Let $Q$ be a manifold and let $\alpha$ be an exact one form on $Q$. Define $$ \phi:T^*Q\to T^*Q,(q,p)\mapsto(q,p+\alpha_q). $$ Then I want to show that $\phi$ is a Hamiltonian diffeomorphism with respect to the canonical symplectic form $\omega^{can}_Q$ on $T^*Q$.
I defined the isotopy $\Phi:[0,1]\times T^*Q\to T^*Q$ by $\Phi_t(q,p)=(q,p+t\alpha_q)$, so $\Phi_0=\text{id}$ and $\Phi_1=\phi$. Define the time-dependant vector field $$X_t=\bigg(\frac{d}{dt}\Phi_t\bigg)\circ\Phi_t^{-1},$$ and I want to show that this is a Hamiltonian vector field, (so $X_t=X_{H_t}$ for some $H:[0,1]\times T^*Q\to\mathbb{R}$) so that we'll obtain that $\phi$ is the time-1 flow of a Hamiltonian vector field, hence a Hamiltonian diffeomorphism. I have that $$\omega^{can}_Q (X_t,-)=(-d\lambda^{can}_Q)(X_t,-)=d(\lambda^{can}(X_t))-\mathcal{L}_{X_t}\lambda^{can},$$ so we are done if $\mathcal{L}_{X_t}\lambda^{can}=0.$ We have that $$ \Phi_t^*\mathcal{L}_{X_t}\lambda^{can}=\frac{d}{dt}\Phi_t^*\lambda^{can}=\frac{d}{dt}(?), $$ How do I compute $\Phi_t^*\lambda^{can}$?