Is there some sort of natural Hilbert space structure on $\mathbb{C}[z]$ so that $\{\frac{z^k}{\sqrt{k!}}\}$ are orthonormal? Can this structure be extended to $\mathbb{C}[z_1]\otimes \mathbb{C}[z_2]\otimes\cdots$ ? Is there any physical context where this structure on $\mathbb{C}[z]$ ( or $\mathbb{C}[z_1]\otimes \mathbb{C}[z_2]\otimes\cdots$ ) is commonly seen?
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The usual definition of "Hilbert space" requires completeness. However, by Baire category, there are no Banach spaces with a numerable basis, and a fortiori no Hilbert spaces with numerable basis. So the answer to your question is no. – Olivier Bégassat Aug 08 '13 at 00:21
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This is more of a comment than an answer.
There is a fun-looking metric that makes your basis orthonormal. Let $\delta$ be the Dirac-delta at $0$, so $\delta(P) = P(0)$ for any polynomial $P$. Let $\overline \delta$ be its conjugate, so $\overline \delta(P) = \overline{P(0)}$. Now define your metric by $$ h(P,\overline Q) = \exp(\delta \overline\delta)(P,\overline Q) := \sum_{k \geq 0} \frac1{k!} P^{(k)}(0) \overline Q^{(k)}(0). $$ Since $P$ and $Q$ are polynomials this is well defined and we quickly check that $(z^k / \sqrt{k!})$ is an orthonormal basis for this metric.
You can extend this metric to the tensor product you mention, but I have no idea if it has any physical significance there (or here).
Gunnar Þór Magnússon
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