Zero point fluctuation of an harmonic oscillator In a paper, I ran into the following definition of the zero point fluctuation of our favorite toy, the harmonic oscillator:
$$x_{ZPF} = \sqrt{\frac{\hbar}{2m\Omega}} $$
where m is its mass and $\Omega$ its natural frequency.
However, when I try to derive it with simple arguments, I think of the equality:
$$E = \frac12 \hbar\Omega=\frac12 m \Omega^2 x_{ZPF}^²$$
(using the energy eigenvalue of the $n=0$ state)
giving me:
$$x_{ZPF} = \sqrt{\frac{\hbar}{m\Omega}} $$
differing from the previous one by a factor $\sqrt2$. I am just puzzled, is it a matter of conventions or is there a fundamental misconception in my (too?) naive derivation? 
 A: I think this is a combination of both a convention and a physical problem. You are equating the energy eigenvalue (ie, the total energy) to an expression that contains only $x_{ZPF}$, and does not contain $p$ at all. In other words, you are equating the total energy to a potential energy. This would be analogous to equating $E_\mathrm{total} = \frac{1}{2}kA^2$ to find the amplitude $A$ of a classical harmonic oscillator. The result is that you are using $x_{ZPF}$ to mean the "amplitude" of the zero-point fluctuation. The true result, as Ondrej Cernotik's answer derives, uses the rms value $x_{ZPF} = \sqrt{\langle\hat x^2\rangle}$. So that's the sense in which it is a convention.
The sense in which it is a real physical problem is that the "amplitude" of a quantum oscillator isn't really a well-defined, measurable thing. The quantum oscillator has a non-zero probability amplitude going all the way out to infinity. The rms value is well-defined and easy to measure. So that's the preferred definition.
A: You can find the value of zero point fluctuations just by calculating the variance $\langle(\Delta\hat{x})^2\rangle = \langle\hat{x}^2\rangle$ in the vacuum state. You can do this either using the $x$-representation or expressing the $\hat{x}$ operator using creation and annihilation operators. These are usually introduced by
$$
\hat{a} = \sqrt{\frac{m\Omega}{2\hbar}}\left(\hat{x}+i\frac{\hat{p}}{m\Omega}\right),
$$
so that you get
$$
\hat{x} = \sqrt{\frac{\hbar}{2m\Omega}}(\hat{a}+\hat{a}^\dagger).
$$
Using this to calculate $\langle\hat{x}^2\rangle = \langle 0|\hat{x}^2|0\rangle$ indeed gives you $$x_{ZPF} = \sqrt{\langle\hat{x}^2\rangle} = \sqrt{\frac{\hbar}{2m\Omega}}.$$
