What remains unchanged after a process in Quantum Mechanics?

Question

Consider the following simple problem -

An electron is moving freely inside a one-dimensional infinite potential box with walls at x = 0 and x = a. If the electron is initially in the ground state (n = 1) of the box and if we suddenly quadruple the size of the box (that is, the right hand side of the wall is moved instantaneously from x = a to x = 4a), calculate the probability of finding the electron in the ground state of the new box.

In this question, the initial wavefunction of the particle is :

$$\psi(x)=\begin{cases} \sqrt{\frac{2}{a}}\sin\left(\frac{\pi x}{a}\right) & x\in[0,\,a] \\ 0 & \text{elsewhere} \end{cases} $$

and its energy should be

$$E=\frac{ℏ^2 π^2}{2m a^2}$$

Now the particle is in the new box. If we assume the energy to remain constant (and I think we should because energy should be conserved) then we have to change the wavefunction of the particle. The wavefunction in the new box corresponding to the energy $$E=\frac{ℏ^2 π^2}{2m a^2}$$

will be

$$\psi'(x)=\begin{cases} \sqrt{\frac{1}{2a}}\sin\left(\frac{\pi x}{a}\right) & x\in[0,\,4a] \\ 0 & \text{elsewhere} \end{cases} $$

Now if we find the probability for the particle to be in the new ground state, it comes out to be 0 since this new wavefunction is the 3rd excited state of the new 1D box.

But, if we had considered that the wave function would remain the same then we would get the answer as provided in the following post -

Frankie S. Palmer , Changing widths of potential wells

A detailed solution for the second method was found here - Check the answer to question 1 in this pdf file

As we can clearly see, in both cases the answer is quite different. So which approach is correct and why? Everywhere I have seen that they use the second approach to keep the wavefunction unchanged but why is that. Shouldn't energy be the quantity that remains unchanged?

Answer 1

Per the comment, a time dependent system does not conserve energy. In the initial well, you have eigenstates

$\psi_n(x)$, and the final well has different eigenstates $\psi'_n(x)$.

If you are in an initial state $\psi_n(x)$ and then the $V(x) \rightarrow V'(x)$ change happens, the wave function does not change, but it is no longer an eigenstate. Rather, it is in a super position of eigenstates:

$$ \psi_n(x) = \sum_{m=1}^{\infty} c_{nm}\psi'_m(x) $$

were the coefficients are found by projection:

$$ c_{nm} = \int_0^{4a}\psi^*_n(x)\psi'_m(x)dx $$

The probability of being in the final state with energy $E=E'_m$ is:

$$ p_{n\rightarrow m} = c^*_{nm}c_{nm} $$

Answer 2

Actually, energy should not be conserved. In the situation you are considering, you initially have a particle in box of side length $L$. Then, you increase the side length of the box, instantaneously, to $4L$. This mathematically corresponds to a time-dependent Hamiltonian $H(t)$. In particular, $$H(x, t) = \begin{cases} H_1(x) & t \leq t_{\text{change}} \\ H_2(x) & t > t_{\text{change}} \end{cases}$$ where $t_\text{change}$ is the time that the box's side length was instantaneously increased. Then, you have that $$\frac{\partial H(x,t)}{\partial t} \neq 0$$ so that $$\frac{d \langle H \rangle}{dt} = -\frac{i}{\hbar}\langle [H, H]\rangle + \biggl \langle \frac{\partial H(x,t)}{\partial t} \biggr \rangle = \biggl \langle \frac{\partial H(x,t)}{\partial t} \biggr \rangle \neq 0$$ which is the quantum mechanical statement that energy is not conserved.

The second method of solution is the so-called "sudden approximation" which reasons based off the physical heuristic that instantaneously changing the Hamiltonian does not give the state enough time to update, so the state immediately post-change is the same as the state immediately pre-change.