Since the Hermite Polynomials are orthogonal, could one state that they span all polynomials [tex]f[/tex] where [tex]f : R \rightarrow R[/tex]? This would be EXTREMELY useful for the harmonic oscillator potential in quantum mechanics...
No. If you think of the two vectors: (1, 0, 0) and (0, 1, 0) They are orthogonal to each other, but they don't span. on the other hand, consider the three vectors: (1, 0, 0), (1, 1, 0), (1, 1, 1) They span, but they are not orthogonal to each other. Therefore, knowing that the Hermite Polynomials are orthogonal is not enough to show that they span. You would need to prove that.
Fair enough... okay so I guess a better argument would be that the Hermite polynomials are infinite in number and are all orthogonal, so they span all functions?
They are useful to express any function defined on the range on which they are orthogonal. Other orthogonal polynomials will be useful to express functions on the range on which they are orthogonal (e.g. Legender polynomials on [-1,1].
I'm using the word span in the sense that any polynomial of the form [tex]f(\xi) = a_n \xi^n + a_{n-1} \xi^{n-1} + a_{n-2} \xi^{n-2} + \cdots + a_1 \xi + a_0[/tex] can be written as a linear combination of [tex]H_0, H_1, \cdots, H_n[/tex]. I see I had a typo in my last post... I meant that the Hermite polynomials could span all polynomials (real).
No. For instance, the following are mutually orthogonal and infinite in number, but they don't span, (0,1,0,0,0,...) (0,0,1,0,0,...) (0,0,0,1,0,...) ...
True. But it's pretty clear the Hermite polynomials span. H_n(x) contains a term containing x^n and none of the preceding H's do. You can formally prove it by induction, but it's actually pretty obvious if you imagine how you would go about expressing a given polynomial in terms of Hermite polynomials.