# Is a projection operator hermitian?

1. Dec 23, 2009

### krishna mohan

I was reading Lie Algebras in Physics by Georgi.......................second edition...

Theorem 1.2: He proves that every finite group is completely reducible.

He takes

$$PD(g)P=D(g)P$$

..takes adjoint...and gets..

$$P{D(g)}{\dagger} P=P {D(g)}{\dagger}$$

So..does this mean that the projection operator P is hermitian?

Last edited: Dec 23, 2009
2. Dec 23, 2009

### vertices

I think he is assuming P is Hermitian (as it must be; if you want to think of this in terms of QM, it leaves all states unchanged, so the eigenvalue associated with the operator is '1', a real number - any operator which outputs a real eigenvalue is Hermitian)

3. Dec 23, 2009

### Fredrik

Staff Emeritus
All projection operators are hermitian. You might want to review the properties of projection operators here.

4. Dec 24, 2009

### krishna mohan

Thanks a lot!!!

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