Maximum of oscillator displacement due to one phonon let's say I have a mechanical mode with frequency $\omega$ and effective mass $m$. I put one phonon in that mode and want to get an estimate (or ideally) calculate the maximum displacement of my mode (i.e. my oscillator) from its equilibrium position. So basically the QM equivalent of $x_{max}$ in a classical description.
Intuitively, the max. displacement should scale with the phonon number, for $|n\rangle$ I should get a larger displacement than for $|0\rangle$. It should also scale with the zero-point fluctuations $x_{ZPF}$
However, I am lacking the tools to formally show that. I ofc. know that I can rewrite $\hat{x} = x_{zpf}(\hat{b} + \hat{b}^{\dagger})$, but if I apply $\hat{x}$ to my Fock state I just get a weird expression.
 A: The expectation value of $\hat{x}$ in the Fock state $\lvert n \rangle$ is of course zero, because $\hat{b}$ and $\hat{b}^{\dagger}$ both connect only Fock states that satisfy $\Delta n = \pm1$.  However, one might define an effective classical maximum displacement as the root-mean square, i.e.,
$$
x_{\textrm{max}} :=\sqrt{\left\langle \hat{x}^2 \right\rangle}\,.
$$
In this case, then, we get
$$
\left\langle \hat{x}^2 \right\rangle =
\langle n\rvert \hat{x}^2 \lvert n \rangle
= x_{zpf}^2\left(
0+
\langle n \rvert\hat{b}\hat{b}^{\dagger}\lvert n \rangle
+\langle n \rvert\hat{b}^{\dagger}\hat{b}\lvert n \rangle
+0
\right)
=
x_{zpf}^2(2n+1)\,,
$$
(using linearity of the inner product and the fact that, e.g. $\langle n \lvert \hat{b}^2 \rvert  n \rangle=0$)
so that
$$
x_{\textrm{max}} = x_{zpf}\sqrt{2n+1}\,.
$$

An alternative here is to consider the fact that for the harmonic oscillator,
$$
\langle \hat{x} \rangle (t) = \langle \hat{x} \rangle (0)\,\cos(\omega t) + \frac{1}{m\omega}\langle \hat{p} \rangle (0)\,\sin(\omega t)\,.
$$
This is true for any state $\lvert\Psi(x,t)\rangle$. For Fock states, both $\langle \hat{x} \rangle (0)$ and $\langle \hat{p} \rangle (0)$ are zero, and so the expectation value is zero. However, we can consider the (non-stationary) states that are considered to be the "most classical", i.e., the coherent states $\lvert \alpha\rangle$, given by
$$
\lvert \alpha\rangle = e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\lvert n \rangle\,.
$$
For coherent states,
$$
\langle \hat{x} \rangle (0) = \frac{\alpha+\alpha^*}{2}\,,~~~~~~~
\langle \hat{p} \rangle (0) = \frac{\alpha-\alpha^*}{2i}\,.
$$
The time-dependence of the coherent states is simple, i.e.,
$$
\lvert \alpha (t) \rangle = \lvert e^{-i\omega t}\alpha(0) \rangle\,,
$$
and this leads exactly to the expressions above.
Finally, the coherent states are eigenstates of $\hat{a}$, i.e.
$$
\hat{a} \lvert \alpha \rangle = \alpha \lvert \alpha \rangle\,,
$$
and so there's a sense in which $\alpha$ (the effective amplitude of the oscillatory motion) "scales" with $\hat{a}$, which is sort of like a square-root of $\hat{N} = \hat{a}^{\dagger}\hat{a}$.  So in this picture, there's still a scaling of the amplitude with $\sqrt{n}$, in a sense.
