Change in potential of quantum harmonic oscillator

Question

We know that for the 1D time-independent SE with potential $V(x)=\frac{1}{2} m \omega^2 x^2$, the solutions have energies $E_n = (n + \frac{1}{2}) \hbar \omega$. I've been attempting the following question:

Suppose a particle of charge $q$ and mass $m$ is subjected to the potential $V_0 = x^2 / 2$, and at time $t = t_0$ a constant electric field is turned on to produce an additional potential $V_1 = -qEx$. If the wave function just before is the ground state $$\psi_0 = \left( \frac{1}{\pi \hbar} \right)^{1/4} e^{-x^2 / (2\hbar)},$$ what are the energy levels of the new potential? Show that the probability of obtaining the ground state immediately afterwards is $e^{-q^2 E^2 / (2\hbar)}$.

My main issue here is that the 'show that' part of the question implies the wave function immediately changes to become a linear combination of the eigenstates of the new potential $V_0 +V_1$. Hence I'm not sure how to write down a time-independent SE which encodes the 'initial state' $\psi_0$ and its energy $\frac{1}{2} \hbar$ - no examples of this were given in my QM course.

Answer 1

There's a neat trick to this one. Note that the new (full) potential $\tilde V$ is given by $$\tilde V = V_0 + V_1 = \frac{1}{2} \left( x^2 - 2qEx \right) = \frac{1}{2} \left( (x-qE)^2 - (qE)^2 \right)$$ This is just a shift $x \mapsto x - qE$ (including an overall shift in energy levels), so we can immediately write down our new ground state $\tilde \psi_0$ (if you like, think of a change of variables $u = x-qE$ giving a SE of the same form) $$\tilde \psi_0(x) = \psi_0(x - qE) = \left( \frac{1}{\pi \hbar} \right)^{1/4} \exp\left( -\frac{(x-qE)^2}{2 \hbar} \right)$$ We're interested in the probability of observing $\tilde \psi_0$ given that our state is still in $\psi_0$ (since it hasn't evolved - we're "immediately" after the potential has changed). Our desired probability amplitude is then $$A = (\tilde \psi_0, \psi_0) = \frac{1}{\sqrt{\pi \hbar}} \int_{-\infty}^\infty \exp \left( -\frac{(x-qE)^2+x^2}{2 \hbar} \right) \; dx$$ We expand and re-complete the square for the numerator in the exponential, $$(x-qE)^2+x^2 = 2 \left(x^2 - xqE + \frac{(qE)^2}{2}\right) = 2 \left[ \left(x - \frac{1}{2} qE\right)^2 + \frac{(qE)^2}{4} \right]$$ So upon substituting $u = x - \frac{1}{2} qE$, we get \begin{align*} A &= \frac{1}{\sqrt{\pi \hbar}} e^{-(qE)^2/(4 \hbar)} \underbrace{\int_{-\infty}^\infty e^{-u^2/\hbar} \; du}_{\sqrt{\pi \hbar}} \\ &= e^{-(qE)^2/(4 \hbar)} \end{align*} So our probability $P$ is given by $$P = \lvert A \rvert^2 = e^{-(qE)^2/(2 \hbar)}$$