# Uncertainty principle clarification

Suppose a mechanical quantum duck is confined in a pond 1m wide. So, what is its uncertainty in position, 0.5m or 1m?

As I think, the uncertainty is 0.5m. The reason is that if we choose the middle as a footprint, displacements at two extreme are 0.5 and -0.5. But others say that 1m is its uncertainty in position? Can anyone help me please?

The question is ill-posed. The uncertainty of a quantum mechanical object for any observable is not determined by where it is "confined". Given a quantum state $\lvert\psi\rangle$, the uncertainty of $A$ is the standard deviation from the expected value of $A$, i.e. $$\sigma_\psi(A) = \sqrt{\langle \psi \vert A^2 \vert \psi \rangle - \langle\psi \vert A \vert \psi \rangle^2}$$ and the expectation values for the position operator $x$ and its square are not, in general, related to the area a quantum mechanical object is "confined" to. They are entirely properties of the specific state the object is in, so the uncertainty is also a property of the specific state. That the position operator might only take values between $-0.5$ and $0.5$ does not directly lead to any specific value of $\sigma_\psi(A)$.