Eigenvalues and Eigenvectors

As a general rule, an Operator $A : V \to V$ changes a vector to another vector, $A | α ⟩ = | β ⟩$ . I.e. $| β ⟩ \propto̸ | α ⟩$ .

For eigenvectors of $A$ , $A$ scales the vector, $A | a_{i} ⟩ = a_{i} | a_{i} ⟩$ .

Definition

An eigenvector of $A$ is a vector $| v ⟩$ such that $A | v ⟩ = α | v ⟩$ . $α$ is called the eigenvalue.

The set of eigenvalues, $Σ (A)$ is the spectrum of $A$ .

Examples

Angular momentum tensor: $\vec{L} = I \cdot \vec{ω}$ .

Scale Factor

Every eigenvector has an arbitrary scale factor $c = N \exp (i θ)$ , which comprises an arbitrary scale and phase.

Phase Factor

All eigenvectors have an arbitrary phase, $\exp (i θ)$ . The phase is generally set so that the first non-zero component is a positive real number.

Determining

$A | a ⟩ = a I | a ⟩ = (A - a I) | a ⟩ = 0$ . Finding $a$ such that $det (A - a I) = 0$ . So, the non-trivial solutions of $| a ⟩$ only exist when $A - a I$ is not invertible. Thus, it must have a zero determinant.

When $a$ s are obtained, Solve for vectors that solve $A - a I = 0$

Degeneracies

If $a$ is degenerate, find any linearly independent set and use $G r a m - S c h m i d t$ to orthogonalize.

Theorem

For any Hermitian operator on a finite dimensional vector space,

The eigenvalues of $H$ are real
Eigenvectors that belong to distinct eigenvalues are orthogonal
The eigenvectors can always be chosent to be an Orthonormal basis for the vector space (for $C$ )
There exists an orthonormal basis of simultaneous eigenvectors of two Hermitian operators $A, B$ iff $[A, B] = A B - B A = 0$

Proof of 4

Assume $A$ and $B$ have simultaneous eigenvectors. Then let $| a, b ⟩$ be any eigenvector for both $A$ and $B$ . So, $A B | a, b ⟩ = B A | a, b ⟩ = a b | a, b ⟩$ , thus, $(A B - B A) | a, b ⟩ = 0$ . Then, any linear combination of eigenvectors gives zero. Thus, $A B - B A = 0$ .

Spectral Theorem for Hermitian Operators

Given a Hermitian operator $H$ on a finite-dimensional, $n = \dim V$ , vector space, $V$ , choose an orthonormal basis of $H$ ’s eigenvectors: $H | h_{i} ⟩ = h_{i} | h_{i} ⟩, ⟨ h_{i} | h_{j} ⟩ = δ_{j}^{i}$ . Then the matrix of $H$ in the basis ${| h_{i} ⟩}$ is diagonal, $H_{j}^{i} = ⟨ h_{i} | H | h_{j} ⟩ = h_{i} δ_{j}^{i}$ .

If we are in another basis, when we express the eigenvectors in this basis we can construct the unitary matrix with the column vectors $| h_{i} ⟩$ , $U = (| h_{1} ⟩ | h_{2} ⟩ \dots | h_{n} ⟩)$ . So, $H = \sum_{i} h_{i} P_{i}$ with $P_{i} = | h_{i} ⟩ ⟨ h_{i} |$ . This is the spectral theorem.

Then the matrix of $H$ is given by $H = U^{†} H_{D} U$ .