# 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,

1. The eigenvalues of $$H$$ are real
2. Eigenvectors that belong to distinct eigenvalues are orthogonal
3. The eigenvectors can always be chosent to be an Orthonormal basis for the vector space (for $$\mathbb{C}$$)
4. 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 $$H=U^\dagger H_DU$$.

Created: 2023-06-25 Sun 02:24

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