Physics Problems & Solutions: Quantum Mechanics

The eigenvalues of a Hermitian operator are always

A. Real
B. Imaginary
C. Degenerate
D. Linear
E. Positive

(GR0177 #27)

Solution:

The eigenvalues are always real and hence observable.
Proof:
$\begin{aligned} A|\psi\rangle&=a|\psi\rangle \\A^\dagger|\psi\rangle&=a^*|\psi\rangle \\(A-A^\dagger)|\psi\rangle&=(a-a^*)|\psi\rangle \end{aligned}$

Hermitian operator: $A^\dagger=A$ .
So, $a-a^*=0\rightarrow a=a^* \rightarrow$ real

Answer: A

Physics Problems & Solutions

Quantum Mechanics - Hermitian Operator

No comments :