Eigenvalues as Outcomes

Every time you measure a physical quantity in quantum mechanics — position, energy, spin, or any other observable — nature is making a choice from a specific list of allowed values. The rule that defines that list is the eigenvalue postulate: the only possible results of a measurement are the eigenvalues of the operator that represents the observable.

Observables are Hermitian operators

An observable is represented by a Hermitian (self-adjoint) operator $\hat{A}$, meaning $\hat{A}^\dagger = \hat{A}$. The Hermitian condition is not just mathematical tidiness — it guarantees that all eigenvalues are real, which is essential because measurement results are real numbers.

The eigenvalue equation is

$\hat{A}\,|\psi_n\rangle = a_n\,|\psi_n\rangle,$

where $a_n$ is the $n$-th eigenvalue and $|\psi_n\rangle$ is the corresponding eigenvector (eigenstate). The statement of the measurement postulate is:

A measurement of $\hat{A}$ on any state yields one of the values $a_1, a_2, \ldots$ — never anything in between.

Why Hermitian operators have real eigenvalues

Suppose $\hat{A}\,|\psi\rangle = a|\psi\rangle$ with $|\psi\rangle \neq 0$. Taking the inner product with $|\psi\rangle$ from the left:

$a = \frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle}.$

Because $\hat{A}^\dagger = \hat{A}$, the numerator satisfies $\langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{A}|\psi\rangle^*$, so it is real, and therefore $a \in \mathbb{R}$. This proof requires no special assumptions about $|\psi\rangle$ — it works for every eigenvector.

A concrete example: the Pauli-Z operator

The Pauli-Z operator

$\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

represents the $z$-component of spin for a spin-$\tfrac{1}{2}$ particle (up to the factor $\hbar/2$). Its characteristic equation is

$(1 - \lambda)(-1 - \lambda) = 0 \implies \lambda^2 = 1 \implies \lambda = \pm 1.$

Measuring the spin of any qubit state along $z$ can therefore return only $+1$ or $-1$, in units of $\hbar/2$. This is the origin of the binary nature of spin measurement: the operator has exactly two eigenvalues.

The general recipe for finding eigenvalues

For a finite-dimensional operator with matrix representation $A$, eigenvalues satisfy the characteristic equation

$\det(A - \lambda I) = 0.$

For a $2 \times 2$ Hermitian matrix

$A = \begin{pmatrix} a & b \\ b^* & d \end{pmatrix}$

the characteristic equation expands to

$\lambda^2 - (a + d)\,\lambda + (ad - |b|^2) = 0,$

which is a quadratic in $\lambda$ with real coefficients (since $a, d \in \mathbb{R}$ for Hermitian $A$). Its discriminant is $(a - d)^2 + 4|b|^2 \geq 0$, confirming the two roots are always real.

From eigenvalues to probabilities

Knowing the eigenvalues tells you what can happen; the Born rule tells you how often. If the system is in state $|\Psi\rangle$ just before measurement, the probability of obtaining eigenvalue $a_n$ is

$p_n = |\langle\psi_n|\Psi\rangle|^2,$

where $|\psi_n\rangle$ is the normalized eigenstate. The probabilities sum to 1 because the eigenstates form a complete orthonormal basis (a consequence of the spectral theorem for Hermitian operators). After measurement, the state collapses to $|\psi_n\rangle$.

Try it

This is a numerical exercise — your code should return a number.

Consider the Hermitian operator with matrix representation

$A = \begin{pmatrix} 5 & 3 \\ 3 & 5 \end{pmatrix}.$

Its measurement outcomes are its eigenvalues. The characteristic equation is

$\det(A - \lambda I) = (5 - \lambda)^2 - 9 = 0.$

Expanding, $(5 - \lambda)^2 = 9$, so $5 - \lambda = \pm 3$, giving $\lambda = 8$ or $\lambda = 2$. Both are real, confirming $A$ is Hermitian (it equals its own transpose, and all entries are real). A measurement of this observable can return only $8$ or $2$ — no other value is possible.

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