# Simple operator identity (1 Viewer)

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#### quasar987

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Can someone explain to me why for any operator A and functions f(x), g(x),

$$(Af(x))^*g(x) = f^*(x)A^+g(x)$$

Where "^+" denotes the hermitian conjugate of A.

I went to see the demonstrator about it and he couldn't explain/prove that result. Thx.

#### CarlB

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I think that this is true only if you integrate over all x. That is, you need to have a definite integral.

The reason for this is that the usual example of A is $$i\partial_x$$. To convert from the LHS to the RHS requires an integration by parts. In doing this integration, you will not get the right result unless you cancel off the integrated part using something like f(infinity) = g(infinity) = 0.

Carl

#### HallsofIvy

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How is your innerproduct defined? For functions spaces it is normally the L2 innerproduct
$$<f,g>= \int_a^b f(x)g^*(x)dx$$
for some a,b.

In that case,
$$<Af,g>= \int_a^b(Af(x))g^*(x)dx$$
and
$$<f,Bg>= \int_a^bf(x)(Bg)^*(x)dx$$

Comparing that with the definition of "Hermitian conjugate" should make it clear that those at equal if and onlyif B= A+.

#### quasar987

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Yesterday, I found a solution to this. It's very simple. First, I prove that for a function f, $f^+ = f^*$. Since Af is a function, $(Af)^* = (Af)^+=f^+A^+=f^*A^+$. Voilà!

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