In[143]:=Clear@n, m, aDE p x averages.nb 1Particle in a 1-D box (con't)What are expectation (average) values <E>, <p>, and <x> when system is in stationary state yn(x)?1. <a> = ŸynA`yn „xÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅŸynyn„x so must know corresponding operators A`2. <a> = eigenvalue of A` only if yn = eigenfunction of A`3. <a> ∫ eigenvalue of A` if yn ∫ eigenfunction of A`Rationale: If yn ∫ eigenfunction of A`, then yn can be written as a linear combination of mutually orthogonaleigenfunctions (see orthogonality & completeness theorems of Chapter 2). Integral shown in #1 above can be expanded interms of these eigenfunctions, and leads to a weighted average of different eigenvalues rather than a single eigenvalue.Suppose yn = ⁄i=1¶ci fi where A`fi= aifi and ci= constantsThen<a> = ŸH⁄i=1¶ci fiL A`H⁄j=1¶cj fjL „xÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅŸH⁄l=1¶cl flLH⁄m=1¶cm fmL„xWorking on numerator only:= ‡H⁄i=1¶ci fiLJ‚j=1¶A`cj fjN„ x = ŸH⁄i=1¶ci fiL H⁄j=1¶ajcj fjL= ⁄i=1¶⁄j=1¶ajcicj Ÿfifj „ x = ⁄i=1¶aici2The final summation assumes the fi form an orthonormal set of functions (see Ch. 2)Working on denominator gives same result without the ai= ŸH⁄l=1¶cl flLH⁄m=1¶cm fmL„ x = ⁄l=1¶cl2Combining it all gives:<a> = ⁄i=1¶aici2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ⁄l=1¶cl2= weighted average of aiThis formula gives us a second definition of the expectation value that is consistent with the one given above. Thefirst definition is more useful when we have a system's wave function and we know the relevant operator. The seconddefinition is more useful when we know the eigenfunctions and eigenvalues of the operator and we know how to expandthe wave function as a linear combination of eigenfunctions. In fact, we rarely have all of this information, so we mostlyuse the second definition to prove various theorems about expectation values.E p x averages.nb 2<E> = En because wave function is eigenfunction of H`This just repeats what we stated above. Since yn is an eigenfunction of H` the definition of <E> is also the formula forcalculating En.<p> ∫ pn because wave function is not eigenfunction of p`Recall (Ch. 3) that p`=-i hbar∑ÅÅÅÅÅÅÅÅ∑x. Following shows p` y ∫ const * yIn[144]:=− ∗ hbar ∗∂xikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzzOut[144]=−è!!!2 J1aN3ê2hbar n π CosAn π xaECan use this result to calculate <p> (since wave function is normalized, we only need to calculate the numerator,Ÿynp`yn „ x)In[159]:=ikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz H− ∗ hbarL∗∂xikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzzOut[159]=−2  hbar n π Cos@n π xaDSin@n π xaDa2In[146]:=‡0a−2  hbar n π Cos@nπxaDSin@nπxaDa2  xOut[146]=− hbar Sin@n πD2aIn[147]:=SimplifyA− hbar Sin@n πD2a, 8n ∈ Integers<EOut[147]=0Average value of momentum = 0, but average value is not eigenvalue of momentum. The particle may not actually be atrest (recall KE ∫ 0), and a single measurement of momentum on one system may not give p = 0.E p x averages.nb 3<x> ∫ xn because wave function is not eigenfunction of x`Recall (Ch. 3) that x`= x. Following shows x` y ∫ const * yIn[148]:=xikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzzOut[148]=è!!!2$%%%%%%1ax SinAn π xaECan use this result to calculate <x> (since wave function is normalized, we only need to calculate the numerator,Ÿynx`yn „ x)In[149]:=ikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzz xikjjjjjjj$%%%%%%2a SinAn π xaEy{zzzzzzzOut[149]=2 x Sin@n π xaD2aIn[150]:=‡0a2 x Sin@n π xaD2a  xOut[150]=−a H−1 − 2n2π2+ Cos@2nπD+ 2nπ Sin@2nπDL4n2π2In[151]:=SimplifyA−a H−1 − 2n2π2+ Cos@2nπD+ 2nπ Sin@2nπDL4n2π2, 8n ∈ Integers<EOut[151]=a2Average position of particle is in middle of box for all quantum states.Recall that P(a/2) dx, the probability of finding particle near middle of box, changes with quantum state. Using P(a/2) dx =yHaê2L2 dx, I find:In[152]:=ikjjjjjjj$%%%%%%2a SinAn π Ha ê2LaEy{zzzzzzz ikjjjjjjj$%%%%%%2a