I'm having a little trouble figuring out how to start this. The question is as follows. At time $t<0$ a hydrogen atom is in the rest frame $\mathfrak{R}$. Then at time $t=0$ the atom suddenly starts moving at a speed $V$ which is non-relativistic. It is now described in the rest frame $\mathfrak{R}'$ under the transformation $$t' = t, \, \, \, \vec{r}' = \vec{r}-\vec{V}t$$ The question asks to find the probability that at time $t>0$ in rest frame $\mathfrak{R}'$ that the electron in the hydrogen atom is still in it's ground state.

I know so far that this is a time dependent perturbation theory. I also know the probability can be computed by using $$|C_m^{(1)}(t)|^2$$ where $$C_m^{(1)}(t) = -\frac{i}{\hbar}\int_0^t dt e^{i\omega_{m,i}t}H'_{m,i}(t)$$ and $$\omega_{m,i}= \frac{E_m-E_i}{\hbar}, \, \, \, H'_{m,i} = \langle\psi_m^0|\hat{H}'(t)|\psi_i^0\rangle$$ What I'm not clear on is what $E_m$, $\psi_m^0$ and $\hat{H}'(t)$ should be. Would these just be the Galilean transformations of $E_i$, $\psi_i^0$ and $\hat{H}(t)$ for a hydrogen atom, or is there something more subtle I'm missing here?

  • $\begingroup$ The atom starts to move suddenly, the perturbation will have no effect on the wavefunction (the so-called "sudden approximation applies"). The electron will then remain in the original 1s eigenstate of the Hamiltonian, which is no longer the 1s state of the new Hamiltonian. From the point of view of the proton after it starts to move, the electron suddenly acquired a momentum $\vec{p} = -m_e \vec{v}$, this means that its wavefunction is the new 1s state multiplied by $\exp(i\vec{k}\cdot{\vec{r}})$ with $\vec{k} = \frac{\vec{p}}{\hbar}$. $\endgroup$ – Count Iblis Apr 30 '17 at 22:34

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