really means in the quantum world) isn't in a straight line, it isn't meaningful to specify the velocity or momentum of the electron. Instead we solve the Schrödinger equation (7.1) to find wavefunctions representing various states of the hydrogen atom. As usual, the square of the magnitude of the wavefunction is to be interpreted as a probability density, in this case, probability per unit volume.

By solving Schrödinger's equation (7.1), each hydrogen wavefunction solution is characterized by three quantum numbers \(\left(n, l, m_l\right)\text{,}\) and is written as \(\psi_{nlm}(r,\theta,\phi)\text{.}\) (The quantum number \(m_s\) isn't needed to specify the wavefunction.) For instance, the lowest energy (ground) state wavefunction of hydrogen (\(n=1, l=0, m_l=0\)) is calculated to be

\begin{equation} \psi_{100} = \frac{1}{\sqrt{\pi a_0^3}}\ e^{-r/a_0}\text{.}\label{eq_H100}\tag{7.8} \end{equation}

The excited state (\(n=2, l=1, m_l=0\)) is given by

\begin{equation} \psi_{210} = \frac{1}{\sqrt{32\pi a_0^3}} \left( \frac{r}{a_0} \right) e^{-r/2a_0} \cos{\theta}\label{eq_H210}\tag{7.9} \end{equation}

and the excited states (\(n=2, l=1, m_l=+1\)) and (\(n=2, l=1, m_l=-1\)) are

\begin{equation} \psi_{211} = -\frac{1}{\sqrt{64\pi a_0^3}} \left( \frac{r}{a_0} \right) e^{-r/2a_0} \sin{\theta}\, e^{i\phi}\text{.}\label{eq_H211}\tag{7.10} \end{equation}
\begin{equation} \psi_{21-1} = \frac{1}{\sqrt{64\pi a_0^3}} \left( \frac{r}{a_0} \right) e^{-r/2a_0} \sin{\theta}\, e^{-i\phi}\text{,}\label{eq_H21-1}\tag{7.11} \end{equation}
Figure 7.2. Plot of probability density for an electron in state \(\psi_{100}\) of hydrogen.

respectively. We can use these wavefunctions to calculate their probability densities \(\left|\psi_{nlm_l}\right|^2\) and then plot these probability densities to get a visualization of the “location” of the electron in this state. Figure 7.2 shows such a plot for the ground state \(\psi_{100}\text{.}\) In this plot, the probability density for finding the electron at any location is related to the density of dots in the plot. Thus, the probability density is greatest at the center \(r=0\) and drops off exponentially as you move away from the origin, in agreement with the wavefunction in Eq.(7.8). Also we notice that the probability density is spherically symmetric, implying that it doesn't depend on the direction in space, which follows from (7.8) since there is no dependence on the coordinates \(\theta\) and \(\phi\text{.}\)

In the case of the excited states (\(n=2\text{,}\) \(l=1\text{,}\) \(m_l=1\text{,}\) 0,-1), it is actually real linear combinations of the wavefunctions given in equations (7.10) and (7.11) that make up the familiar atomic \(p\)-orbitals that are important in chemical bonding. For instance, the linear combination

\begin{align*} \psi_{2p_x} \amp = -\frac{1}{\sqrt{2}} \left[ \psi_{211} - \psi_{21-1} \right]\\ \amp = \frac{1}{\sqrt{32 \pi a_0^3}}\ \left( \frac{r}{a_0} \right) \ e^{-r/2a_0} \sin{\theta} \cos{\phi} \end{align*}

corresponds to what is known as the “\(p_x\)” orbital that you might have seen in chemistry textbooks. The probability density is shown in Figure 7.3. In this case you will notice that the probability density is not spherically symmetric as in the ground state, but instead has two lobes that extend outward along the \(x\)-axis.

Figure 7.3. Plot of probability density for an electron in an orbital state \(\psi_{2p_x}\) of hydrogen.

Finally, here are a few rules relating the atomic quantum numbers \(n, l, m_l\text{.}\) We won't derive them here in PHYS 212. You'll have to take PHYS 222 for more details about this.

  • The possible values for the principal quantum number \(n\) are

    \begin{equation} n = \mbox{1, 2, 3, \dots}\tag{7.12} \end{equation}

  • Given a principle quantum number \(n\) for an electron, the orbital quantum number \(l\) can have the following values

    \begin{equation} l= 0,\, 1,\, \dots,\, n-1\text{.}\label{eq_lquantumnumber}\tag{7.13} \end{equation}

    So, for example, if \(n=1\text{,}\) then \(l\) can only be 0. If \(n=2\text{,}\) then \(l\) can be 0 or 1. If \(n=3\text{,}\) then \(l\) can be 0, 1, or 2. Etc.

  • Given an orbital quantum number \(l\text{,}\) the magnetic quantum number \(m_l\) can have the following values

    \begin{equation} m_l=-l,\, -(l-1),\, \dots,\, -1,\, 0,\, 1,\, \dots,\, l-1,\, l\text{.}\label{eq_mlquantumnumber}\tag{7.14} \end{equation}

    So, for example, if \(l=0\text{,}\) then \(m_l\) can only be 0. If \(l=1\text{,}\) then \(m_l\) can be \(-1\text{,}\) 0, or 1. If \(l=2\text{,}\) then \(m_l\) can be \(-2\text{,}\) \(-1\text{,}\) 0, 1, 2, etc.

The rule for the quantum number \(m_s\) is the same as was discussed in Chapter 5 for an electron: \(m_s = -1/2\) or \(+1/2\text{,}\) regardless of the other three quantum numbers \(n, l\) and \(m_l\text{.}\)

Some of the consequences of these rules will be explored in the problems.