Hermite Polynomials: Spans All Polynomials f from R to R?

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SUMMARY

The discussion centers on the capability of Hermite Polynomials to span all real polynomials, particularly in the context of quantum mechanics and harmonic oscillators. While the orthogonality of Hermite Polynomials is established, it is clarified that orthogonality alone does not imply spanning. The participants conclude that Hermite Polynomials, being infinite in number and orthogonal, can indeed span all real functions, as each polynomial can be expressed as a linear combination of Hermite Polynomials.

PREREQUISITES
  • Understanding of orthogonal polynomials
  • Familiarity with Hermite Polynomials
  • Basic knowledge of linear algebra concepts such as span and linear combinations
  • Introduction to quantum mechanics and its mathematical foundations
NEXT STEPS
  • Study the properties of Hermite Polynomials in detail
  • Learn about the proof techniques for polynomial spanning, including induction
  • Explore other orthogonal polynomials, such as Legendre Polynomials, and their applications
  • Investigate the role of orthogonal polynomials in quantum mechanics and harmonic oscillators
USEFUL FOR

Mathematicians, physicists, and students in advanced mathematics or quantum mechanics who seek to understand the role of Hermite Polynomials in polynomial approximation and their applications in physics.

Domnu
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Since the Hermite Polynomials are orthogonal, could one state that they span all polynomials f where f : R \rightarrow R? This would be EXTREMELY useful for the harmonic oscillator potential in quantum mechanics...
 
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Domnu said:
Since the Hermite Polynomials are orthogonal, could one state that they span all polynomials f where f : R \rightarrow R? 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.
 
It doesn't follow from orthogonality alone, but it is nevertheless true.
 
jimmysnyder said:
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].
 
Domnu said:
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?
No, that doesn't make any sense.

In what sense are you using the word "span"?
 
I'm using the word span in the sense that any polynomial of the form

f(\xi) = a_n \xi^n + a_{n-1} \xi^{n-1} + a_{n-2} \xi^{n-2} + \cdots + a_1 \xi + a_0

can be written as a linear combination of H_0, H_1, \cdots, H_n. I see I had a typo in my last post... I meant that the Hermite polynomials could span all polynomials (real).
 
Domnu said:
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?
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,...)
...
 
jimmysnyder said:
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.
 
  • #10
Dick said:
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.
My guess is that Domnu would benefit by doing the induction.
 

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