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Quantum Mechanics}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.


See below for user comments and alternate solutions! See below for user comments and alternate solutions!
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 :PAlternate Solution - Unverified
Comments
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.NEC
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
ETScustomer
2017-10-08 20:59:05
I cannot even imagine how hard it is!
poopterium
2017-10-26 02:05:27
booooooooooooooooooo
Alternate Solution - Unverified

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The dot product of orthogonal vectors should be equal to zero which gives x =5. Tell me if i am wrong.

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