For Hermitian matrices and operators, the most "natural" inner product is $f^H \cdot g$ or $\int f^\* g\; dx$. A similar situation holds interpreting Fourier transforms as the inner product of functions with complex exponential functions. My question is, why is this the most "natural" choice? Is there something deeper to this choice (other than making $< f,f>$ a norm) related to some kind of duality of vector spaces?



Why do inner products require conjugation?
Bi- (or sesqui-) linear forms are nicer if they're nondegenerate. But they can always be restricted to subspaces. So, they're even nicer if they're nondegenerate on all subspaces. For symmetric forms on ${\mathbb R}^n$, that forces definiteness (positive or negative).


The usual bilinear form on ${\mathbb C}^n$ doesn't have this inherited-nondegeneracy property, but the Hermitian one does.


Anyway that's a practical issue, rather than a naturality statement. One way to decide it is natural is to embed $M\_n({\mathbb C})$ into $M\_{2n}({\mathbb R})$ by using the obvious $\mathbb R$-basis of ${\mathbb C}^n$. This fills the $2n\times 2n$ matrix with lots of $2\times 2$ real matrices whose transposes correspond to complex conjugate. Transposes come up in dot products if you notice that $\langle v,w\rangle$ is the unique entry in the $1\times 1$ matrix $v^T w$.


