Completeness of derivatives of Hilbert basis with respect to a parameter Let us take a Hilbert basis $\left|x_\lambda\right >$ in a Hilbert space $\mathcal{H}$, i.e. the $\left|x_\lambda\right >$ are a complete, orthonormal set of vectors. The subscript indicates that they depend parametrically on a parameter $\lambda$. 
Consider now the new family of vectors $\left|\partial_\lambda x_\lambda\right >$ that is obtained by taking the derivative with respect to $\lambda$ of $\left|x_\lambda\right >$. The notation simply means 
$$\left|\partial_\lambda x_\lambda\right >=\sum_i \partial_\lambda\left<i, x_\lambda\right>\left |i \right>$$ 
with $\left|i\right>$ some fixed, i.e. parameter-independent, basis. 
My question: Under what conditions is this new family complete? 
To be clear, an example may be the following. The eigenfunctions of the harmonic oscillator hamiltonian are a basis for $L^2(\mathbb C)$, which depends parametrically on the frequency $\omega$ of the oscillator. Are derivatives with respect to $\omega$ of Hermite polynomials times Gaussian still a complete family?
 A: I assume you mean you have a family $(e_k(t))_{k\in\mathbb N}$ of orthonormal bases of a Hilbert space $\mathcal H$, depending differentiably on a parameter $t\in\mathbb R$. Let us try to decide whether $(e_k'(t_0))_{k\in\mathbb N}$ is complete in $\mathcal H$ for some fixed $t_0\in\mathbb R$. For this, we assume that there is a differentiable function $U : \mathbb R\to L(\mathcal H)$ such that $e_k(t) = U(t)e_k$, where $U(t_0) = I$ (the identity operator). This should be the case in your harmonic oscillator example. Obviously, $U(t)$ is a unitary operator for each $t$. Now, if $(e_k'(t))_{k\in\mathbb N}$ is not complete, there exists a non-zero vector $x\in\mathcal H$ such that $\langle x,e_k'(t_0)\rangle = 0$ for all $k\in\mathbb N$. Hence, $0 = \langle x,U'(t_0)e_k\rangle = \langle U'(t_0)^*x,e_k\rangle$ for all $k$, meaning that $U'(t_0)^*x = 0$. Actually, you can remove the star here, since $UU^* = I$ implies $U'U^* + U(U')^* = 0$ and thus $(U')^* = -U^*U'U^*$, which, in $t = t_0$, is $U'(t_0)^* = -U'(t_0)$ (in other words, $U'(t_0)$ is skew symmetric). So, the system $(e_k'(t_0))_{k\in\mathbb N}$ is not complete if and only if there exists a non-zero vector in the kernel of $U'(t_0)$.
