Checkpoint: Diagonalize an Observable

This checkpoint pulls the whole module together. You will take a Hermitian observable, find its spectrum and eigenvectors, and rebuild it from its spectral decomposition — the round trip that the spectral theorem promises always works. If your eigenvalues and eigenvectors are right, the rebuilt operator reproduces the original exactly.

The task

Diagonalize the observable

A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}.

It is real and symmetric, hence Hermitian, so the spectral theorem guarantees real eigenvalues and an orthonormal eigenbasis.

Step 1 — eigenvalues

Use the trace–determinant form of the characteristic equation. Here $\operatorname{tr}A = 2$ and $\det A = (1)(1) - (2)(2) = -3$ , so

\lambda^2 - 2\lambda - 3 = (\lambda - 3)(\lambda + 1) = 0 \quad\Longrightarrow\quad \lambda_1 = 3,\ \ \lambda_2 = -1.

Both real, as required. Their sum is $2 = \operatorname{tr}A$ and product is $-3 = \det A$ — a quick consistency check.

Step 2 — eigenvectors

Solve $(A - \lambda\mathbb{1})|v\rangle = 0$ for each eigenvalue.

For $\lambda_1 = 3$ : $\begin{pmatrix} -2 & 2 \\ 2 & -2\end{pmatrix}|v\rangle = 0$ gives $v_1 = v_2$ , so $|v_1\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1\end{pmatrix}$ .
For $\lambda_2 = -1$ : $\begin{pmatrix} 2 & 2 \\ 2 & 2\end{pmatrix}|v\rangle = 0$ gives $v_1 = -v_2$ , so $|v_2\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1\end{pmatrix}$ .

The two eigenvectors are orthogonal, $\langle v_1|v_2\rangle = \tfrac12(1 - 1) = 0$ , exactly as the spectral theorem demands for distinct eigenvalues of a Hermitian operator. Each is normalized to unit length.

Step 3 — reconstruct from the spectral decomposition

The spectral theorem says

A = \lambda_1\,|v_1\rangle\langle v_1| + \lambda_2\,|v_2\rangle\langle v_2|.

The two outer products are

|v_1\rangle\langle v_1| = \tfrac12\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}, \qquad |v_2\rangle\langle v_2| = \tfrac12\begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix}.

Weighting by the eigenvalues and adding,

3\cdot\tfrac12\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} + (-1)\cdot\tfrac12\begin{pmatrix} 1 & -1 \\ -1 & 1\end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 1\end{pmatrix} = A.

The reconstruction matches $A$ entry by entry. In particular the $(0,0)$ entry is $3\cdot\tfrac12 + (-1)\cdot\tfrac12 = 1$ , which equals $A_{00}$ — the value you will return as a check.

Try it

Carry out the three steps above in code: find the eigenvalues, build normalized eigenvectors, form the spectral decomposition, and return the reconstructed $(0,0)$ entry of $A$ . A correct diagonalization returns $1$ .

Run your code to see the quantum state.