# Why uncertainity is minimum for coherent states?

While reading for quantum damped harmonic oscillator, I came across coherent states, and I asked my prof about them and he said me it is the state at which $\Delta x\Delta y$ is minimum. I didn't quite understand why it is minimum.

As you probably know, for any particle the product of the uncertainties in the position, $\Delta x$, and the momentum $\Delta p$ (not $\delta y$ as you state) is bounded below by a positive constant; $$\Delta x\, \Delta p\geq\frac{\hbar}{2}.$$ (If this doesn't ring a bell, you need to read up on Heisenberg's uncertainty principle.) For general states, the uncertainty product will probably be quite larger than $\hbar$, and for classical objects it will be very much larger. If we want to be very precise about a measurement, though, we would want the particle to have minimal uncertainty: that is, we'd want to impose the condition $$\Delta x\, \Delta p=\frac{\hbar}{2}.\tag{1}$$ The states that obey this condition are called coherent states.
A bit more technically, the general solutions to equation (1) are called squeezed coherent states, essentially because we can "squeeze" the uncertainty from $x$ into $p$ or vice versa. If the particle is in a harmonic-oscillator potential then we can choose a unique way to "split" the uncertainty product into "minimal" position and momentum parts, $$\Delta x=\sqrt{\frac{\hbar}{2m\omega}},\;\Delta p=\sqrt{\frac{1}{2}m\omega\hbar },$$ using the dimensional information contained in $\omega$ and $m$.