Eigenvalue Equations

An eigenvalue equation asks a sharp question of an operator: is there a direction the operator leaves unchanged except for stretching? Those special directions — eigenvectors — and their stretch factors — eigenvalues — are where the physics lives. The possible outcomes of a measurement are the eigenvalues of the observable, and the states that give a definite outcome are its eigenvectors.

The equation

A nonzero vector $|v\rangle$ is an eigenvector of the operator $A$ with eigenvalue $\lambda$ if

Applying $A$ does not rotate $|v\rangle$ into a new direction; it merely scales it by the number $\lambda$ (possibly complex). We exclude $|v\rangle = 0$, which would satisfy the equation trivially for any $\lambda$ and tell us nothing. Note that if $|v\rangle$ is an eigenvector, so is any nonzero multiple $c|v\rangle$ — eigenvectors come as whole directions (one-dimensional subspaces, or larger if the eigenvalue is degenerate).

Finding eigenvalues: the characteristic equation

Rewrite the equation as $(A - \lambda\mathbb{1})|v\rangle = 0$. A nonzero solution $|v\rangle$ exists precisely when the operator $A - \lambda\mathbb{1}$ is not invertible, which happens when its determinant vanishes:

This is the characteristic equation, a polynomial of degree $n$ in $\lambda$ for an $n\times n$ matrix. Its roots are the eigenvalues.

For a $2\times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$, expanding the determinant gives the compact form

so the two eigenvalues are

A useful built-in check: the eigenvalues always satisfy $\lambda_+ + \lambda_- = \operatorname{tr}A$ and $\lambda_+ \lambda_- = \det A$.

Worked example

Take the real symmetric matrix

Here $\operatorname{tr}A = 6$ and $\det A = 9 - 1 = 8$, so the characteristic equation is $\lambda^2 - 6\lambda + 8 = 0 = (\lambda - 4)(\lambda - 2)$. The eigenvalues are $\lambda_+ = 4$ and $\lambda_- = 2$, both real — as guaranteed for a Hermitian (here symmetric) operator. Their sum is $6 = \operatorname{tr}A$ and product is $8 = \det A$, confirming the answer.

Finding the eigenvectors

Once an eigenvalue is known, solve the linear system $(A - \lambda\mathbb{1})|v\rangle = 0$. For $\lambda_+ = 4$,

and the equation $-v_1 + v_2 = 0$ forces $v_1 = v_2$. A normalized eigenvector is $|v_+\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda_- = 2$ the same method gives $|v_-\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. The two eigenvectors are orthogonal, as they must be for distinct eigenvalues of a Hermitian operator.

Try it

Solve $A|v\rangle = \lambda|v\rangle$ for $A = \begin{pmatrix} 3 & 1 \\ 1 & 3\end{pmatrix}$ using the trace–determinant formula, and return the larger of the two eigenvalues.

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