Coherent states - operator algebra problem with physics motivation Motivation: I have a mathematical problem motivated by quantum field theory in physics. It should be quite easy to prove, but for some reason I can't do it.
Intro:
Let there be operators $\hat{a_i}$ and $\hat{a_i}^{\dagger}$ for some integer $i$. Let there be the state (an element of vector space) $\left | 0 \right >$. A commutation relation holds such that 
$$
\left [ \hat{a_i} , \hat{a_j}^{\dagger}  \right ] = \left ( \hat{a_i} \hat{a_j}^{\dagger} - \hat{a_j}^{\dagger}  \hat{a_i} \right ) =  \delta_{ij}
$$
It holds that $a_i \left | 0 \right > = 0$ for all $i$.
Problem: Let there be a state (an element of vector space)
$$
\left | \psi \right > = e^{\sum_i \phi_i \hat{a_i}^{\dagger}} \left | 0 \right > \, ,
$$
where I define the exponential via its Taylor series as $e^{\hat{A}} = 1 + \hat{A} + \hat{A}\hat{A}/2! + \hat{A}\hat{A}\hat{A}/3! + \dots$ .
I want to show that
$$
\hat{a_x} \left | \psi \right > = \phi_x \left | \psi \right >  \, .
$$
Thanks a lot!
 A: First, we establish a simple result. Suppose we have two operators $A,B$ with $[A,B]=1$. Then one can easily prove via induction that more generally $[A,B^k]=k B^{k-1}$ for any natural $k$. Therefore the map $d_B$ defined as $d_B(\cdot)=[A,\cdot]$ acts as a derivative on all such terms $B^k$, and since this map is linear we furthermore have that this extends to any function $f(B)=\sum_k f_k B^k$. In particular, it applies to the exponential $\displaystyle e^{t B}=\sum_k \frac{t^k}{k!}B^k$ and so $d_B e^{t B}=[A,e^{t B}]= t e^{tB}$.
With this, the problem is straightforward. First, the coherent state can be written as $|\psi\rangle = \prod_i e^{\phi_i \hat{a}_i^\dagger}| 0\rangle$ since all the creation operators commute. But if $i\neq x$ then $\hat{a}_x$ also commutes with $\hat{a}_i^\dagger$ (and therefore $e^{\phi_i a_i^\dagger}$); otherwise, the formula shown above gives $\hat{a}_x e^{\phi_x \hat{a_x}^{\dagger}} = e^{\phi_x \hat{a_x}^{\dagger}}(\hat{a}_x+\phi_x).$ Consequently $$\hat{a}_x |\psi\rangle =a_x\left(\prod_i e^{\phi_i \hat{a}_i^\dagger}\right)| 0\rangle=\left(\prod_i e^{\phi_i \hat{a}_i^\dagger}\right)(\hat{a}_x+\phi_x)| 0\rangle=\phi_x |\psi\rangle$$ since $\hat{a}_x$ annihilates the ground state.
