How do you test if two functions are parallel (?)? Question:
There is a definition of function orthogonality: that the integral of the product of functions is zero. Is there a notion of functions being parallel? If so, what is the condition for this? I have tried to find this condition, to no avail.
Background:
The context of this question is this: In griffiths' quantum mechanics, there is a statement in the section on ladder operators that says the following:
$(a_+)^2\psi_n$ is orthogonal to $\psi_n$, and $(a_-)^2\psi_n$ is as well, so when we take the expectation of $x^2$, which is $$\frac{\hbar}{2m\omega}\int \psi (a_++a_-)^2\psi^*dx$$ these two terms (the ones with $a_{+/-}$) cancel out. I was trying to understand this, and came across the idea that maybe if a function (say $f$) is orthogonal to another function (say $g$), then maybe this means that $f$ is orthogonal to $g^*$. I was trying to prove this by showing that a function and its complex conjugate are parallel, then I arrived at the question here.
 A: So let's bear in mind that these notions of orthogonality come from those we use when studying vectors in $\newcommand{\R}{\mathbb{R}}
\newcommand{\ip}[1]{\left\langle #1 \right\rangle}
\newcommand{\n}[1]{\left\| #1 \right\|}
\newcommand{\abs}[1]{\left| #1 \right|}
\R^n$. Namely: under the usual norm, inner product,

*

*$x,y \in \R^n$ are orthogonal ($x \perp y$) if $\ip{x,y}_{\R^n} = 0$

*$x,y \in \R^n$ are parallel ($x \parallel y$) if $\abs{\ip{x,y}_{\R^n}} = \n{x}_{\R^n}\n{y}_{\R^n}$
The latter comes as $\ip{x,y}_{\R^n} = \n{x}_{\R^n} \n{y}_{\R^n} \cos \theta$, for $\theta$ the angle between them. If they're parallel, that angle is $0$ or $\pi$ radians.
Naturally, sure, we could extend this to the $L^2$ inner product:
$$\ip{f,g}_{L^2(\Omega)} := \int_\Omega f(x) \overline{g(x)} \, \mathrm{d}x$$
and say $f,g$ are parallel by that definition, i.e.
$$\abs{ \ip{f,g}_{L^2(\Omega)}} = \n{f}_{L^2(\Omega)} \n{g}_{L^2(\Omega)}$$
and since $\n{x} = \sqrt{\ip{x,x}}$ in inner product spaces (for $\n \cdot$ the induced norm),
$$\int_\Omega f(x) \overline{g(x)} \, \mathrm{d}x
=
\left( \int_\Omega \abs{f(x)}^2 \, \mathrm{d}x \right)^{1/2} 
\left( \int_\Omega \abs{g(x)}^2 \, \mathrm{d}x \right)^{1/2}$$

An equivalent definition is the simpler "$x,y$ are parallel if there is a scalar $a$ such that $x=ay$". To see why this works, note the sesquilinearity of inner products and the homogenity property of norms:
$$\ip{x,y} = \ip{ay,y} = a \ip{y,y} = a \n{y}^2$$
and
$$\n{x} \n{y} = \n{ay} \n{y} = \abs{a} \n{y}^2$$
