Clear@n, m, aD<x> ∫ xn because wave function is not eigenfunction of x`Recall (Ch. 3) that x`= x. Following shows x` y ∫ const * yxikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzzè!!!2$%%%%%%1ax SinAn π xaECan use this result to calculate <x> (since wave function is normalized, we only need to calculate the numerator,Ÿynx`yn „ x)ikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz xikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz2 x Sin@n π xaD2a‡0a2 x Sin@n π xaD2a  x−a H−1 − 2n2π2+ Cos@2nπD+ 2nπ Sin@2nπDL4n2π2SimplifyA−a H−1 − 2n2π2+ Cos@2nπD+ 2nπ Sin@2nπDL4n2π2, 8n ∈ Integers<Ea2Average position of particle is in middle of box for all quantum states.Recall that P(a/2) dx, the probability of finding particle near middle of box, changes with quantum state. Using P(a/2) dx =yHaê 2L2 dx, I find:ikjjjjjjj$%%%%%%2a SinAn π Ha ê2LaEy{zzzzzzz ikjjjjjjj$%%%%%%2a SinAn π Ha ê2LaEy{zzzzzzz2 Sin@n π2D2aA E p x averages wrapup.nb 1PlotA2 SinAnπ2E2, 8n, 1, 6<E2 3 4560.511.52 Graphics P(a/2) oscillates between large value & 0.How do the probabilities of finding particle between 0Ø0.05a and 0.45aØ0.5a behave?P05 =‡00.05 aikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz ikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz  xP50 =‡0.45 a0.5 aikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz ikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz  x0.05 −0.159155 [email protected] nDn1n H0.05 n + 0.159155 [email protected] nD− 0.159155 [email protected] nDLPlot@8P05, P50<, 8n, 1, 50<, PlotStyle → [email protected], [email protected]<D10 20 30 40 500.020.040.060.080.1 Graphics Both probabilities converge on 0.05 as n grows!A E p x averages wrapup.nb 2RecapMeasurement produces eigenvalue of A` if made only once- does not require y to be eigenfunction of A`- if y = e-fn, then observe e-value of this e-fn, but...- if y ∫ e-fn, then result is unpredictable, but can obtain rel. probabilities of different results by expanding y aslinear combination of e-fnsMeasurement produces expectation value <a> if repeated on many identical systems- does not require y to be eigenfunction of A`- if y = e-fn, then <a> = e-value of this e-fn- if y ∫ e-fn, then result obtained from Ÿy A` y „ xMost probable location x of finding particle ∫ <x>- <x> = Ÿy x` y„x- max(P(x)) = max( y2 )A E p x averages wrapup.nb