Why is the word 'simultaneously' important in stating Heisenberg's uncertainty principle? The Heisenberg's uncertainty principle states that a particle cannot have a precise value of its position and conjugate momentum simultaneously. 
If these uncertainties are intrinsic properties of a state why is the word 'simultaneously' important? Is this important only for those states which have non-trivial time-dependence? For trivial time-dependence i.e., energy eigenstates $\Delta x$ and $\Delta p_x$ are fixed in time and it appears that one can measure them at separate times. But this is not true for other states. Am I correct?
 A: The word "simultaneously" is there because you can obtain a precise measurement of an particle's position and afterward obtain a precise measurement of the momentum. But you can't do both at the same time. Further, when you make the second measurement (the measurement of the momentum), you disturb the object's position, so it in fact no longer has a definite position, and the position measurement you had made at first no longer contains any meaningful information about the particles position.
For a concrete example, consider a particle in the ground state of the quantum harmonic oscillator. In this state, $\Delta x$ and $\Delta p$ are both non-zero and constant in time, so neither the position nor momentum is known. Now if you measure the position, the particle's wave function collapses to a delta function, and you observe a precise position for the particle. But now when you measure the momentum, the wavefunction collapses into a plane wave state, which has infinite extent. So while you obtain a precise value for the momentum, you have lost all the information about the particle's position.
A: Just look at the formal version of the Heisenberg uncertainty principle:
$$ \sigma_x(\psi) \sigma_p(\psi) \geq \hbar/2,$$
where $\sigma_A(\psi) = \sqrt{\langle\psi\vert A^2\vert \psi\rangle - \langle \psi \vert A\rvert \psi\rangle^2}$ is the standard deviation of an operator $A$ for the state $\vert \psi \rangle$.
When we say a state has a "well-defined" or "precise" value of the observable $A$, we mean it is an eigenstate. It is straightforward to check that $\sigma_A(\psi) = 0$ in an eigenstate. So no state can be both an eigenstate of $x$ and of $p$, since that would mean $0\geq \hbar / 2$, which is clearly false.
The "simultaneously" means precisely that: A (time-dependent) state $\psi(t)$ may be an eigenstate of $x$ in one instance, and an eigenstate of $p$ in another, but it is impossible that $\psi(t_0)$ for any fixed $t_0$ is both.
A: In principle it holds that when you decide to measure the momentum the more uncertain of the position and vice versa but surely this leads to there is a simultaneity in these states but this principle means we are unable to measure the two occuring variables without interference. it doesn't detract from the actual physical occurrence of the two, we just ain't able to measure the two simultaneously in principle.
Well that's what I think it means
