Problem 4.10 (Engel)Part A. Are the eigenfunctions of H` for the particle in a one-dimensional box also eigenfunctions of the momentumoperator px`?SolutionStrategy. The eigenfunctions of H` are yHxL ="#####2ÅÅÅÅa sin@n p xÅÅÅÅÅÅÅÅÅÅÅaD. If these are also eigenfunctions of px`, they will satisfypx` y=const *y, so let's evaluate the left side of this equation and see what happens.Execution.px`=-i hbar∑ÅÅÅÅÅÅÅÅ∑xpx` y = - i hbar∑yÅÅÅÅÅÅÅÅÅ∑xThe derivative can be evaluated as follows:∂xikjjjjjjj$%%%%%%2ÄÄÄÄÄa SinAn p xÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄaEy{zzzzzzzè!!!2 J1aN3ê2n π CosAn π xaEThe result is a cosine function, which is obviously not a constant multiple of the original sine function. y is not aneigenfunction of px`.Part B. Calculate the average value of px for the case where n = 3 and 5, i.e., yHxL ="#####2ÅÅÅÅa sin@3 p xÅÅÅÅÅÅÅÅÅÅÅaD or "#####2ÅÅÅÅa sin@5 p xÅÅÅÅÅÅÅÅÅÅÅaDSolutionStrategy. The average value of px is also called the expectation value, <px>, and is obtained as the value of thefollowing integral:<x> = Ÿ0ay px`y „ xNote: this integral relies on these facts: 1) y is a real function, and 2) y has been normalized.Execution.n = 83, 5<ψ=$%%%%%%2a SinAn π xaE83, 5<9è!!!2$%%%%%%1aSinA3 π xaE,è!!!2$%%%%%%1aSinA5 π xaE=pψ= H− hbarL ∂xψ9−3 è!!!2 J1aN3ê2hbar π CosA3 π xaE, −5 è!!!2 J1aN3ê2hbar π CosA5 π xaE=ψpψ=ψ∗pψ9−6  hbar π Cos@3 π xaDSin@3πxaDa2, −10  hbar π Cos@5πxaDSin@5 π xaDa2=‡0aψpψ x80, 0<The average momentum of the particle for both the n = 3 and n = 5 states is zero. You may have noticed that both ofthe expressions that needed to be integrated contained cos*sin and was an odd function, so the integrals were guaran-teed to vanish.Comment. The text would like us to generalize this result to the following: the average momentum of the particle in abox is zero regardless of n. You might be reluctant to make this generalization, however, because the two examplesthat were chosen both involved odd values of n. To guarantee the result, we calculate the average value for an arbitrarycase of n = m2P4.10.nbψ=$%%%%%%2a SinAm π xaEè!!!2$%%%%%%1aSinAm π xaEpψ= H− hbarL ∂xψ−è!!!2 J1aN3ê2hbar m π CosAm π xaEψpψ=ψ∗pψ−2  hbar m π Cos@m π xaDSin@mπxaDa2‡0aψpψ x− hbar Sin@m πD2aSimplify@%, m ∈