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I learned that a Hermitian matrix $A$ is defined as a matrix that satisfies $$A^\dagger=(A^*)^\intercal=A,$$ i.e. its Hermitian conjugate $A^\dagger$ is the same as the original matrix $A$.

I also learned that in QM, a Hermitian operator $H$ is defined as an operator that satisfies $$ \langle f|Hg\rangle=\langle Hf|g\rangle,$$ where $f$ and $g$ are vectors.

Since operators and matrix can be represented by matrices in a particular basis, how can it be shown that a Hermitian matrix with the property $(A^*)^\intercal=A$ also satisfies $ \langle f|Ag\rangle=\langle Af|g\rangle$?

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$\langle f|Ag\rangle=\langle f|A|g\rangle$.

$\langle Af|g\rangle$:

  • $(\langle Af|) = (|Af\rangle)^\dagger =(A|f\rangle)^\dagger = \langle f |A^\dagger$,
  • so $\langle Af|g\rangle = \langle f |A^\dagger|g\rangle$

If $A = A^\dagger$, then $\langle f|Ag\rangle =\langle Af|g\rangle$.

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    $\begingroup$ in infinite dimensional spaces there are subtleties of domain etc... $\endgroup$ – ZeroTheHero Jun 21 '20 at 2:24
  • $\begingroup$ Hi :-) I think that there is a missing symbol here: $\langle f |A^\dagger$ into your answer. Best regards. $\endgroup$ – Sebastiano Jun 21 '20 at 10:35
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In matrix form, $$\langle f|Ag\rangle = f^\dagger A g,$$ $$\langle Af|g\rangle= (Af)^\dagger g.$$

Using the matrix property of $(AB)^\dagger=B^\dagger A^\dagger$ on the latter expression, we get $$\langle Af|g\rangle = (Af)^\dagger g=f^\dagger A^\dagger g.$$

Hence if $A=A^\dagger$, then $$\langle Af|g\rangle= f^\dagger A^\dagger g= f^\dagger A g=\langle f|Ag\rangle.$$

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