Uncertainty principle-Exact Measurement of one observable possible? I'm currently learning for my exam in physical chemistry and have the following question:
According to the uncertainty principle the momentum and the location cannot be measured exactly at the same time. But can only ONE be measured exactly? Because this would mean that the deviation equals zero and then the product of the deviations also equals zero, but doesn't this violate the uncertainty principle?
Thanks
 A: There are two questions here, one of theory and one of practice.
In theory, you can definitely imagine measuring a particle's position (or its momentum) exactly.  Its wavefunction $\psi(x)$ is then represented by a Dirac delta function for the position you measured the particle at.  That is, $\psi(x) \propto \delta(x - x_{m})$ in one dimension, where $x_m$ is the measured position of the particle.  This state isn't normalizable since the Dirac delta isn't actually a function (instead it's something called a distribution).  We can still treat it as well-defined and find its momentum-space wavefunction, given by the Fourier transform of its position-space wavefunction.  We then get:
\begin{equation}\tilde{\psi}(p) \propto \int\limits_{-\infty}^{\infty} \psi(x) e^{-ipx/\hbar} = e^{-ipx_{m}/\hbar}\end{equation}
$\tilde{\psi}(p)$ has the same modulus everywhere, that is $\tilde{\psi}(p)^* \,\tilde{\psi}(p) \propto 1$, so the particle is equally likely to have any momentum, and the uncertainty in the momentum is infinite.  If you're careful about questions of normalization, then you indeed get that the product of the uncertainty in position and momentum resolves itself as a constant.
In practice, it's impossible to measure a particle's position with infinitely high precision.  This resolves any of the confusion about normalization in the above argument, and sure enough if we replace the Dirac delta function above with a very narrow Gaussian distribution, we find that no matter how narrow we take the Gaussian to be, the product in the uncertainties in position- and momentum-space can never be less than $\hbar/2$, the value predicted by the Heisenberg Uncertainty Principle.
A: Of course you can and we do it all the time, for all practical purposes. Spin eigenstates exist everywhere, eg, the ground state of a hydrogen atom, for the electron state. That's the angular momentum that Gennaro was referencing (or one example). The angular position is then undetermined, the angular wave function is constant and the angular prob density is angularly symmetric, with variance of 0. The same is true for momentum and position; a plane wave has an exact momentum and completely undetermined position in that direction. Again 0 and infinite variance for the wave functions with absolute value squared interpreted as probability density functions. 
There is no issue in physics. Take a limit of those uncertainties by taking an almost plane wave and always before reaching the 0 and infinite product it satisfies the uncertainty principle. The limit of zero times infinity is not a physical issue. 
It is important in physics to understand what the theory means. It is not that there is anything wrong with the uncertainty principle, but issues with zeros and infinities sometimes don't mean anything you can not deal with. Sometimes they are critical, like the not understood infinities that occur in general relativity - then we know the theory breaks down. Similarly, please Understand that as WetSavanaAnimal writes, there is the practical measurement, though quantum theory having eigenstates is a different thing. We can indeed determine whether a state is, or is placed in, a certain eigenstate. Finally, the classical measurement errors usually are noise and other factors we can't control, but we can do more and more accurate measurements, including quantum measurement apparatus to measure, or sometimes more accurately, select or place something in certain  quantum states. Physicists deal with that all the time in for instance researching entanglement and quantum computing.   
