Introduction: Finding Eigenvalues Systematically
To find eigenvalues, we need to solve \(\hat{A}|\psi\rangle = \lambda|\psi\rangle\), which rearranges to \((\hat{A} - \lambda I)|\psi\rangle = 0\). For a non-trivial solution, the matrix \((A - \lambda I)\) must be singular—its determinant must be zero.
The Characteristic Equation
The eigenvalues of matrix \(A\) are the roots of the characteristic polynomial:
\[\det(A - \lambda I) = 0\]For an \(n \times n\) matrix, this gives a polynomial of degree \(n\) in \(\lambda\).
The Algorithm
- Form the matrix \(A - \lambda I\)
- Compute \(\det(A - \lambda I)\)
- Set the determinant equal to zero and solve for \(\lambda\)
- For each \(\lambda\), solve \((A - \lambda I)\vec{v} = 0\) to find eigenvectors
Worked Examples
Example 1: 2×2 Matrix
Find eigenvalues of \(A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\):
\[A - \lambda I = \begin{pmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{pmatrix}\] \[\det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 \cdot 1 = \lambda^2 - 7\lambda + 10 = 0\] \[(\lambda - 5)(\lambda - 2) = 0 \Rightarrow \lambda = 5 \text{ or } \lambda = 2\]Example 2: Finding Eigenvector for λ = 5
\[(A - 5I)\vec{v} = \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0\]This gives \(-v_1 + 2v_2 = 0 \Rightarrow v_1 = 2v_2\)
Eigenvector: \(\vec{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\) (or any scalar multiple)
Example 3: Checking with Trace and Determinant
Useful checks:
- Sum of eigenvalues = Trace: \(5 + 2 = 7 = 4 + 3\) ✓
- Product of eigenvalues = Det: \(5 \times 2 = 10 = 4 \cdot 3 - 2 \cdot 1\) ✓
The Quantum Connection
In quantum mechanics, the characteristic equation gives the energy spectrum when applied to the Hamiltonian. For the harmonic oscillator, solving \(\det(H - E I) = 0\) yields \(E_n = \hbar\omega(n + \frac{1}{2})\). The characteristic polynomial encodes all possible measurement outcomes of an observable.