# Eigenstate

## Definition

An eigenstate for an operator $$\hat{A}$$ is a state $$|a\rangle$$ such that $$A|a\rangle = a|a\rangle$$.

## Eigenvalues and Eigenvectors

A state vector $$|\psi\rangle$$ is an eigenvector of $$\hat{A}$$ iff $$\hat{A}|\psi\rangle = \alpha|\psi\rangle$$. $$\alpha$$ is the eigenvalue of the operator.

### Theorem

The eigenvalues of a Hermitian operator are real and eigenstates of $$\hat{A}$$ correspond to different eigenvalues are orthogonal.

Proof.

$$A|\varphi_n\rangle = a_n|\varphi_n\rangle$$.

(1) $$\langle\varphi_m|A|\varphi_n\rangle = a_n\langle\varphi_m|\varphi_n\rangle$$.

$$\langle\varphi_m|A^\dagger = a_m^*\langle\varphi_m|$$.

(2) $$\langle\varphi_m|A^\dagger|\varphi_n\rangle = a_m^*\langle\varphi_m|\varphi_n\rangle$$.

(1) - (2): $$a_n\langle\varphi_m|\varphi_n\rangle - a_m^*\langle\varphi_m|\varphi_n\rangle = (a_n-a_m^*)\langle\varphi_m|\varphi_n\rangle = 0$$

For $$m=n$$, $$a_n-a_m^*=0$$, hence $$a_n=a_n^*$$. Therefore, the eigenvalues are real.

For $$a_m\neq a_n$$ (different eigenvalues), $$\langle\varphi_m|\varphi_n\rangle = 0$$, thus they are orthogonal.

## Finding Probabilities of Measurements

Measuring $$A$$. $$\psi(\vec{r},t)=\sum_n c_n|\varphi_n\rangle$$, $$a_n\to|c_n|^2$$, $$c_n = \langle\varphi_n|\psi\rangle$$.

Created: 2023-06-25 Sun 02:29

Validate