More Electron Traps — Question 4

Question 4

A constant force shifts a harmonic trap without changing its level spacing

An electron is modeled by $H=p^2/(2m)+Kx^2/2-Fx$, with positive K and constant force F. Set $\omega=\sqrt{K/m}$ and retain the original energy zero.

1. Complete the square and find the shifted equilibrium position.

2. Find all energy levels and the change in ground-state energy.

Question 4 - Solution

A constant force shifts a harmonic trap without changing its level spacing

(a) Method and calculation

$$\frac12Kx^2-Fx=\frac12K\left(x-\frac FK\right)^2-\frac{F^2}{2K}.$$$$\boxed{x_c=F/K}.$$

(b) Interpretation and further calculation

$$\boxed{E_n(F)=(n+\tfrac12)\hbar\omega-F^2/(2K)}.$$$$\boxed{E_{n+1}-E_n=\hbar\omega},\quad\boxed{E_0(F)-E_0(0)=-F^2/(2K)}.$$

Physical check

The force translates the wave functions and lowers the minimum of the potential. The curvature, and hence spacing, stays fixed.