# Eigenvalues and Eigenvectors

As a general rule, an Operator \(A:\mathcal{V}\to\mathcal{V}\) changes a vector to another vector, \(A|\alpha\rangle = |\beta\rangle\). I.e. \(|\beta\rangle \not\propto |\alpha\rangle\).

For eigenvectors of \(A\), \(A\) scales the vector, \(A|a_i\rangle = a_i|a_i\rangle\).

## Definition

An eigenvector of \(A\) is a vector \(|v\rangle\) such that \(A|v\rangle = \alpha|v\rangle\). \(\alpha\) is called the eigenvalue.

The set of eigenvalues, \(\Sigma(A)\) is the spectrum of \(A\).

### Examples

Angular momentum tensor: \(\vec{L}=I\cdot \vec{\omega}\).

## Scale Factor

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

## Phase Factor

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

## Determining

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

When \(a\) s are obtained, Solve for vectors that solve \(A-a\mathbb{I}=0\)

## Degeneracies

If \(a\) is degenerate, find any linearly independent set and use \(Gram-Schmidt\) 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 \(\mathbb{C}\))
- There exists an orthonormal basis of simultaneous eigenvectors of two Hermitian operators \(A,B\) iff \([A,B]=AB-BA=0\)

### Proof of 4

Assume \([A,B]=0\). Let \(|a\rangle\) be an eigenvector for \(A\). Then \(BA|a\rangle = a(B|a\rangle) = A(B|a\rangle)\). Thus, \(B|a\rangle\) is also an eigenvector for \(A\). Vice-versa for \(B\).

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

## Spectral Theorem for Hermitian Operators

Given a Hermitian operator \(H\) on a finite-dimensional, \(n=\dim\mathcal{V}\), vector space, \(\mathcal{V}\), choose an orthonormal basis of \(H\)’s eigenvectors: \(H|h_i\rangle = h_i|h_i\rangle, \langle h_i|h_j\rangle = \delta^i_j\). Then the matrix of \(H\) in the basis \(\{|h_i\rangle\}\) is diagonal, \(H^i_j=\langle h_i|H|h_j\rangle=h_i\delta^i_j\).

If we are in another basis, when we express the eigenvectors in this basis we can construct the unitary matrix with the column vectors \(|h_i\rangle\), \(U=(|h_1\rangle\:|h_2\rangle\:\cdots\:|h_n\rangle)\). So, \(H=\sum_i h_iP_i\) with \(P_i=|h_i\rangle\langle h_i|\). This is the spectral theorem.

Then the matrix of \(H\) is given by \(H=U^\dagger H_DU\).