GREPhysics.NET: GR0177 #27

GR0177 #27

Problem

GREPhysics.NET Official Solution

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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

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

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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. Reply to this comment
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 Reply to this comment

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