More Electron Traps — Question 10

Question 10

A sudden increase of trap frequency excites the electron above the new ground state

An electron begins in the ground state of a harmonic trap of frequency $\omega$. The frequency suddenly changes to $\omega\prime=2\omega$, leaving the wave function unchanged at that instant. Initially $\langle K\rangle=\langle V\rangle=\hbar\omega/4$. For two oscillator ground states, use $|\langle0_{\omega\prime}|0_\omega\rangle|^2=2\sqrt{\omega\omega\prime}/(\omega+\omega\prime)$.

1. Find the mean energy immediately after the change and its excess over the new ground energy.

2. Find the probability of the new ground state and the total excited-state probability.

Question 10 - Solution

A sudden increase of trap frequency excites the electron above the new ground state

(a) Method and calculation

$$\langle H\prime\rangle=\langle K\rangle+4\langle V\rangle=\boxed{5\hbar\omega/4}.$$$$E_0\prime=\hbar\omega\prime/2=\hbar\omega\Rightarrow\boxed{\langle H\prime\rangle-E_0\prime=\hbar\omega/4}.$$

(b) Interpretation and further calculation

$$\boxed{P_0\prime=\frac{2\sqrt2}{3}=.942809}.$$$$\boxed{P_{\mathrm{excited}}=1-2\sqrt2/3=.0571910}.$$

Physical check

The frequency change is sudden, so adiabatic ground-state following does not apply. A high ground-state probability can coexist with a nonzero mean excitation energy.