The superposition shown---both Yn1n2 and Yn1n2*Yn1n2---is an equal mix of the two states n1 and n2 for the infinite square well, Yn1n2(x,t)=(1/2)-1/2 [fn1 (x,t) + fn2 (x,t)]. The wave function evolves with time according to the TDSE. You may change state by choosing an n1 and n2. Time is shown in units of the revival time for the ground state wave function of a particle in an infinite square well. In other words it is the time for the wave function to undergo a phase change of 2p.
1) For the wave function to "revive" or return to it's original state- it takes 1 unit- meaning the same revival time for the ground state wave function of a particle in an infinite square well.
2)For the probability density to repeat it takes .33 units or 1/3 of the revival time for the grounds state wave function of a particle in an infinite square well.
3)As for thinking of a reason, I can't really. For the first one, my only guess is that the amount of time it takes for the wave function is 1 because the time for the wave function to "revive" for the ground state is 1. So, because wave function 1 will only return to it's original state after 1 unit of time, if you superimpose another wave function on top of this it also will not go back to it's original state until one unit of time is up.
As for the question 2, maybe it has something to do with frequency of probability changes= En1-En2/h, where En is dependent on n^2, So f is proportional to 3, which means the revival time is 1/3 of the one in the ground state.
1. 1 full time unit
2. 0.33 time units
3. well, the revival time of the wave function is 3x the probability fn repeat time -- perhaps this is because there are 3 n's total: one n=1 and one n=2. The prob density doesn't care which part of the wave function is where - just cares about the square of the magnitude.
3. Because the top diagram includes real and imaginary energies and it takes longer to go through all the stages whist the probability is squared and it repeats faster.