# How to prove or disprove a type of states form an overcomplete basis in the Hilbert space?

I am a PhD student in Physics. Let us consider a vector in an infinite dimensional Hilbert space as

\begin{equation} |f\rangle\equiv \begin{bmatrix} 1 \\ z \\ z^2 \\ \vdots \end{bmatrix}, \end{equation}

where $z$ is a complex number. The norm of the vector $|f\rangle$ is \begin{equation} \left\lVert\langle f|f\rangle\right\rVert=1+|z|^2+|z|^4+\cdots=\frac{1}{1-|z|^2}. \end{equation}

Hence, the vector $|f\rangle$ is normalizable only when $|z|<1$, and the normalized vector should be $|z\rangle=\sqrt{1-|z|^2}|f\rangle$. The overlap between any two distinct vectors is \begin{equation} \langle z_2|z_1\rangle = \frac{\sqrt{1-|z_2|^2}\sqrt{1-|z_1|^2}}{1-z^*_2z_1}, \end{equation} which yields \begin{equation} |\langle z_2|z_1\rangle|^2=\frac{(1-|z_2|^2)(1-|z_1|^2)}{|z_1-z_2|^2+(1-|z_2|^2)(1-|z_1|^2)}\leq 1, \end{equation} Hence, any two distinct vectors are non-orthogonal, and the overlap of the two vectors is always small than one.

Now, my question is: can these vectors form a complete/overcomplete basis in the infinite dimensional Hilbert space?

For any two arbitrary vectors $|\psi\rangle$ and $|\phi\rangle$, the above statement is true when the following relation is valid \begin{equation} \langle\psi|\phi\rangle=\frac{1}{\pi}\int \langle\psi|z\rangle\langle z|\phi\rangle g(z)d^2z, \end{equation} where $g(z)$ is an unknown integration kernal. If we denote $z=re^{i\phi}$, the integration becomes \begin{align} \langle\psi|\phi\rangle&=\frac{1}{2\pi}\sum_{n,m}\langle\psi|m\rangle\langle n|\phi\rangle\int_0^1 g(r)r^{n+m+1}dr\int_0^{2\pi}e^{i(n-m)}d\phi\\ &=\frac{1}{2\pi}\sum_{n,m}\langle\psi|m\rangle\langle n|\phi\rangle\int_0^1 g(r)r^{n+m+1}dr\int_0^{2\pi}e^{i(n-m)}d\phi\\ &=\sum_{n}\langle\psi|n\rangle\langle n|\phi\rangle\int_0^1 g(r)r^{2n+1}dr, \end{align} where we have selected a kernal function $g(r)$ that is independent of the polar angle $\phi$. Hence, if we can prove that for any positive integer $n$ \begin{equation} \int_0^1 g(r)r^{2n+1}dr=1, \end{equation} then $|z\rangle$ for $|z|<1$ will form a overcomplete basis in the infinite dimensional Hilbert space. However, I can't find such $g(r)$. Therefore, I doubt that $|z\rangle$ do not form a complete basis.

• You are not really stating a clear-cut mathematical question, but I think what you meant to ask is: Is the linear span of $(z^n)_{n\ge 0}$, $|z|<1$, dense in $\ell^2$? The answer to this is yes, because if $y\in\ell^2$ is orthogonal to all these vectors, then letting $z\to 0$ shows you that $y_0=0$, next $y_1=0$ in the same style etc. – Christian Remling Sep 21 '17 at 21:48
• The much better site for questions of this type is math.stackexchange.com though they too have rather strict standards about presentation of questions. – Christian Remling Sep 21 '17 at 21:49
• @ChristianRemling I don't think the OP is asking about monomials, I think he is asking (in effect) about normalized Cauchy kernels in $H^2(D)$, and whether they form a total set in this Hilbert space – Yemon Choi Sep 21 '17 at 22:03
• @ChristianRemling Also, complete and overcomplete are terms in frame theory, IIRC. I don't think that the presentation of this question is deficient, even if one might argue that its level is more suited to MSE – Yemon Choi Sep 21 '17 at 22:05

As suggested by Yemon Choi, the mapping $\hat{\psi}(z) = \langle \psi | f \rangle$ is an analytic function on the unit disc, a member of the $H^2$ complex Hardy space. Even though $f$ is not a member of the $\ell^2$ Hilbert space when $|z| = 1$, the mapping $\hat{\psi}$ extends to the boundary of the unit disk (in the sense of almost everywhere limit along radii, for example), and by Parseval's identity, \begin{aligned}\langle \psi | \phi \rangle & = \frac{1}{2 \pi} \int_0^{2\pi} \hat{\psi}(e^{i t}) \overline{\hat{\phi}(e^{i t})} dt \\ & = \frac{1}{2 \pi} \int_0^{2\pi} \langle \psi | f_{e^{it}} \rangle \langle f_{e^{it}} | \phi \rangle dt .\end{aligned} In other words, your "function" $g(z)$ should be the limit of $\tfrac{1}{2 \pi} (1 - r^2)^{-1/2}$ times the Lebesgue measure on the boundary of the disk of radius $r$ as $r \to 1^-$.