Hermitian operator

1. Sep 21, 2008

kthouz

How to check if an operator is hermitian? I mean what is the condition
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. Sep 21, 2008

Fredrik

Staff Emeritus
Using the (x,y) notation for the scalar product, the definition of $H^\dagger$ is

$$(H^\dagger x,y)=(x,Hy)$$

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

3. Sep 21, 2008

atyy

Fredrik's definition is correct, and it is also true that an operator thus defined has real eigenvalues, as you said.

4. Sep 22, 2008

clem

For an hermitian, ALL of the evs must be real.
Just checking one ev isn't enough.

5. Sep 22, 2008

fermi

True, this is necessary, but not sufficient. It is possible that all eigenvalues are
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.