Quantum state with zero standard deviation of position operator Is any quantum state $|\psi\rangle$  possible such that the standard deviation $\sqrt{\langle\psi|(\Delta\hat{x})^2|\psi\rangle}$ of the position operator $\hat{x}$ is zero? If not, why? 
 A: The answer is negative. If there were such a state, $\psi$, it would satisfy 
$$\langle \psi| X^2 \psi \rangle - \langle \psi| X \psi \rangle^2= 0 \quad (1)$$ 
namely
$$\langle \psi| X^2 \psi \rangle = \langle \psi| X \psi \rangle^2$$ 
that is, since $X=X^\dagger$ and $||\psi||=1$:
$$\langle \psi| X \psi \rangle^2 = \langle X\psi| X \psi \rangle \langle \psi |\psi \rangle\:.\quad (2)$$
On the other hand,  Cauchy-Schwarz' inequality implies 
$$\langle \psi| X \psi \rangle^2 \leq  \langle X\psi| X \psi \rangle \langle \psi |\psi \rangle\:,$$
where, as is known $=$ holds if and only if the two vectors in the LHS scalar product are proportional. Since it is the case due to (2), we conclude that:
$$X\psi = \lambda \psi$$
for some $\lambda \in \mathbb C$. However, as is known,the operator  $X$ has no eigenvectors, its spectrum being purely continuous: $\sigma(X)= \mathbb R$. Therefore $\psi$ does not exist. 
Nevertheless, formally speaking, you could think of distributions $\psi_{x_0}(x) =\delta(x-x_0)$ as formal eigenvectors of $X$, with eigenvalue $x_0$.  These formal states, indicated by $|x_0\rangle$ with the suggestive (and very useful) Dirac notation, do not belong to the Hilbert space of the theory $L^2(\mathbb R)$, since they are not (equivalence classes of) functions. 
It is possible to extend the formalism (as done by Gelfand) to include these distributions as defining generalized states into a rigorous mathematical sense. In that case, however,  (1) cannot be intepretated literally for these generalized states, since it would involve  products of $\delta$ functions which cannot be defined.  
