Heisenberg picture of displacement

In a certain model the displacement operator for the normal modes in a lattice is given by $$u_{s}=\sum_{\textbf{k}}\left(\frac{\hbar}{2mn\omega_{k}}\right)^{1/2}(a_{k}e^{iksb}+a_{k}^{+}e^{-iksb}).$$ Here, $$a$$ and $$a^{+}$$ denote the usual creation and annihilation operators for bosons and $$b$$ is a real constant. If the Hamiltonian for this dynamical system takes the form $$H=\sum_{k}\hbar\omega_{k}(a_{k}^{+}a_{k}+1/2),$$ which can be thought of as a set of uncoupled harmonic oscillators, then the Heisenberg picture for this same operator is $$u_{s}(t)=e^{iHt}u_{s}e^{-iHt}.$$ This is the quantity I'm trying to find. However, upon computing this last expression one finds $$u_{s}(t)=e^{it\sum_{k}\hbar\omega_{k}(a_{k}^{+}a_{k}+1/2)}\sum_{\textbf{k}}\left(\frac{\hbar}{2mn\omega_{k}}\right)^{1/2}(a_{k}e^{iksb}+a_{k}^{+}e^{-iksb})e^{-it\sum_{k}\hbar\omega_{k}(a_{k}^{+}a_{k}+1/2)},$$ which I do not know how to handle.

Any ideas as to how to keep calculating?

An easy way to this result is to use the basis in which the operator is diagonal. So you have states such that $$\hat a^\dagger \hat a |n\rangle = n |n\rangle,$$and in this basis one can choose phase factors so that $$\hat a = \sum_{n=0}^\infty \sqrt{n+1}~ |n\rangle\langle n+1|.$$Because the operator is diagonal in this basis, $$\exp\left(i \omega t~\hat a^\dagger \hat a\right) = \sum_{n=0}^\infty e^{in\omega t}|n\rangle\langle n|,$$ and so for example\begin{align} \tilde a(t) &= \exp\left(i \omega t~\hat a^\dagger \hat a\right) ~\hat a~ \exp\left(-i \omega t~\hat a^\dagger \hat a\right)\\ &=\sum_{\ell m n}e^{i(\ell - n)\omega t}\sqrt{m+1} ~ |\ell\rangle\langle \ell|m\rangle\langle m + 1|n\rangle \langle n |\\ &=\sum_{n=1}^\infty e^{-i\omega t}\sqrt{n}|n-1\rangle\langle n|\\ &=e^{-i\omega t}\hat a. \end{align} Indeed once you have worked this out without all of the other complexity, it is not hard (but it takes some experience!) to see that all of the complexity “commutes” with this essential complexity of figuring out how one of these things evolves, so that you can just substitute in the answer, $$\tilde u_{s}(t) =\sum_{\textbf{k}}\left(\frac{\hbar}{2mn\omega_{k}}\right)^{1/2}(a_{k}e^{iksb - i \omega_k t}+a_{k}^{+}e^{i\omega_k t-iksb}).$$
$$e^{H}ae^{-H}= a+[H,a]+\frac{1}{2!}[H,[H,a]]+\frac{1}{3!}[H,[H,[H,a]]]+...$$
Here $$H=ita^\dagger a$$. You also need the commutation of $$[H,a]=it[a^\dagger a, a]$$. You can calculate it using the commutator relation of:
$$it[a^\dagger a, a]=ita^\dagger[ a, a]+ it[a^\dagger, a]a=-ita$$