Physics 202H - Introductory Quantum Physics I Homework #08 - Solutions Fall 2004 Due 5:01 PM, Monday 2004/11/15

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1 Physics H - Introductory Quntum Physics I Homework #8 - Solutions Fll 4 Due 5:1 PM, Mondy 4/11/15 [55 points totl] Journl questions. Briefly shre your thoughts on the following questions: Of the mteril tht hs been covered in the course up to the mid term test, wht hs been the most difficult for you to understnd? Wht mteril hs been the most interesting? Wht mteril hs been the most surprising? Is there ny mteril tht you thought you understood before this course tht you now hve drsticlly different understnding of? Wht ws is nd wht hs chnged? Any comments bout this week s ctivities? Course content? Assignment? Lb? 1. Plese complete the nonymous mid-course survey online on WebCT. Erly feedbck will hopefully llow us to hve the best possible course this semester rther thn just hving next yer s students benefit. In ddition to the bonus ssignment mrks, survey prticiption my count towrds overll clss prticiption scores. [5.1-bonus] Solution: Do the survey - get the bonus mrks.. (From Eisberg & Resnick, Q 5-15, pg 168 Wht is the bsic connection between the properties of wve function nd the behviour of the ssocited prticle? Limit your discussion to bout 5 words or so. [1] Solution: The wve function determines the probbility of finding the ssocited prticle in given region during given time period. More specificlly, the probbility density is given by P (x, t Ψ (x, tψ(x, t. Not only does the wve function determine the probbility density, but llows us to clculte the expecttion vlue of ny dynmicl function of position, time, nd momentum of the ssocited prticle (s long s we use the momentum opertor in our clcultion, thus we cn clculte x, x, p, etc. From Eisberg & Resnick: The wve function contins ll the informtion tht the uncertinty principle will llow us to lern bout the ssocited prticle.

2 3. (From Eisberg & Resnick, P 5-9, 5-1, 5-11, 5-1, pg 168 ( Following the procedure of Exmple 5-9, verify tht the wve function (, x < /, πx Ψ (x, t e iet/, / < x < +/,, +/ < x. is solution to the Schroedinger eqution in the region / < x < +/ for prticle which moves freely through the region but which is strictly confined to it nd determine the vlue of the totl energy E of the prticle in this excited stte of the system, nd compre with the totl energy of the ground stte E 1 found in Exmple 5-9. [1] Solution: The Schroedinger eqution is Ψ(x, t Ψ(x, t m x + V (x, tψ(x, t i. For the region between x > 1/ nd x < +/, the potentil energy is zero V (x, t, so the Schroedinger eqution reduces to Ψ(x, t Ψ(x, t m x i Ψ(x, t x im Ψ(x, t Ψ(x, t Ψ(x, t x ik, k Ψ (x, t x A π cos ( A 4π πx sin A ie sin m. The vrious derrivtives of Ψ (x, t in the region / < x < +/ re: e ie t/ Ψ (x, t x Ψ (x, t e ie t/ e ie t/ (.1 (.3 (.4 Solving (.3 nd (.4 for like terms nd setting them equl to ech other gives us ( Ψ (x, t πx 4π x e iet/ Ψ (x, t ie Ψ (x, t x i 4π Ψ (x, t E (.5 Compring (.1 nd (.5, we see tht Ψ (x, t is solution to the Schroedinger eqution in the region / < x < +/ provided tht m 4π E E 4π m π m h m. (.6 Thus the given wve function Ψ (x, t is solution to the Schroedinger eqution for E (h /(m. Eqution (.6 shows us tht E (4π /(m, while Exmple 5-9 gve us E 1 (π /(m, thus E 4E 1.

3 (b Plot the spce dependence of this wve function Ψ (x, t. Compre with the ground stte wve function Ψ 1 (x, t of Figure 5-7, nd give qulittive rgument relting the difference in the two wve functions to the difference in the totl energies of the two sttes. [5] Solution: The wve function of the ground stte Ψ 1 (x, t hs spce dependence which is one hlf of complete sin cycle. In figure 1 we hve plotted the normlized wve functions, nticipting the result of the next problem, with 1. We used Mple with the following commnds to generte the plot. > restrt; with(plots; > plot([(sqrt(*cos(pi*x,(sqrt(*sin(*pi*x], x-(.5..(.5, colour[nvy, blue], legend["psi_1", "psi_"], linestyle[3,]; 1.5 x psi_1 psi_ Figure 1: ψ 1 (x nd ψ (x with 1 As we cn see in figure 1, the wve function of the excited stte Ψ (x, t hs spce dependence which is complete sin cycle. Thus the wvelength of Ψ 1 (x, t is twice tht of Ψ (x, t, which s we expect, corresponds with greter energy for Ψ (x, t compred with Ψ 1 (x, t. Ψ (x, t lso hs more shrply curved shpe, lso corresponding to hving greter energy.

4 (c Normlize the wve function Ψ (x, t bove by djusting the vlue of the multiplictive constnt A so tht the totl probbility of finding the ssocited prticle somewhere in the region of length equls one. Compre with the vlue of A obtined in Exmple 5-1 by normlizing the ground stte wve function Ψ 1 (x, t. Discuss the comprison. [1] Solution: To normlize wve function we integrte P (x, t Ψ (x, tψ(x, t over ll vlues of x, nd set tht equl to one. For the wve function in question, it hs vlue of zero for ny x outside of x < / so: 1 + +/ / A +/ A π 1 A P (x, t / sin A +/ / e iet/ sin u du A π. Ψ (x, tψ (x, t e iet/, du π [ ] u sin (u +π 4 A [ π π + π ] With this vlue for A, the complete normlized wve function is, x < /, ( πx Ψ (x, t sin e iet/, / < x < +/,, +/ < x. The normliztion for Ψ 1 (x, t is identicl with the normliztion for Ψ (x, t, in ech cse A /., x < /, Ψ 1 (x, t ( πx cos e iet/, / < x < +/,, +/ < x. (d Clculte the expecttion vlue of x, the expecttion vlue of x, the expecttion vlue of p, nd the expecttion vlue of p for the prticle ssocited with the wve function Ψ (x, t bove. [1] Solution: The expecttion vlue of x is clculted by integrting Ψ xψ nd the expecttion vlue of x is clculted by integrting Ψ x Ψ. x +/ / +/ / +/ / Ψ (x, txψ(x, t x sin even e iet/ x odd {}}{{}}{ π } u {{ sin u} du odd x. e iet/, du π

5 x +/ / +/ / +/ / Ψ (x, tx Ψ(x, t x sin e iet/ x u} sin {{ u} du π 3 even e iet/, du π u sin u du We cn use the trig identity sin u (1 cos (u/ nd the integrl formul u cos (u du (u/ cos (u + (u / 1/4 sin (u to simplify things little bit: x π 3 [ u 3 1 cos (u u du (u u cos (u du 3 u [ ] π 3 3 π x ( ( u cos (u 1 4 ( π ] +π sin (u To clculte the expecttion vlue for momentum, we need to use the momentum opertor i ( / x. p +/ / +/ i π p. / +/ ( Ψ (x, t i x Ψ(x, t ( ( πx e iet/ i x ( i π πx sin cos / odd +π even {}}{{}}{} sin u{{ cos u} du odd e iet/, du π

6 To clculte the expecttion vlue for p we need to use ( / x. p +/ / +/ / +/ 8π Ψ (x, t ( x / +/ 3 4π Ψ(x, t e ( iet/ x ( ( 4π πx sin / p 4π sin sin u du 4π ( π h, du π [ u sin (u 4 ] +π e iet/ 4π [ π + π ] The expecttion vlue of x is zero, nd the expecttion vlue of x is bout (.77. The expecttion vlue of p is zero, nd the expecttion vlue of p is h /. Note tht the squre root of the expecttion vlue of p, nmely p h/ (clled the root-mensqure vlue of p is mesure of the fluctutions round the verge p. This RMS vlue is lso mesure of the uncertinty in p.

7 4. (From Eisberg & Resnick, P 5-16, pg 169 Show by direct substitution into the Schroedinger eqution tht the wve function Ψ(x, t ψ(xe iet/ stisfies tht eqution if the eigenfunction ψ(x stisfies the time-independent Schroedinger eqution for potentil V (x. [1] Solution: The Schroedinger eqution is Ψ(x, t Ψ(x, t m x + V (x, tψ(x, t i. (4.1 If we mke the substitution Ψ(x, t ψ(xe iet/ nd V (x, t V (x we hve [ ψ(xe iet/ ] m d ψ(x x + V (x, t [ ψ(xe iet/ ] i [ ψ(xe iet/ ] m e iet/ + V (x, tψ(xe iet/ i ψ(x de iet/ dt d ψ(x m e iet/ + V (x, tψ(xe iet/ i ψ(x ie e iet/ ( d ψ(x m + V (x, tψ(x e iet/ (Eψ(x e iet/. (4. We cn divide (4. through by e iet/ to get d ψ(x m + V (x, tψ(x Eψ(x. (4.3 If ψ(x stisfies (4.3, then working upwrds from line (4.3 we hve tht Ψ(x, t stisfies the Schroedinger eqution (4.1. Eqution (4.3 is the time-independent Schroedinger eqution for potentil V (x, nd we were given tht ψ(x is solution to the time-independent Schroedinger eqution for potentil V (x, therefore, Ψ(x, t is solution to the Schroedinger eqution. Hedstrt for next week, Week 9, strting Mondy 4/11/15: Red Chpter 5 Schroedinger s Theory of Quntum Mechnics in Eisberg & Resnick Section 5.6 Required properties of Eigenfunctions Section 5.7 Energy Quntiztion in the Schroedinger Theory Section 5.8 Summry