Problem 4.7 (Engel)Use the eigenfunction yHxL = A' ‰+Âkx+ B' ‰-Âkxto apply the boundary conditions for the particle in a box.Part A. How do the boundary conditions restrict the acceptable choices for A', B' and k?SolutionExecution.Both exponentials equal 1 at x = 0 but the boundary condition requires yH0L = 0.yH0L = 0 = A' + B'Which leads to B' =-A', and simplifies yHxL as follows:yHxL = A' H‰+Âkx-‰-ÂkxLThe second boundary condition yHaL = 0 leads to:yHaL = 0 = A' H‰+Âka-‰-ÂkaL‰+Âka=‰-ÂkaThis says that k must be chosen so that ‰+Âka is real (then it will equal its complex conjugate). In other words, sin(k a)= 0. This will happen ifka = np where n = integerork =n pÅÅÅÅÅÅÅÅayHxL = A' H‰+Ân p xêa-‰-Ân p xêaLPart B.Do this eigenfunction, and the traditional A sinHnpxêaL, give different probability densities when each function isnormalized?SolutionExecution.First we determine the values of the normalization constants, A and A'. P4.6 showed us that A ="#####2ÅÅÅÅa. The probability density for this eigenfunction is:= @A sinHn p xÄÄÄÄÄÄÄÄÄÄÄÄaLD2=2ÄÄÄÄa@sinHn p xÄÄÄÄÄÄÄÄÄÄÄÄaLD2Now working on A'yHxL = A' H‰+Ân p xêa-‰-Ân p xêaLy*HxL = A' H‰-Ân p xêa-‰+Ân p xêaLy*HxL yHxL = HA'L2 H‰+Ân p xêa-‰-Ân p xêaL H‰-Ân p xêa-‰+Ân p xêaLFullSimplify@H nπxêa−−nπxêaL H−nπxêa−nπxêaLD4 SinAn π xaE2y*HxL yHxL = HA'L2 4ASinJnpxÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅaNE24 ‡0aSinAn p xÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄaE2 ‚ xa J2 −Sin@2nπDn πNSimplify@%, n ∈ IntegersD2aThe normalization integral equals one, so A' =1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅè!!!!!!!2 aThe probability density becomes:y*HxL yHxL = I1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅè!!!!!!!2 aM2 4@SinHn p xÅÅÅÅÅÅÅÅÅÅÅaLD2=2ÅÅÅÅa@SinHn p xÅÅÅÅÅÅÅÅÅÅÅaLD2This result is identical to the previous one, so both ways of writing the eigenfunction are