# 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

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 $z=0$.

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: $$\left|z=0\right\rangle = \left|n=0\right\rangle$$

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.

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".

• This is right of course, and I was about to upvote, but isn't the OP asking for more than just a mathematical definition? In particular, the OP uses the words "intuition" and "significance" which is why I say this. Nov 9, 2013 at 1:41
• The OP is indeed asking for more, but (s)he also asks: "Is ground state |0⟩ of harmonic oscillator a coherent state just because it minimize the uncertainty product?" I am not sure answers must be "all or nothing":-) "Don't shoot the pianist, he's doing the best he can":-) Nov 9, 2013 at 1:51
• Haha ok fair enough. +1 since its right, but I don't think you deserve the checkmark; no offense :) Nov 9, 2013 at 1:57
• @joshphysics: Thank you. We often get what we don't deserve, we often don't get what we deserve... Please forgive me some cynicism:-) If you expect life to be fair, you are in for disappointment:-) So no offense taken:-) Nov 9, 2013 at 2:10

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.