More Electron Traps — Question 1

Question 1

An evenly spaced trap spectrum determines curvature and zero-point energy

A one-dimensional harmonic trap has $V(x)=m\omega^2x^2/2$ and $E_n=(n+1/2)\hbar\omega$, $n=0,1,2,\ldots$. Adjacent levels are separated by $.0800\,\mathrm{eV}$. Use $\hbar=6.582\times10^{-16}\,\mathrm{eV\,s}$ and $m_e=9.11\times10^{-31}\,\mathrm{kg}$.

1. Infer $\omega$ and the ground-state energy.

2. Find the spring constant $K=m_e\omega^2$ and the energy of mode $n=3$.

Question 1 - Solution

An evenly spaced trap spectrum determines curvature and zero-point energy

(a) Method and calculation

$$\Delta E=\hbar\omega\Rightarrow\boxed{\omega=1.21544\times10^{14}\,\mathrm{s}^{-1}}.$$$$\boxed{E_0=\tfrac12\Delta E=.0400\,\mathrm{eV}}.$$

(b) Interpretation and further calculation

$$\boxed{K=m_e\omega^2=.0134581\,\mathrm{N/m}}.$$$$\boxed{E_3=\tfrac72\Delta E=.280\,\mathrm{eV}}.$$

Physical check

The mode index starts at zero for this spectrum, while the energy remains positive because of zero-point motion.