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Actualy, i am using the principe that say that the eigenvalue associated with the operator must be a REAL NUMBER.That is to say that i work out to that eigenvalue and see if it is a real number. Am i right?

- Thread starter kthouz
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- #1

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Actualy, i am using the principe that say that the eigenvalue associated with the operator must be a REAL NUMBER.That is to say that i work out to that eigenvalue and see if it is a real number. Am i right?

- #2

Fredrik

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[tex](H^\dagger x,y)=(x,Hy)[/tex]

and the definition of "hermitian" is [itex]H^\dagger=H[/itex]. So you usually don't have to think about eigenvalues.

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atyy

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- #4

Meir Achuz

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For an hermitian, ALL of the evs must be real.

Just checking one ev isn't enough.

Just checking one ev isn't enough.

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True, this is necessary, but not sufficient. It is possible that all eigenvalues areFor an hermitian, ALL of the evs must be real.

Just checking one ev isn't enough.

real for a non-Hermitian operator. A very simple example is the 2x2 matrix

((1, 3), (2, 2)) which is not Hermitian, but it has two distinct real eigenvalues,

namely 4 and -1.

On the other hand, Fredrik's definition is a good one, stick to it. It is valid for

all finite dimensional Vector spaces with an inner product. In QM, the vector

space can be infinite dimensional of course. In that case, Fredrik's definition

requires some supplementary conditions about bounded operators.

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