# Eigenfunctions orthogonal in Hilbert space

Gold Member
Hello,

I am having a question regarding how eigenfunctions are orthogonal in Hilbert space, or what does that even mean (other than the inner product is zero). I mean, I know in ##\mathbb {R^{3}}##, vectors are orthogonal when they are right angles to each other.

However, how can functions be "orthogonal", in the sense of being perpendicular, and does Hilbert Space have infinite dimensions?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I am having a question regarding how eigenfunctions are orthogonal in Hilbert space, or what does that even mean (other than the inner product is zero).

There is no other meaning to it, it simply means the inner product between the functions is zero. The geometrical interpretation for ##\mathbb R^n## is that two orthogonal vectors are at right angle to each other, but really this is also a matter of definition of orthogonality.

and does Hilbert Space have infinite dimensions?

It can have infinite dimensions, yes. It can even be non-separable. It does not have to be infinite dimensional.

Gold Member
But how can you even take the inner product of functions? I thought this was something you did with vectors. For example, whats the inner product of ##f_{1}(x) = x## and ##f_{2}(x) = x^{2}##? This doesn't mean anything to me

DrClaude
Mentor
But how can you even take the inner product of functions? I thought this was something you did with vectors. For example, whats the inner product of ##f_{1}(x) = x## and ##f_{2}(x) = x^{2}##? This doesn't mean anything to me
For functions, the inner product is usually defined (in physics) as
$$\langle f, g \rangle \equiv \int_a^b f^* g \, d\tau$$
where ##a## and ##b## are appropriate limits and the integration element ##d\tau## will depend on how the function is expressed. In 1D, ##d\tau## will usually be ##dx## or ##dp##.

Orodruin
Staff Emeritus