Why are coherent states not linearly independent?

Question

From the completeness relation one can see that, $$|\psi \rangle = \int \frac{d^2 \alpha}\pi \langle \alpha | \psi \rangle |\alpha\rangle.$$ And if $|\psi\rangle = |\beta \rangle$ (which is another coherent state), then $$|\beta \rangle = \int \frac{d^2 \alpha}\pi \langle \alpha | \beta \rangle |\alpha\rangle = \int \frac{d^2 \alpha}\pi \exp(-\frac{|\alpha|^2}2-\frac{|\beta|^2}2+\alpha^{*}\beta) |\alpha\rangle.$$

From this above expression how can one come to the conclusion that the coherent states are not linearly independent and thus over-complete.

Answer 1

The linear independence means that you can't represent any state of the set as the superposition of other states.

Howevee the second formula means exactly that - you can represent any coherent state as a superposition of other coherent states. Of course in the rhs the state $|\beta\rangle$ gives a contribution as well. However you may transfer it to the lhs earning a delta contribution in the integral, \begin{equation} \Big(1-\frac{1}{\pi}\Big)|\beta\rangle=\frac{1}{\pi}\int d^2\alpha \left[\exp\Big(-\frac{|\alpha|^2}{2}-\frac{|\beta|^2}{2}+\alpha^\ast\beta\Big)-\delta^{(2)}(\alpha-\beta)\right]|\alpha\rangle \end{equation} The key here is the coefficient $\frac{1}{\pi}$ in the completness relation, otherwise the $|\beta\rangle$ would cancel in the lhs