# Countable basis of coherent states used to express coherent states

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?

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 $$\pi$$!

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.

• @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 Commented Mar 5, 2020 at 20:36
• @NorbertSchuch Indeed. I will clearly investigate, since this answer is related with my last paper! Commented Mar 5, 2020 at 20:51
• 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? Commented Mar 6, 2020 at 11:06
• @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 ... Commented Mar 6, 2020 at 11:12
• 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 Commented Mar 6, 2020 at 16:13