Hermitian conjugate of a Hermitian Conjugate

teaJ
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I know that (A\mp )\mp =A . Where A is an Hermitian operator How does one go about proving this through the standard integral to find Hermitian adjoint operators?

I should mention, I don't want anyone to just flat out show me step by step how to do it. I'd just like a solid starting place.

\int (A\mp \psi*) \varphi dx = \int (\psi*)A\varphi dx
 
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Edited with the question. Forums are difficult sometimes :)
 
teaJ said:
I know that (A\mp )\mp =A . Where A is an Hermitian operator How does one go about proving this through the standard integral to find Hermitian adjoint operators?

I should mention, I don't want anyone to just flat out show me step by step how to do it. I'd just like a solid starting place.

\int (A\mp \psi*) \varphi dx = \int (\psi*)A\varphi dx

I prefer \dagger for Hermitian conjugate. Anyway...

Writing \langle \psi|\phi \rangle for \int \psi^* \phi dx, we have the definition of A^\dagger:

Equation 1: \langle A^\dagger \psi | \phi \rangle = \langle \psi | A \phi \rangle

But we also have a fact about inner products:

Equation 2: \langle X | Y \rangle^* = \langle Y | X \rangle.

Now, apply equation 2 to both sides of equation 1 to get another fact about Hermitian conjugates.
 
teaJ said:
I know that (A\mp )\mp =A . Where A is an Hermitian operator
That's not true. An operator is hermitian or synonymously symetric, if ##\langle A^\dagger \phi|\psi\rangle =\langle \phi| A\psi \rangle ## for ## \phi,\; \psi \in \mathcal{D}(A)##. However, the domain of ##A^\dagger ## may be larger than that of A, ##\mathcal{D}(A)\subset \mathcal{D}(A^\dagger)## and then also ##\mathcal{D}(A)\subset \mathcal{D}(A^{\dagger \dagger})##. However if A is self adjoint, i.e. if additionally ##\mathcal{D}(A)=\mathcal{D}(A^\dagger)## then also ##\mathcal{D}(A)=\mathcal{D}(A^{\dagger \dagger})##.
 
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