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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 ($\dim l^2>3$)

BigbearZzz
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  • Just a clarification. Was "a function is nothing like an arrow" a complaint they had? – kmeis Oct 09 '15 at 14:57
  • Yeah, something along that line. It's not like they don't ACCEPT that functions are vectors but more like they can't SEE how functions can be treated in a similar manner as those "genuine" vectors. – BigbearZzz Oct 09 '15 at 15:01
  • So they do agree that, in similar fashion to $\mathbb{R}^3$, a vector can be described by its components for each of its basis vectors and so functions are just returning components for an infinite basis? Yet, they are struggling to visualize this as "an arrow"? – kmeis Oct 09 '15 at 15:13
  • Perhaps my answer to a similar question may be of use: http://math.stackexchange.com/questions/1176941/what-does-orthogonality-mean-in-function-space/1176956#1176956 – Gyu Eun Lee Oct 10 '15 at 02:34

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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.

David
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  • I really appreciate your answer but as I stated in the section "things that didn't work" on #2, they just cannot see how functions are related to 3D vectors. – BigbearZzz Oct 09 '15 at 14:55
  • I'm not sure there is any other way around this concept really, functions are related to vectors because we map them to vectors USING the properties of orthogonality, simply because it makes life easier. – David Oct 09 '15 at 16:54