# Temporal stability of multimode coherent states

For the standard quantum harmonic oscillator, the coherent states $$\{|\alpha\rangle, \alpha\in \mathbb{C}\}$$ are temporally stable. That is, $$e^{-iH t}|\alpha\rangle = |e^{-i t} \alpha\rangle,$$ up to phase, where $$H$$ is the Hamiltonian for the standard QHO. (See, e.g., Sec 3.3.4 of Coherent States in Quantum Physics by Gazeau.)

For a system of $$k$$ coupled QHOs, the state space is spanned by the multimode coherent states $$|\vec{\alpha}\rangle = |\alpha_1, \dots, \alpha_k\rangle = |\alpha_1\rangle\cdots|\alpha_k\rangle,$$ where $$\vec{\alpha} = (\alpha_1, \dots, \alpha_k)\in \mathbb{C}^k$$. If the oscillators are uncoupled, then these multimode coherent states are again temporally stable, simply by considering the systems independently.

I was wondering whether the following is true: If the oscillators are coupled with Hamiltonian $$H$$, are the multimode coherent states again temporally stable, in the sense that there is some (necessarily unitary, see below) $$U(t)\in \mathbb{C}^{k\times k}$$ such that (up to phase) $$e^{-iH t}|\vec{\alpha}\rangle = |U(t) \vec{\alpha}\rangle ?$$

I'm new to QM, so maybe this is an easy exercise and I apologize in advance if it is trivial. Any references would be appreciated.

If not true for a system of coupled QHOs, is there some natural system/Hamiltonian for which this is indeed true?

=====

Note the matrix $$U(t)$$ is necessarily unitary. To see why, assume the multimode states are temporally stable, and let $$|\vec{\alpha}'\rangle = e^{-iH t}|\vec{\alpha}\rangle$$ and $$|\vec{\beta}'\rangle = e^{-iH t}|\vec{\beta}\rangle$$, both up to phase, then

$$e^{-|\vec{\alpha}'-\vec{\beta}'|^2}=|\langle \vec{\alpha}'|\vec{\beta}'\rangle|=|\langle \vec{\alpha}|\vec{\beta}\rangle|=e^{-|\vec{\alpha}-\vec{\beta}|^2}.$$

So, the map $$\vec{\alpha}\to \vec{\alpha}'$$ is an isometry on $$\mathbb{C}^k$$.

=====

Editing to add: After some thought, it seems that multimode coherent states are stable (in the sense noted above) if (and only if) we are dealing with a Hamiltonian of the form: $$H = \sum_{i,j} A_{i,j} \hat{a}_i^{\dagger}\hat{a}_j + c,$$ where $$c$$ is a constant, and $$(A_{ij})\subset\mathbb{C}$$ are such that $$A_{ij}=A_{ji}^*$$ for $$1\leq i,j\leq k$$. Here, $$\hat{a}_i$$ and $$\hat{a}_i^{\dagger}$$ denote the annihilation and creation operators on the $$i$$th factor of the state space.

Clearly decoupled QHOs have a Hamiltonian of this form. However, I think that Hamiltonians that are quadratic in the ladder operators seem to fall under the umbrella of so-called Andersen Hamiltonians (see, e.g., here) or quadratic Fermionic Hamiltonians. This is the first I've heard of them. I'll try following this up on my own, but if anyone has any simple explanation of their relevance, I'd be appreciative.

My approach to see why this is the case is as follows: The $$U(t)$$ noted above must be a unitary semigroup, so of the form $$U(t) = e^{i t A}$$ for some Hermitian $$A\in \mathbb{C}^{k\times k}$$. Similarly, any phase factor out front must be of the form $$e^{i t \theta}$$. So, you write $$e^{-i t H}|\vec{\alpha}\rangle = e^{i t \theta} | e^{i t A}\vec{\alpha}\rangle$$ for a multimode coherent state $$|\vec{\alpha}\rangle$$. When you differentiate both sides with respect to $$t$$ and set $$t=0$$, you reach the desired conclusion after using the various properties involving ladder operators and coherent states.

The disclaimer is that I didn't work out all the details carefully, but I believe it is correct.