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 $x$?


1 Answer 1


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 $x$).


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