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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.

Relativisticcucumber
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    In the sense of Hilbert spaces, two functions are parallel if they are constant multiples of each other. – eyeballfrog Jul 05 '22 at 20:01
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    In this setting we have a vector space whose elements are functions satisfying some condition---probably $L^2$ or similar in the context of your problem. Two vectors (i.e., functions) in this space are parallel if one is a scalar multiple of the other. – Travis Willse Jul 05 '22 at 20:01
  • A way to get parallel curves parametrically – Тyma Gaidash Jul 05 '22 at 20:07
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    You have defined that two functions are orthogonal iff the integral of their product is zero. That can define an inner product and a norm, if some conditions are met. E.g. for continuous functions from $\mathbb{R}$ to $\mathbb{R}$, integral of the squared function will be null iff the function is null. Then, two functions are parallel iff their inner product is equal to the product of their norms. For continuous functions this can probably be proven although I am too tired right now to think about it... – Jean-Armand Moroni Jul 05 '22 at 20:37

1 Answers1

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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

$$\n{x} \n{y} = \n{ay} \n{y} = \abs{a} \n{y}^2$$

PrincessEev
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