A particle is in its ground state of an infinite square well of width a <xl i>=√2/a*sin(πx/a) and since it's an eigenstate of the Hamiltonian it will evolve as <xlα(t)>=√2/a*sin(πx/a)e^(-iE1t/ħ) where E=π2ħ2/2ma2 If the well now suddenly expands to witdh 2a If the well suddenly expands to 2a the particle's state remains the same and the wavefunction is given by <xli>=2/a*sin(πx/a) if 0 < x< a and 0 for x>a which is not the ground state of the new Hamiltonian i.e. √1/a*sin(πx/2a) compute <xlα(t)> for this new system. Is there a time-dependende and/or conservation laws? Okay so this is a really hard question for me, I know how to solve the infinite square well for the first case but I don't really have any intuition when it comes to the second part of this problem. My initial thought was to calculate <xlα(t)> using the new Hamiltonian stated above and the time dependence e^(-iE1t/ħ) but this I feel like is way too easy. Then I read somewhere that the wavefunction is the sum over all states and so I thought that maybe I should sum over the two states (the old and new Hamiltonian) times the time-dependence but it feels wrong aswell.. Could anyone give me a push in the right direction with this problem? I'm thankful for any answers.