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Suppose a particle’s wavefunction satisfies the 1d time-independent Schrodinger equation with potential $U(x)$ and that its ground state is known to be $\psi_0(x)$.

The particle is in the state $\psi_0$ when at time $t = t_0$ the potential is suddenly changed from $U(x)$ to $V(x)$ (e.g., this could correspond to a potential well doubling in size). Suppose I know the stationary states of this new potential and their corresponding energy eigenvalues; call them $\phi_n$ and $E_n$ respectively (say).

What is the expression for the probability of finding a particle in the old ground state $\psi_0$ immediately after time $t=t_0$?

I know the new wavefunction is

$\phi(x,t) = \sum_{n=0}^\infty c_n\phi_n(x)e^{-iE_nt/\hbar}$ where the $c_n$'s are determined from the normalisation condition using Fourier.

Not sure how to take it from here. Any help appreciated.

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  • $\begingroup$ I found this example useful for getting some intuition about the above $\endgroup$
    – Vadim
    May 31 at 13:58
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The probability of the particle being in the ground state of $U$ at $t=t_0$ is $1$, and the evolution of the wave function is continuous. So in the limit $\lim_{\Delta t\rightarrow 0}t = t_0 + \Delta t$, the probability of finding it in the ground state of $U$ tends to $1$.

What is happening is that at $t=t_0$, the wave function is the ground state eigenfunction of the Schrödinger equation for potential $U$, but when the potential is changed to $V$, that will in general not be an eigenfunction of the new Schrödinger equation. So to get the time evolution for $t > t_0$, you have to write the old eigenfunction $\Psi(t_0, x) = \psi_0(x)$ as a linear combination of the new eigenfunctions such that $\Psi(t_0, x) = \sum_n c_n \phi_n(x)$, with $c_n = \langle\phi_n | \psi_0\rangle$, and then each of the $\phi_n$ time evolves with a phase factor of $e^{-iE_n (t - t_0)/\hbar}$, so for $t>t_0$ we get $\Psi(t, x) = \sum_n c_n\phi_n(x)\,e^{-iE_n (t - t_0)/\hbar}$. In the limit of $t\rightarrow t_0$ this obviously recovers $\psi_0(x)$ by construction.

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  • $\begingroup$ I see! Thank you:^) Do you mean $\psi_0(x) = \sum_{n} c_n\phi_n(x)$ though? (the eigenfunctions are stationary states i.e. functions of $x$ only) $\endgroup$
    – Vadim
    May 18 at 11:24
  • $\begingroup$ and is the probability just $|c_0|^2$ then? $\endgroup$
    – Vadim
    May 18 at 11:38
  • $\begingroup$ @Vadim Ah yes, abused the notation a bit too much. Should be clearer now. $\endgroup$
    – noah
    May 18 at 11:42
  • $\begingroup$ @Vadim You were asking about the probability of finding it in the old eigenstate $\psi_0$, that is just 1 as described in the answer. $|c_0|^2$ is the probability to find it in the new ground state $\phi_0$. $\endgroup$
    – noah
    May 18 at 11:43
  • $\begingroup$ thank you so much for elucidating, I get it now! $\endgroup$
    – Vadim
    May 18 at 11:45

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