What is the probability to find the system in the ground state? [closed]

Question

I previously posted a question related to this Hamiltonian, but the original concern was different:

We examine the following Hamiltonian:

\begin{equation} H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2 - \theta(t)qEx \end{equation}

In this case, $\theta(t)$ equals 0 for $t < 0$ and 1 for $t \geq 0$. For $t < 0$, we find the system in the ground state. The question at hand is: What is the probability of the system being in the ground state for $t > 0$? The given integral is useful:

\begin{equation} \int_{-\infty}^{\infty} \exp\left(-ax^2\right) \exp\left(-a(x-y)^2\right)dx = \sqrt{\frac{\pi}{2a}}\cdot \exp\left(-ay^2/2 \right) \end{equation}

I am aware that the probability can be computed as

\begin{equation} p_n = |\langle\Psi_{\mathrm{old}}|\Psi_{\mathrm{new}}\rangle|^2. \end{equation}

Nonetheless, obtaining a definitive probability seems elusive. I can rework the Hamiltonian to derive the equation for the new eigenfunctions, which are merely shifted by a constant. Consequently, they are given by

\begin{equation} \Psi_{\mathrm{new}}(x) = \Psi_{\mathrm{old}}\left(x - \frac{qE}{m\omega^2}\right). \end{equation}

It seems logical to apply them in the calculation of probability, as inferred from the integral where the argument contains $(x-y)^2$. However, I discovered that for $t > 0$, the state doesn't transition to the new ground state, implying I cannot simply calculate the integral of both ground states. The new state can be defined by a linear combination of the eigenfunctions from the new Hamiltonian. However, this implies that I would need the constant factors. Given that this was a part of an examination question, I am pondering over how to identify these factors or avoid the need for them altogether.

Answer 1

Being careful about normalization, the ground state of the Harmonic oscillator is $$ \Psi_{HO}(x) = \left(\frac{2a}{\pi}\right)^{\frac{1}{4}} e^{- a x^2} $$ where $a=\frac{m \omega}{2 \hbar}$. In your notation, you then have $$ \Psi_\mathrm{old}(x) = \Psi_{HO}(x) \quad \quad \Psi_\mathrm{new}(x) = \Psi_{HO}\left(x-\frac{q E}{m \omega^2}\right) $$

where $\Psi_{old}$ is the previous ground state, and $\Psi_{new}$ is the new ground state. Expanding in the eigenbasis $\psi_k(x)$, of the $t>0$ Hamiltonian, we have $$ \Psi_\mathrm{old}(x) = \sum_{k=0}^\infty c_k \psi_k$$ where $c_k = \langle \psi_k, \Psi_\mathrm{old}(x) \rangle = \int_{-\infty}^\infty \psi^*_k(x) \Psi_\mathrm{old}(x)$ So if the state is $\Psi_{old}$ at $t=0$, we solve for the time evolution and get $$\Psi(x,t) = \sum_{k=0}^\infty c_k \psi_k(x) e^{i \frac{E_k}{\hbar} t}, \quad t>0 $$

If you want the probability for this state to still be in the old ground state, this is given by

$$ |\langle \Psi_{old} | \Psi(t) \rangle|^2 = |\sum_{k=0}^\infty |c_k|^2 e^{i \frac{E_k}{\hbar} t}|^2 $$

This seems hard to compute, because $c_k$ is nontrivial in general. However, if you're interested in the probability for the wave function to be measured in the new ground state $\Psi_{new} = \psi_0$, this is given by

$$ |\langle \Psi_{new} | \Psi(t) \rangle|^2 = |c_0|^2$$ and remember $$c_0 = \langle \psi_0 | \Psi_{old} \rangle = \langle \Psi_{new} | \Psi_{old} \rangle = \sqrt{\frac{2 a}{\pi}} \int_{-\infty}^\infty e^{- a x^2} e^{- a \left(x-\frac{qE}{m\omega^2}\right)} dx = e^{-a y^2} $$ so that the proba is $$ |\langle \Psi_{new} | \Psi(t) \rangle|^2 = e^{-2 a y^2} = \exp\left(-\frac{m \omega}{\hbar} \frac{q^2 E^2}{m^2 \omega^4}\right) = \exp \left(- \frac{q^2 E^2}{m \omega^3 \hbar} \right) $$