More Electron Traps — Question 2

Question 2

The ground-state rms size is smaller than the classical turning radius

A one-dimensional harmonic trap has $V(x)=m\omega^2x^2/2$ and $E_n=(n+1/2)\hbar\omega$, $n=0,1,2,\ldots$. The ground state has $\langle x^2\rangle=\hbar/(2m\omega)$ and zero mean position. For an electron use $\hbar\omega=.0800\,\mathrm{eV}$ and $\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}$.

1. Find the ground-state position standard deviation.

2. Find the classical turning position at energy $E_0$ and compare it with the standard deviation.

Question 2 - Solution

The ground-state rms size is smaller than the classical turning radius

(a) Method and calculation

$$\sigma_x^2=\frac{\hbar^2/(2m_e)}{\hbar\omega}=\frac{.0381}{.0800}\,\mathrm{nm^2}.$$$$\boxed{\sigma_x=.690109\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\tfrac12m\omega^2x_t^2=\tfrac12\hbar\omega\Rightarrow\boxed{x_t=\sqrt{\hbar/(m\omega)}=\sqrt2\,\sigma_x}.$$$$\boxed{x_t=.975961\,\mathrm{nm}}.$$

Physical check

A quantum ground state has tails beyond the classical turning positions. The rms size is a moment of the full density, not a sharp edge.