[Physics] Why is ground state $| 0 \rangle$ of harmonic oscillator a coherent state

quantum mechanics

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

Best Answer

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.