For the sake of concreteness, let's say that our Hilbert space is the beloved $\mathscr L^2(\Bbb R)$. Suppose that we have $\psi,\phi\in\mathscr L^2(\Bbb R)$, what's the intuitive meaning to a semi-layperson when we say that $\langle\psi,\phi\rangle=0$? (or, for a physicist, $\langle\phi|\psi\rangle=0$)
I should make it clear here that I am very much comfortable with the word orthogonality and its use in Hilbert space theory in general. However, my friends who study physics seem to have a lot of problem trying to understand the "abstract right-angle" in $\mathscr L^2(\Bbb R)$, the fact that $\mathscr L^2(\Bbb R)$ is an infinite dimensional space only makes matter worse. I want to give a satisfactory answer to my friends but I am at lost. Can anyone give me an "intuitive" way to visualize when 2 "states" (I mean functions in $\mathscr L^2(\Bbb R)$) are "at the right-angle" to each other?
Things that I have tried (and why they didn't work):
1.) Give the definition of inner product (added more confusion)
2.) Compare the situation to $\Bbb R^3$ (a function is nothing like an arrow!)
3.) Use $l^2$ as and example ($\dim l^2>3$)
Usually matrices are the best way to visualise quantum states in a finite Hilbert space. In this way states are vectors, "bra" states row vectors and "ket" states column vectors and operators are matrices.
For the right angle I guess you could point out that for vectors we define an angle based on the dot product $\cos(\theta)=\mathbf{v}_1 \cdot \mathbf{v}_2/\vert \mathbf{v}_1 \vert \vert \mathbf{v}_2 \vert $ and which we can do completely analogously here where the space just has more dimensions (or less if there are only 2 states) $\cos(\theta)=$Re$(\langle \phi \vert \psi \rangle)/\sqrt{\langle \phi \vert \phi \rangle \langle \psi \vert \psi \rangle}$
Essentially compare to how the angle represents a "real" angle between vectors with the dot product and we can always map the system back to this representation. If you want something more visual I suggest looking up the Bloch sphere, we physicists like that.