If A and B are Hermitian operators is (i A + B ) Hermitian?

[Settho]

If A and B are Hermitian operators is (i A + B ) a Hermitian operator?
(Hint: use the definition of hermiticity used in the vector space where the elements are quadratic integrable functions)

I know an operator is Hermitian if:
- the eigenvalues are real
- the eigenfunction is orthonormal
- the eigenfunctions form a complete basis set.

I know how to prove if A + B is a Hermitian operator, but because of the i in front of A I still am a little bit confused.

So I know to prove if the eigenvalues are real. You have to use these integrals first and prove both sides are the same so that λ = λ*

So I did this:

I have doubts about the last one, because I think that is now how it is suppose to go, but I don't know. And then I thought because both sides aren't the same the eigenvalues aren't real and it isn't Hermitian.

Is this correct? If it isn't how am I suppose to handle this problem?

 

[PeroK]

You could always look for a counterexample.

 

[DrClaude]

Hermitian operators also have the property $A = A^*$.

 

[strangerep]

[...] how am I suppose to handle this problem?

Also trying looking up the properties of the "conjugate transpose" operation on Wikipedia. E.g., what is $(AB)^* = \;?$

 

[hilbert2]

A clue: It will be hermitian in the special case where A=0.

One example of this kind of operator are the raising and lowering operators of a harmonic oscillator, where the A and B are multiples of the hermitian position and momentum operators.