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08 Apr 2018 | Seattle, WA

It's not immediately apparent, but this linear algebra problem is very clever:

Find the eigenvalues of matrix \(A\),
$$
A=\begin{pmatrix} \frac{x}{2}+\frac{y}{2}+w & 0 & \frac{x}{2}-\frac{y}{2} \\ 0 & x+y & 0 \\ \frac{x}{2}-\frac{y}{2} & 0 & \frac{x}{2}+\frac{y}{2}+w \end{pmatrix}.
$$

While not a particularly unique problem, it holds an inherent lesson about eigenvalues which it teaches by pushing the student to step back and question what they are doing.

Students oftentimes try to outsmart their homework, especially in math. It is necessary to reinforce key concepts through repetition which can lead students to focus on how to solve a problem rather than why it is mathematically valid. This problem pushes back against that tendency. It is easier to solve by confronting its underlying meaning than by rotely repeating known methods.

If a student attempts to solve the problem rotely, they will be faced with an ensnaring tangle of variables and operators. The approach I would expect from a student starts like this

We want to solve for \(\lambda\) in
\(|A -\lambda \mathbb{I}| = 0\)
$$\longrightarrow
\begin{vmatrix}
\frac{x}{2}+\frac{y}{2}+w-\lambda & 0 & \frac{x}{2}-\frac{y}{2} \\
0 & x+y-\lambda & 0 \\
\frac{x}{2}-\frac{y}{2} & 0 & \frac{x}{2}+\frac{y}{2}+w-\lambda
\end{vmatrix} = 0
$$

$$\longrightarrow (\frac{x}{2}+\frac{y}{2}+w-\lambda)(x+y-\lambda)(\frac{x}{2}+\frac{y}{2}+w-\lambda) \\ -(\frac{x}{2}-\frac{y}{2})(x+y-\lambda)(\frac{x}{2}-\frac{y}{2}) = 0 $$

$$\longrightarrow (\frac{x}{2}+\frac{y}{2}+w-\lambda)(x+y-\lambda)(\frac{x}{2}+\frac{y}{2}+w-\lambda) \\ -(\frac{x}{2}-\frac{y}{2})(x+y-\lambda)(\frac{x}{2}-\frac{y}{2}) = 0 $$

The student might notice the shared \(x+y-\lambda\) in that jumble and pull it out. If they don't, a brute-force attempt will yield yet more opportunities for missing details as the tangled mess grows.

Instead, maybe they rub their temples, get up, and go for coffee. While waiting, they might start thinking about why the determinant of a matrix would be zero. Under what conditions does that occur? If a row or column of a matrix is all zeroes, its determinant is necessarily zero. If we were to set \(\lambda=x+y\) in this problem, the determinant would be zero because you have a row (or column) of all zeroes!

$$
\begin{vmatrix}
\frac{x}{2}+\frac{y}{2}+w-x+y & 0 & \frac{x}{2}-\frac{y}{2} \\
0 & 0 & 0 \\
\frac{x}{2}-\frac{y}{2} & 0 & \frac{x}{2}+\frac{y}{2}+w-x+y
\end{vmatrix} = 0
$$

This implies that this eigenvalue has a corresponding eigenvector that points along the secondary axis.

Let's try and see this in a different way. If we think of the matrix \(A\) as transforming a vector in Cartesian coordinates (\(a\), \(b\), \(c\)), what is the effect of the transform on a vector that only points in the \(b\)-direction?

$$
\begin{pmatrix}
\frac{x}{2}+\frac{y}{2}+w & 0 & \frac{x}{2}-\frac{y}{2} \\
0 & x+y & 0 \\
\frac{x}{2}-\frac{y}{2} & 0 & \frac{x}{2}+\frac{y}{2}+w
\end{pmatrix}
\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix}
=
\begin{pmatrix}
0 \\
x+y \\
0
\end{pmatrix}
=
(x+y)
\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix}
$$

Here, truly, \(x+y\) must be an eigenvalue that simply scales its eigenvector. In this context of this problem, the student may see that she can separate the task of solving for \(\lambda\) in \(|A -\lambda \mathbb{I}| = 0\) into two parts:

$$ x+y -\lambda = 0 $$
and
$$
\begin{vmatrix}
\frac{x}{2}+\frac{y}{2}+w-\lambda & \frac{x}{2}-\frac{y}{2} \\
\frac{x}{2}-\frac{y}{2} & \frac{x}{2}+\frac{y}{2}+w-\lambda
\end{vmatrix} = 0
$$

In doing so, she has simplified the problem and seen an example of separating eigenvalues into sets in orthogonal spaces. We know by definition that eigenvectors for a given matrix must each be orthogonal. In this case, we have separated our eigenvalues into a set (of one) with a corresponding eigenvector along the \(b\) axis and a set (of two) with corresponding eigenvectors in \(a\)-\(c\) plane. The latter set has orthogonality, of course, but more work will be needed to solve for the eigenvalues.

The importance of decomposing matricies by identifying orthogonal sets really hit home for me with quantum mechanics. For a quantum system, defined by a vector, the eigenvalues of the Hamiltonian are energy levels that can be occupied by that system. The number of energy levels increases with the degrees of freedom in the system.

As an example, let's solve the following quantum mechanics problem:

Determine the energy levels of the Stark Hamiltonian for the \(n = 3\) level of hydrogen.

The relevant quantum numbers are the principle (\(n\)), angular (\(\ell\)), and magnetic (\(m\)). These represent the state of the hydrogen atom under the influence of a magnetic field. Notationally, we write each state as \(\lvert n,\ell,m \rangle\). For instance, the state where \(n=3\), \(l=2\), and \(m=1\), would be written as \(\lvert 3, 2, 1 \rangle\).

We first find every state the system can occupy given the rules that govern the relationships between quantum numbers,

The possible states for the system are

\( \lvert 3, 0, 0 \rangle, \lvert 3, 1, 0 \rangle, \lvert 3, 2, 0 \rangle \\ \lvert 3, 1, 1 \rangle, \lvert 3, 2, 1 \rangle \\ \lvert 3, 1, -1 \rangle, \lvert 3, 2, -1 \rangle \\ \lvert 3, 2, 2 \rangle \\ \lvert 3, 2, -2 \rangle \\ \)

\( \lvert 3, 0, 0 \rangle, \lvert 3, 1, 0 \rangle, \lvert 3, 2, 0 \rangle \\ \lvert 3, 1, 1 \rangle, \lvert 3, 2, 1 \rangle \\ \lvert 3, 1, -1 \rangle, \lvert 3, 2, -1 \rangle \\ \lvert 3, 2, 2 \rangle \\ \lvert 3, 2, -2 \rangle \\ \)

then form a matrix from each combination of states.

We cross our fingers and hope we didn't make a mistake somewhere. Calculating every combination, we get the following matrix multiplied by a constant \(-qE_0a_0\)

$$
\begin{pmatrix}
0 &0 & -3\sqrt{6} &0 &0 &0 &0 &0 &0 \\
0 &0 &0 &0 &0 &-\frac{9}{2} &0 &0 &0 \\
-3\sqrt{6}a_0 &0 &0 &0 &0 &0 &-3\sqrt{3} &0 &0 \\
0 &0 &0 &0 &0 &0 &0 &-\frac{9}{2} &0 \\
0 &0 &0 &0 &0 &0 &0 &0 &0 \\
0 &-\frac{9}{2} &0 &0 &0 &0 &0 &0 &0\\
0 &0 &-3\sqrt{3} &0 &0 &0 &0 &0 &0 \\
0 &0 &0 &-\frac{9}{2} &0 &0 &0 &0 &0 \\
0 &0 &0 &0 &0 &0 &0 &0 &0\\
\end{pmatrix}
$$

To get the energy levels, we need to find the eigenvalues by solving for \(\lambda\) in

$$
\begin{vmatrix}
-\lambda &0 & -3\sqrt{6} &0 &0 &0 &0 &0 &0 \\
0 &-\lambda &0 &0 &0 &-\frac{9}{2} &0 &0 &0 \\
-3\sqrt{6}&0 &-\lambda &0 &0 &0 &-3\sqrt{3} &0 &0 \\
0 &0 &0 &-\lambda &0 &0 &0 &-\frac{9}{2} &0 \\
0 &0 &0 &0 &-\lambda &0 &0 &0 &0 \\
0 &-\frac{9}{2} &0 &0 &0 &-\lambda &0 &0 &0\\
0 &0 &-3\sqrt{3} &0 &0 &0 &-\lambda &0 &0 \\
0 &0 &0 &-\frac{9}{2} &0 &0 &0 &-\lambda &0 \\
0 &0 &0 &0 &0 &0 &0 &0 &-\lambda\\
\end{vmatrix} = 0
$$

This certainly has the potential to be tedious. But if we think of the orthogonality of the spaces within the matrix, we can make the whole task easier.* Using the same technique from above, we can separate this into five sub-sets to solve.

$$
\begin{vmatrix}
-\lambda &-3\sqrt{6} &0 \\
-3\sqrt{6} &-\lambda &-3\sqrt{3} \\
0 &-3\sqrt{3} &-\lambda
\end{vmatrix} = 0
$$
$$
\begin{vmatrix}
-\lambda &-\frac{9}{2} \\
-\frac{9}{2} &-\lambda
\end{vmatrix} = 0
$$
$$
\begin{vmatrix}
-\lambda &-\frac{9}{2} \\
-\frac{9}{2} &-\lambda
\end{vmatrix} = 0
$$
$$
\begin{vmatrix}
-\lambda &0 \\
0 &-\lambda
\end{vmatrix} = 0
$$
Solving for \(\lambda\) yields the eigenvalues (the energy levels)
$$
V = -qEa_0(-9, 9, \frac{9}{2}, \frac{9}{2}, -\frac{9}{2}, -\frac{9}{2}, 0, 0, 0)
$$

While I hope you can see how useful this technique is for finding eigenvalues, there is also a lesson that applies beyond mathematics. Whenever you find yourself repeating an action, it is useful to stop and ask how it can be subdivided into simpler tasks. Most often, the path to discovering this subdivision comes though understanding the underlying meaning of the action.

* If you want to dig into this problem more, it is worthwhile to note that the orthogonality arises from the separate magnetic number manifolds (0, 1, -1, 2, -2). Why do you think this is?