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Let $|\alpha \rangle$ be coherent state in Fock space. According to the paper "Coherent-state representation for the photon density operator" by Cahill (Phys. Rev. 138, B1566 (1965), §VII), every convergent series $\{\alpha_j\}$ on the complex plane generates a complete set $\{|\alpha_j \rangle \}$ of the Fock space. That means, every vector $| \psi \rangle$ can be written as super position $| \psi \rangle = \sum_j g_j |\alpha_j\rangle$.

That also means, that any coherent state can be written as $$ | \alpha \rangle = \sum_j c_j |\alpha_j \rangle . $$ Let us suppose that $|\alpha \rangle \notin \{|\alpha_j \rangle \}$. That means at least two coefficients $c_j$ are non-zero. Now, let us apply the annhilation operator $a$ to the coherent state. By definition we get $a |\alpha \rangle = \alpha |\alpha \rangle$. Using the expansion, we get $$ a | \alpha \rangle = \sum_j c_j \alpha_j |\alpha_j \rangle \stackrel{!}{=} \alpha \sum_j c_j |\alpha_j \rangle . $$ Doesn't that imply, that $\alpha_j = \alpha$, which contradicts our early assumption, that $|\alpha \rangle \notin \{|\alpha_j \rangle \}$? Or it means that the set $\{|\alpha_j \rangle \}$ is overcomplete, which means the super position is not unique. But is there a way to remove elements, to make it a non-overcomplete bases such that the super position is unique? Wouldn't we then get a contradiction as indicated above?

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A more detailed investigation of the completeness of an arbitrary countable subset $\lvert \alpha_i\rangle$ of coherent states can be found in "On the completeness of a system of coherent states" by Perelomov.

Notable findings from this paper:

  1. Indeed, any set of coherent states obtained from a convergent sequence of distinct $\alpha_i$ is supercomplete, and cannot be made complete by removal of a finite number of states.

  2. For arbitrary sequences $\alpha_i$ one may look at the state defined by $$ \lvert \psi\rangle = \frac{1}{\pi}\int \mathrm{e}^{-\lvert \alpha\rvert^2 / 2}\psi(\alpha^\ast)\lvert \alpha\rangle\tag{1}$$ for an entire analytic function $\psi$ for which $$ \frac{1}{\pi}\int\mathrm{e}^{-\lvert \alpha\rvert^2}\lvert \psi(\alpha)\rvert ^2 <\infty. \tag{2}$$ Conversely, any given state $\psi$ defines such a function by $$ \psi(\alpha) = \sum_n \frac{\alpha^n}{\sqrt{n!}}\langle n\vert \psi\rangle,\tag{3}$$ i.e. there is a bijection between entire analytic functions fulfilling eq. (2) and states in the Hilbert space. For any set of $\lvert \alpha_i\rangle$, one can now construct entire functions $\psi(\alpha)$ with zeros at all the $\alpha_i^\ast$, and the corresponding state, if it exists, i.e. fulfills eq. (2), will be orthogonal to all the $\lvert\alpha_i\rangle$ since $\psi(\alpha_i) = \mathrm{e}^{\lvert \alpha\rvert^2 / 2}\langle \alpha_i\vert\psi\rangle$. So completeness of an arbitrary set of coherent states comes down to the question whether an entire function with zeros at the $\alpha_i$ exists for which eq. (2) is true.

    Turns out that the growth behaviour (and hence convergence of the integral in eq. (2)) of an entire function with known zeros $\alpha^\ast_i$ is tied to the exponent of convergence of the sequence $\alpha_i$, which is the infimum of all $\lambda$ such that $\sum_i\lvert a_i\rvert^{-\lambda}$ converges. For $\lambda > 2$, our set of coherent states is still supercomplete (removing a single item does not change the exponent of convergence) and for $\lambda < 2$, it is not complete.

    For $\lambda = 2$, it depends on the convergence behaviour of $\sum_i \lvert a_i\rvert^{-\lambda}$ and I urge you to read the paper for all the details and citations. As an incentive to do so, let me just quote that there are indeed complete but not over-complete sets of coherent states: For any lattice $\alpha_{ij} = i \omega_1 + j \omega_2$ spanned by two complex numbers $\omega_1, \omega_2$, the set of coherent states $\lvert \alpha_{ij}\rangle$ is overcomplete and becomes merely complete when a single state is removed - but only when the cell area of the lattice is exactly $\pi$!

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Nice question! It would surprise me that any coherent state basis would not be overcomplete, and I read your expansion a a very elegant proof of it. [Erratum: such bases exist, see the answer by ACuriousMind.] Since the size of the set has to be infinite, “there [is actually no] way to remove elements to make it non-overcomplete”, to take your formulation.

A simple way to see that, is starting with the fact that every convergent series $\alpha:=\{\alpha_i\}_{i≥0}$ generate a complete set $\{\left|\alpha_i\right>\}_{i≥0}$ of the Fock space. Consider now the series $\alpha':=\{\alpha_i\}_{i≥1}$, that is the same series without its first term. It also converges, and one can therefore also express any state as a combination of $\{\left|\alpha_i\right>\}_{i≥1}$. In particular, there exist $\{h_i\}_{i≥1}$ such that $$\left|\alpha_0\right> = \sum_{i=1}^{\infty}h_i \left|\alpha_i\right>,$$ giving us two different decompositions of $\left|\alpha_0\right>$ in the basis generated by $\alpha$.

Of course, one could say that one just needs to remove $\alpha_0$, since it’s not needed, and work with the series $\alpha'$ instead. But the same problem applies and one could remove $\alpha_1$, and so on. One can remove any subset of $\alpha$ which keeps the series both infinite and convergent, it would still keep the set overcomplete.

This reasoning does not exclude the existence of another way to construct a complete but not overcomplete set, with e.g. a divergent series. However, your reasoning with annihilation operators indeed shows such feat is impossible. Edit: it is actually possible, see the other answer. I guess we can then not write $a\left|\alpha\right>=\sum_j c_j \alpha_j \left|\alpha_j\right> $ because the sum is divergent.

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    $\begingroup$ @NorbertSchuch : Actually, if you look at the answer of ACuriousMind (published after your question), there are indeed complete but not overcomplete bases. They look like the constituents of a GKP states, with one removed $\endgroup$ – Frédéric Grosshans Mar 5 '20 at 20:36
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    $\begingroup$ @NorbertSchuch Indeed. I will clearly investigate, since this answer is related with my last paper! $\endgroup$ – Frédéric Grosshans Mar 5 '20 at 20:51
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    $\begingroup$ I can not quite wrap my head around the fact that the state vector is divergent for the non-overcomplete bases. Is the representation of $| \alpha \rangle$ then also divergent, or is it just $a | \alpha \rangle$? For the bases I considered all representations are convergent, right? $\endgroup$ – yasalami Mar 6 '20 at 11:06
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    $\begingroup$ @yasalami $a|\alpha\rangle=|\alpha\rangle$, so both should have the same issue. But if the expansion does not converge nicely then it is not even clear if $a$ commutes with the sum ... $\endgroup$ – Norbert Schuch Mar 6 '20 at 11:12
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    $\begingroup$ I've just spent too much time with Ulysse Chabaud on a whiteboard discussing this. The problem is that $a$ is not continuous, and that we cannot push a into the infinite sum sign, as said by @NorbertSchuch $\endgroup$ – Frédéric Grosshans Mar 6 '20 at 16:13

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