Why is ground state $| 0 \rangle$ of harmonic oscillator a coherent state? Is ground state $| 0 \rangle$ of harmonic oscillator a coherent state just because it minimize the uncertainty product?
What is the intuition of this. I don't quite understand the significance of the ground state being a coherent state.
Any explanation would be appreciated
 A: Mathematical answer #1
Mathematically, you can use the definition $a|z\rangle \equiv z|z\rangle$ of coherent states to explicitly construct them in terms of energy eigenkets. Up to a normalization constant, the result is:
$$\left|z\right\rangle = \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}}\left|n\right\rangle$$
Now, what happens if you let $z \rightarrow 0$ in the expression above? The only surviving term in the right-hand sum will be the one with $n=0$, and we get the result:
$$\left|z=0\right\rangle = \left|n=0\right\rangle$$
So the ground state of the system is a coherent state with parameter $z=0$.
Mathematical answer #2
You could also see this directly from definitions. Coherent states are defined as $a|z\rangle \equiv z\left|z\right\rangle$. Per definition, the coherent state with $z=0$ will then have the following property:
$$a\left|z=0\right\rangle = 0$$
The definition of the ground state of the harmonic oscillator is:
$$a\left|n=0\right\rangle = 0$$
Putting the two definitions together, we get the same result as earlier:
$$\left|z=0\right\rangle = \left|n=0\right\rangle$$
Physical answer
One of the interesting properties of the coherent states $\left|z\right\rangle$, is that the expectation values of position and momentum follow the equations of a classical harmonic oscillator:
$$\langle x \rangle = \sqrt{\frac{2\hbar}{m\omega}}\Re\left\{z \exp(-i\omega t)\right\}$$
$$\langle p \rangle = \sqrt{2\hbar m\omega}\Im\left\{z \exp(-i\omega t)\right\}$$
It might look a bit more familiar if we choose a real value of $z$:
$$\langle x \rangle = \sqrt{\frac{2\hbar}{m\omega}} \; z \cos(\omega t)$$
$$\langle p \rangle = -\sqrt{2\hbar m\omega} \; z \sin(\omega t)$$
Now, what would happen if you let $z\rightarrow 0$? Then you basically get a state with $\langle x \rangle = \langle p \rangle = 0$.
So if you think of the coherent states as the states where expectation values behave like a classical harmonic oscillator, then the ground state is the state with zero energy, i.e. the state with expectation values that doesn't oscillate at all.
A: It's an eigenstate of the annihilation operator.
EDIT (11/09/2013): As I explained in a comment, the above was meant to answer the first OP's question: "Is ground state |0⟩ of harmonic oscillator a coherent state just because it minimize the uncertainty product?" However, I should have added that not all states that "minimize the uncertainty product" are "coherent states" under the standard definition: "squeezed states" (http://cua.mit.edu/8.422/Reading%20Material/PHYSICS-henry-glotzer-a-squeezed-state-primer-am-j-phys-v56-p318-1988-AJP000318.pdf , Am. J. Phys., v.56, p.318 (1988); they are also called "squeezed coherent states" - http://en.wikipedia.org/wiki/Squeezed_coherent_state) minimize the uncertainty product, but the set of squeezed states is much larger than the set of "coherent states". So, again, the answer to the first OP's question is: No, it's not "just".  
A: One of the most important features of coherent states is that they minimise the Heisenberg uncertainty principle, just as the ground state $|0\rangle$ does. In that respect, the ground state is a coherent state. 
