# Probability of finding particle in ground state of old potential immediately after potential is changed?

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.

• I found this example useful for getting some intuition about the above May 31 '21 at 13:58

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.

• 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) May 18 '21 at 11:24
• and is the probability just $|c_0|^2$ then? May 18 '21 at 11:38
• @Vadim Ah yes, abused the notation a bit too much. Should be clearer now.
– noah
May 18 '21 at 11:42
• @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 $\phi_0$.
– noah
May 18 '21 at 11:43
• thank you so much for elucidating, I get it now! May 18 '21 at 11:45