Is a projection operator hermitian?

[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?

 

[vertices]

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?

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)

 

[Fredrik]

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

 

[krishna mohan]

Thanks a lot!