# Equivalence of Hermitian operator and Hermitian matrix in Quantum Mechanics

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 $$\langle f|Ag\rangle=\langle Af|g\rangle$$?

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

$$\langle Af|g\rangle$$:

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

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

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

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