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Section 7.2 Quantum in 3D

In earlier chapters we have seen how to use the Schrödinger equation to solve for the wavefunctions of a particle in one dimension. In three dimensions, the Schrödinger equation can be written in spherical coordinates \((r, \theta, \phi)\) as:

\begin{equation} -\frac{\hbar^2}{2mr^2} \left[ \frac{\partial}{\partial r} \left( r^2 \frac{\partial \psi}{\partial r} \right) + \frac{1}{\sin{\theta}} \frac{\partial}{\partial \theta} \left( \sin{\theta} \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{\sin^2{\theta}} \frac{\partial^2 \psi}{\partial \phi^2} \right] + U \psi = E \psi\text{.}\label{eq_3Dschrodinger}\tag{7.1} \end{equation}

At this point, you might be hyperventilating, but don't worry — we will restrict ourselves to cases where everything is spherically-symmetric (with no \(\theta\)- or \(\phi\)-dependence), so most of the terms in this equation drop out. And we won't ask you to solve this equation, but will instead use the approach from earlier where we give you a solution and ask for you to test to see if it does or does not satisfy the equation.

Show that the test wavefunction \(\psi(r) = Ae^{-br^2}\) is a solution to (7.1) for the harmonic oscillator with potential energy function \(U =\frac{1}{2}kr^2\text{.}\) If it is a solution (which it is), determine the energy of the quantum state.

Solution.

If you have never seen a partial derivative such as \(\partial \psi/\partial r\text{,}\) fear not — it is the same thing as a regular derivative where everything else (in this case, \(\theta\) and \(\phi\)) is treated as constant. And since the test wavefunction for this example doesn't have any \(\theta\) or \(\phi\)'s in it, we can quickly simplify Schrödinger's equation by dropping out any terms that have derivatives with respect to \(\theta\) or \(\phi\text{.}\) So, we now have:

\begin{equation} -\frac{\hbar^2}{2mr^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial \psi}{\partial r} \right) + \frac{1}{2}kr^2 \psi = E \psi\text{.}\label{eq_ReducedSE}\tag{7.2} \end{equation}

The derivative \(\partial \psi/\partial r = Ae^{-br^2}(-2br)\) and the expression \(\frac{\partial}{\partial r}(r^2\frac{\partial \psi}{\partial r}) = -6br^2Ae^{-br^2} +4b^2r^4Ae^{-br^2}\text{.}\) So, (7.2) becomes:

\begin{equation} -\frac{\hbar^2}{2mr^2} \left[-6br^2Ae^{-br^2}+4b^2r^4Ae^{-br^2}\right] + \frac{1}{2}kr^2 \psi = E \psi\text{.}\tag{7.3} \end{equation}

The terms \(Ae^{-br^2}\) are present in each term, so they cancel. Simplifying,

\begin{gather*} \frac{3b\hbar^2}{m} - \frac{2b^2\hbar^2}{m}r^2 + \frac{1}{2}kr^2 = E\\ \rightarrow r^2\left(\frac{1}{2}k-\frac{2b^2\hbar^2}{m}\right) + \left(\frac{3b\hbar^2}{m} - E\right) = 0 \end{gather*}

and this is satisfied only if both the constant and \(r^2\) terms separately add to zero. Setting the \(r^2\) terms to zero gives us \(b = \frac{\sqrt{mk}}{2\hbar}\) and \(E = \frac{3b\hbar^2}{m} = \frac{3\hbar^2}{m}\frac{\sqrt{mk}}{2\hbar} = \frac{3}{2}\hbar\sqrt{k/m}\text{.}\)

So, yes, the test function \(\psi(r) = Ae^{-br^2}\) satisfies Schrödinger's equation for a harmonic oscillator as long as \(b\) and \(E\) are given by the values above.

A particularly important case for the 3D Schrödinger equation ((7.1)) is that of a hydrogen atom where the potential energy function \(U = -ke^2/r\text{.}\) The complete general solutions for this equation, describing all possible states for an electron in a hydrogen atom, are beyond the scope of this course. 1  We'll be content with checking a few of the simplest cases (you'll be doing this for homework, using an approach very similar to the example above). But certain general observations will be made.

The first observation is that the spatial wavefunction \(\psi(x,y,z)\) for any single-particle, 3D quantum system involves three quantum numbers. For atoms, each of the three quantum numbers (\(n, l\text{,}\) and \(m_l\)) which arise in solving this particular 3D problem can be associated with a particular spherical coordinate. The principal quantum number \(n\) is associated with the radial coordinate \(r\text{,}\) and is related to the number of nodes in the wavefunction as you move outward from the origin on a radial line. The orbital quantum number \(l\) is associated with the polar angle \(\theta\) (measured from the “North pole”). When \(l=0\text{,}\) the associated wavefunction has no dependence on \(\theta\text{.}\) The orbital magnetic quantum number \(m_l\) is associated with the azimuthal angle \(\phi\text{;}\) when \(m_l = 0\text{,}\) there is no \(\phi\)-dependence. 2 

Actually, there is a fourth quantum number \(m_s\) associated with the spin angular momentum of the electron — this is the same quantum number that we discussed in the previous chapter. So, a full description of the state of an electron in an atom requires the specification of four quantum numbers: \(|n,l,m_l,m_s\rangle\text{.}\)

The second observation is that the quantum numbers are directly connected to measurable physical quantities of the atom. For solutions of (7.1), the energy of the electron in hydrogen is determined to a high degree of accuracy by the principal quantum number \(n\text{.}\) 3  The energy formula for an electron in a state of quantum number \(n\) is

\begin{equation} E_n = -\frac{1}{n^2} \frac{m k^2 e^4}{2 \hbar^2} = - \frac{13.6 \Xunits{eV}}{n^2}\text{.}\label{eq_hydrogenE}\tag{7.4} \end{equation}

Another useful physical quantity is the electron's orbital angular momentum \(\vec{L}\text{,}\) whose magnitude is given in terms of the orbital quantum number \(l\) as

\begin{equation} |\vec{L}| = \sqrt{l (l + 1)} \hbar\text{.}\label{eq_Lmag}\tag{7.5} \end{equation}

The \(z\)-component of the orbital angular momentum \(L_z\) is given by

\begin{equation} L_z = m_l \hbar\text{,}\label{eq_Lcomponent}\tag{7.6} \end{equation}

and the \(z\)-component of the spin angular moment \(S_z\) is given by

\begin{equation} S_z = m_s \hbar\text{.}\label{eq_Scomponent}\tag{7.7} \end{equation}
But not beyond the scope of a class in physical chemistry or upper-level quantum mechanics — something to look forward to if you are a chemistry or physics major!
In other math classes you have taken you may have used \(\phi\) for the polar angle and \(\theta\) for the azimuthal angle. This is just a matter of convention. In this course we will consistently use the definitions given in the text above.
For real atoms, there are slight modifications to the energy based on the other quantum numbers, but we won't discuss those slight corrections in this course.