Question

a, b & c ONLY (already calculated the probability)

1. A particle, initially (t +-00) in its ground state in an infinite potential well whose walls are located at x = 0 and x = a, is subject at time t = 0 to a time-dependent perturbation Û (x, t) = € îe-“, where e is a small real number. Calculate the probability that the par- ticle will be found in its first excited state after a sufficiently long time (t + +oo). Write down an expression for the probability as function of time. a) Explain with words the meaning of each factor that involves the perturbation (spatial and time parts). What does their respective magnitude depend on? b) Is it possible for any of them to be zero? Why or why not? c) How would the result of this problem change if the walls of the potential were at x=- a/2 and x=a/2?

here is the probability

Answer 1

The probability from ground state to est excited state ने Pita = 11 S < 424 v It) 14. Et dp/? and ₂ = la sin 272 Where * = ha sin <424V (t) t = agelse 2 TX a sin sin x da Ee-t2 alcos TM 3 ITN COB ST) da Eeth CB TA da - J 3R a los a a & e-t2 a. sin + Ärz dos an sin 31ta 2 3172 BIT Cos 912 - 16€a 1 e-tz 972 Pier 16ça t2 gra int-t2 dt e This is probability a function of time Porta 16 Ea 2 972 (ws 3 a "168a 2mar 2 92A 9t 4+2 8mra 4

It is not possible to become any term to Zerw, reno, then Because If any term is the probability will be zerw. But there is pockbilim the term to which is in exponential be Zeno term. Because But in exponential term, there is term which can be zero. ho 11 22 indt 2 Pita ** <**141291417 Eet 2 sinnt & sin the de (los cos 37 da -/ 2 & e-t292 y la 32 n -012 a/2 11 {ert2 c42 [Ia ceste dom - S al2 & Cos 3th -012 din ] [ as the integral function in odd] so, Probability is zero.