Are the Two Definitions of Hermitian Operators Equivalent?

[AlexChandler]

Homework Statement

This is something I've been trying to prove for a bit today. My quantum mechanics book claims that the following two definitions about hermitian operators are completely equivalent

my operator here is Q (with a hat) and we have functions f,g

\langle f \mid \hat Q f \rangle = \langle Q f \mid \hat f \rangle for any function f in hilbert space

and

\langle f \mid \hat Q g \rangle = \langle Q f \mid \hat g \rangle for any functions f,g in hilbert space

2. Related formulas

\langle f \mid g \rangle = \int f^{\ast} g dx

The Attempt at a Solution

Clearly the second definition implies the first, but I'm having trouble showing that the first implies the second.
My quantum mechanics book has this as an exercise and as a hint it suggests to let f=g+h
and then let f=g+ih with i being the square root of -1. I have done both of these things, expanding the inner products in terms of integrals. If i assume the first definition and let f=g+h, i can get

\langle f \mid \hat Q g \rangle + \langle g \mid \hat Q f \rangle = \langle \hat Q f \mid g \rangle + \langle \hat Q g \mid f \rangle

doing a similar thing with f=g+ih i get the same result, and not sure where to go from here. Anybody have a better way to prove it or any ideas? thanks

 

[vela]

I don't see any definitions in your post.

 

[AlexChandler]

Ah Sorry about that. I've just edited my post and corrected that

 

[morphism]

Start with (Qf,f)=(f,Qf). If you take f=g+h and simplify, you get (g,Qh) + (h,Qh) = (Qg,h) + (Qh,g). And if you take f=g-ih and simplify, you get (g,Qh) - (h,Qg) = (Qg,h) - (Qh,g).

 

[Liquidxlax]

isn't it just that

Q=Q^dagger where dagger is the transpose and complex conjugate

for a matrix to be hermitian

such that <Q| = |Q>