Let $H$ be a (complex) Hilbert space and let $B\colon H\times H\to \mathbb{R}$ be a continuous sesquilinear form (i.e. a continuous function that is linear in one argument and conjugate-linear in the other). Consider the associated quadratic form $Q(\psi)=B(\psi, \psi)$.
Assume that there exists a continuous one-parameter group of unitary operators $\{e^{i\theta T}\}_{\theta\in \mathbb{R}}$ such that $$\begin{array}{cc} Q\left( e^{i\theta T}\psi - \psi \right)=O\left(\lvert \theta\rvert^3\right), & \theta \to 0. \end{array} $$ Here $\psi$ belongs to the domain of the self-adjoint generator $T$ and is fixed.
Question. Is it true that $Q(T\psi)=0$?
Intuitively, this should be true because for very small values of $\theta$ one has $e^{i\theta T}\psi-\psi\approx i\theta T\psi$. Since $Q$ is continuous, one expects that $Q(i\theta T\psi)=O(\lvert \theta\rvert^3)$, which can only happen if $Q$ vanishes along the whole (complex) line generated by $T\psi$.
A special case of this result is used implicitly in a paper which I am reading (reference below). The point of this question is understanding whether the result continues to hold in this more abstract setting.
T. Duyckaerts, F. Merle, S. Roudenko. "Maximizers for the Strichartz norm for small solutions of mass-critical NLS", Annali della Scuola Normale Superiore di Pisa, Vol. X, issue 2 (2011).
The relevant result is used implicitly at the beginning of section 3. Warning: notation there is a bit different. The function $\psi$ of this question corresponds to $G_0$ of the paper.
Actually it is rather easy. Using the homogeneity of $Q$ we may write $$ Q(e^{i\theta T}\psi -\psi)=\lvert \theta\rvert^2 Q\left( \frac{e^{i\theta T}\psi -\psi}{\theta}\right),$$ and the left hand side is $O(\lvert \theta\rvert^3)$ by assumption, so $$ Q\left( \frac{e^{i\theta T}\psi -\psi}{\theta} \right) = O(\theta).$$ Since $Q$ is continuous, taking the limit as $\theta \to 0$ we obtain $$Q(iT\psi)=0, $$ as desired.