# Identity of bosonic coherent states

I have a short question about the meaning of the identity of the bosonic coherent states.

Before I ask the question I will explain some background.

The eigenstate of the bosonic annihilation operator $$\widehat{a}$$ is: $$|\phi\rangle = \sum_{n}\frac{\phi^{n}}{\sqrt{n!}}\frac{\left(\widehat{a}^{\dagger}\right)^{n}}{\sqrt{n!}}|vac\rangle = e^{\phi \widehat{a}^{\dagger}}|vac\rangle$$ It satisfy $$\widehat{a}|\phi\rangle = \phi|\phi\rangle$$.

I assume that we can see the operator $$\widehat{a}$$ as destroying a boson particle at a specific position as an example.

The eigenstate for the set of bosonic annihilation operators $$\left\{\widehat{a}_{x} \right\}$$ is: $$|\vec{\phi}\rangle = e^{\sum_{x}\phi_{x}\widehat{a}^{\dagger}}|vac\rangle$$ It satisfy $$\widehat{a}_{i}|\vec{\phi}\rangle = \phi_{i}|\vec{\phi}\rangle$$. Note $$x$$ represents different positions.

The identity operator is: $$\widehat{I} = \int \left(\prod_{x} \frac{d\Re{\phi_{x}}d\Im{\phi_{x}}}{\pi} \right)e^{-\sum_{x}\phi_{x}^{*}\phi_{x}}|\phi\rangle \langle \phi|$$

My question is, what is the meaning of $$|\phi\rangle \langle \phi|$$ in the integral? Is $$|\phi\rangle$$ in the integral the eigenstate of the annihilation operator at the specific position $$x$$?

The state in the integral is the joint eigenstate of each of the annihilation operators. The annihilation operators are for particular modes labeled here by $$x$$. I will clarify what I think the notation should be, for consistency.
The vector states should be $$|\vec{\phi}\rangle=e^{\sum_x \phi_x \hat{a}^\dagger_x}|\mathrm{vac}\rangle$$ such that they are eigenstates of the annihilation operators $$\hat{a}_x$$ with eigenvalues $$\phi_x$$. It is assumed that operators for different modes commute and that the modes are bosonic: $$[\hat{a}_x,\hat{a}^\dagger_y]=\delta_{xy}$$. Normally we don't use $$x$$ to represent a discrete index but alas (OP has sums over $$x$$, so it is likely to be a discrete parameter).
Then we note that the states are not normalized; the normalized state in this notation is $$e^{-\sum_x |\phi_x|^2/2}|\vec{\phi}\rangle$$. These normalized states are what go into the resolution of identity; hence the factor of $$e^{-\sum_x |\phi_x|^2/2}\times (e^{-\sum_x |\phi_x|^2/2})^*=e^{-\sum_x |\phi_x|^2}$$.
Another way of rewriting the whole thing, making explicit that each mode is independent because one should learn about the single-mode case before the multimode version, is $$I=\bigotimes_x\int \frac{d\Re{\phi_x} d\Im{\phi_x}}{\pi}e^{-|\phi_x|^2}e^{\phi_x \hat{a}_a^\dagger}|\mathrm{vac}\rangle\langle \mathrm{vac}|e^{\phi_x \hat{a}_a}=\bigotimes_x\int \frac{d\Re{\phi_x} d\Im{\phi_x}}{\pi}e^{-|\phi_x|^2}|\phi_x\rangle\langle \phi_x|,$$ where the symbol $$\bigotimes_x$$ implies a tensor product over multiple modes (the single-mode case just removes that symbol and selects one $$x$$).