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GR0177 #27
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Quantum Mechanics$\Rightarrow$}Hermitian Operator

The eigenvalues of a Hermitian operator are always real. This is one of those quantumisms one memorizes after even a semi-decent course in QM.

One can quickly prove it by the other quantumism-proof one memorized for that same course: Suppose one has $A|\psi\rangle = a |\psi\rangle$. Then, $A^\dag |\psi\rangle = a^* |\psi\rangle$.

$A | \psi \rangle - A^\dag | \psi \rangle = (a-a^*)|\psi\rangle$. But, the left side is just 0 from, as $A^\dag = A$ from the definition of Hermitian operator. Then, the right side must be 0, too, since $\langle \psi |\psi \rangle \neq 0$ in general.

Alternate Solutions
 ramparts2009-10-06 18:07:11 Hermitian operators represent observables. Their eigenvalues are things we measure. It is very hard to measure imaginary things :PReply to this comment
chemicalsoul
2009-10-28 10:02:53
The dot product of orthogonal vectors should be equal to zero which gives x =5. Tell me if i am wrong.
ramparts
2009-10-06 18:07:11
Hermitian operators represent observables. Their eigenvalues are things we measure. It is very hard to measure imaginary things :P
 ETScustomer2017-10-08 20:59:05 I cannot even imagine how hard it is!
 poopterium2017-10-26 02:05:27 booooooooooooooooooo

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$