LULEÅ UNIVERSITY OF TECHNOLOGY Division of Physics Solution to written exm in Quntum Physics F0047T Exmintion dte: 06-03-5 The solutions re just suggestions. They my contin severl lterntive routes.. Sme/similr s problem 4.4 in rnsden & Jochin. In the region where the potentil is zero x < 0 the solutions re of the trveling wve form e ikx nd e ikx, where k me/ h. A plne wve ψx Ae ikx ωt describes prticle moving from x towrds x. The probbility current ssocited with this plne wve is j h A mi e ikx x e+ikx e +ikx x e ikx A h k A m v A plne wve ψx e i kx ωt describes prticle moving the opposite direction from x towrds x. The probbility current ssocited with this plne wve is j h mi e +ikx x e ikx e ikx x e+ikx h k m v Solution for the region x > 0 where the potentil is V 0 4.5eV. The potentil step is lrger thn the kinetic energy.0 ev of the incident bem. The prticle my therefore not enter this region clssiclly. It will be totlly reflected. In quntum mechnics we perform the following clcultion: The two solutions for the two regions re: Ψx Ae ikx + e ikx for x < 0 where k me/ h Ce κx + De κx for x > 0 where κ mv 0 E/ h we cn put C 0 s this prt of the solution would diverge, nd is hence not physicl, s x pproches. At x 0 both the wvefunction nd its derivtive hve to be continous functions, s the potentil is everywhere finite. The derivtive is: Ψx x Aike ikx ike ikx Dκe κx At x 0 we rrive t the following two equtions: A + D iak ik Dκ D k A k+κ A k iκ k+iκ i A D A +i V 0 /E +i V 0 /E V 0 /E We cn now clculte the coeficient of reflection, R The coeficients represent the following mplitudes: A is the incident bem, is the reflected bem nd C is the trnsmitted bem. The ssocited probbility currents re denoted j A, j nd j C. Conservtion yields j A j + j C. Hence we cn define the coeficient of reflection s the frction of reflected flux R j j A nd the coeficient of trnsmission s T j C j A R j j A k A k This is esily seen from the rtio /A being the rtio of two complex number where one is the complex conjugte of the other nd therefore hving the sme bsolute vlue. Imiditely follows tht T 0 s the currents hve to be conserved.
b+c Solution for the region x > 0 where the potentil is V 0 4.5eV. The potentil step is smller thn the kinetic energy 7.0eV or 5.0eV of the incident bem. The prticle my therefore enter this region clssiclly. It will however lose some of its kinetic energy. In quntum mechnics there is probbillity for the wve to be reflected s well. The two solutions for the two regions re: Ψx Ae ikx + e ikx for x < 0 where k me/ h Ce ik x + De ik x for x > 0 where k me V 0 / h whe cn put D 0 s there cnnot be n incident bem from x. At x 0 both the wvefunction nd its derivtive hve to be continous functions. The derivtive is: Ψx Aike ikx ike ikx x Cik e ik x At x 0 we rrive t the following two equtions: A + C Ak k Ck C A k k+k A k k k+k C A E E+ E V0 A E E V 0 E+ E V0 The coeficients represent the following mplitudes: A is the incident bem, is the reflected bem nd C is the trnsmitted bem. The ssocited probbility currents re denoted j A, j nd j C. Conservtion yields j A j + j C. Hence we cn define the coeficient of reflection s the frction of reflected flux R j j A nd the coeficient of trnsmission s T j C j A For the two cses in prt b nd c the coeficients re: R j T j C C k j A A k k j A A k A C A E E V0 E+ E V0 5.0 5.0+ 0.6987 E V0 E E E+ E V0 E V0 E 5.0 5.0+ 5.0 0.7303 R j T j C C k j A A k k j A A k A C A E E V0 E+ E V0 7.0 7.0+ 0.063437 E V0 E E E+ E V0 E V0 E 7.0 7.0+ The lst result could lso be reched by T + R. 7.0 0.936563. This is dimensionl problem with Schrödinger eqution where V x, y 0 like h h Ψx, y Ψx, y EΨx, y m dx m dy This eqution is seprble nd the nstz Ψx, y ψx ψy gives the following result h m dx ψ xx h m dy ψ yy E x ψ x x + E y ψ y y ie two independent one dimensionl Schrödinger equtions one for the vrible x nd on for y. We therefor solve the one dimensionl problem first nd fter tht we construct the two
dimensionl solution. To find the eigenfunctions we need to solve the Schrödinger eqution which is in the region where V x is zero Solutions re of the kind: h m dx Ψ EΨ d dx Ψ + k Ψ 0 where k me h Ψx A cos kx + sin kx Now we need to tke the boundry conditions for the wve function Ψ Ψ Ψ 0 into ccount. A cos k + sin k 0 nd A cosk + sink 0 Adding the two conditions gives: cos k k 0 nd subtrcting them gives sin 0. These two conditions cnnot be fulfilled t the sme time, so either A or hs to be zero. We strt with A 0 nd we get the following solution: The normlising constnt you get from the condition / / Ψ dx. The condition sin k k 0 gives π even integer. The solution is: ψ n x sinnπx with eigenenergys E n n π h where n, 4, 6,... M In similr wy the other function is nlysed A 0 which gives: The condition cos k 0 gives k π odd integer. The solution is: ψ n x cosnπx with eigenenergys E n n π h where n, 3, 5,... M The eigenfunctions in the y direction re the sme s for the x direction s the potentil is similr for this direction. Now we hve the eigenfunctions of the one dimensionl problem nd the solution to the dimensionl problem is redily produced. The eigenfunctions re: Ψ n,m x, y ψ n x ψ m y eigenenergys E n,m E n + E m where n,,,. nd m,,,. 3 In the re where the potentil is infinite the wve function is equl to zero. An lterntive route tken by mny students hs been to present clcultion with the following boundry conditions: Ψ Ψ0 Ψ 0 into ccount. In this cse the solution is for these boundry conditions: ψ n x sinnπx with eigenenergys E n n π h M where n,, 3,... 4 This solution hs to be dpted to the boundry conditions relted to this exm problem: ψ n x sinnπ x + with eigenenergys E n n π h M where n,, 3,... 5 sin nπx ψ n x nπx sin + nπ cos nπ the solution in eq, nd 3 s we let n run from to. 3 nπx + cos sin nπ. We see tht we recover
b The ground stte eigenfunction is given by using eq. Ψ n,m x, y ψ x ψ y cosπx cosπy 6 The next lowest stte eigenfunction is given by using eq. nd. Note there re two eigenfunctions with the sme energy Ψ n,m x, y you my use either one of them. Orthogonlity is defined s by explicit clcultion / x / / Ψ n,m x, y ψ x ψ y y / x y sinπx cosπy 7 Ψ n,m x, yψ n,m x, y δ n,n δ m,m 8 cosπx cosπy sinπx cosπy clcultions 0 9 this is seprble integrl in x nd y, suggestion do the integrl in x first s this will be zero s they belong to different eigenvlues. Thus the clcultion ends with zero s it should. 3. The eigenfunctions nd eigenvlues of the free-prticle Hmiltonin re found by solving the time-independent Schrödinger eqution h ux + V xux Eux, m dx with V x zero everywhere. Thus, the eigenvlue eqution reds ux dx + k ux 0, where k me/ h. The eigenfunctions re given by the plne wves e ikx nd e ikx, or liner combintions of these, s e.g. sin kx nd cos kx. The wve function of the prticle t t 0 is given by ψx, 0 cos 3 kx + sin 3 kx. This is not n eigenfunction in itself but it cn be written s sum of eigenfunctions using the Euler reltions e ikx + e ikx 3 e ikx e ikx 3 ψx, 0 + 0 i e i3kx + 3e ikx + 3e ikx + e i3kx e i3kx 3e ikx + 3e ikx e i3kx 8 8i 3 4 coskx + 4 cos3kx + 3 4 sinkx sin3kx 4 Thus, ψx, 0 cn be written s superposition of plne wves with two different vlues of k k nd k 3k. 4
b The energy of plne wve e ikx is given by E h k /m. Thus, the energy of e ik x or e ik x is E h k /m nd the energy of e ik x or e ik x is E h k /m 9 h k /m. c The function ux e ikx is solution to the the time-independent Schrödinger eqution. The corresponding solutions to the time-dependent Schrödinger eqution re given by uxt t,with T t e iet/ h. Therefore, uxt t e ikx Et/ h. A sum of solutions of this form is lso solution, since the Schrödinger eqution is liner. This mens tht if ψx, 0 is given by eqution, then the time dependent solution is given by where ψx, t e i3kx + e i3kx e iet/ h + 3 e ikx + e ikx e iet/ h + 8 8 3 e i3kx e i3kx e iet/ h 3 e ikx e ikx e ie t/ h 8i 8i 4 E h k m nd E 9 h k m 5 4. A mesurement of the spin in the direction ˆn sin πê 4 y + cos πê 4 z ê y + ê z. The spin opertor Sˆn is Sˆn S y + S z h i i The eigenvlue eqution is Sˆn χ λχ h i i b λ b 6 We find the eigenvlues from h h λ i h λ i h 0 λ ± h The eigenspinors to S n corresponding to the + h we get from h i + h i b b ib ib let b nd hence i This gives the unnormlised spinor i nd fter normlistion we hve χˆn+ + i Now we cn expnd the initil eigenspinor χ + in these eigenspinors to S n, the second eigenspinor you cn get from orthogonlity to the first one. i A 0 + + + i 5
The coefficients re subjected to the normlistion condition A +. The coefficient A cn be obtined by multiplying the previous eqution from the left with χ ˆn+. A + i The probbility to get + h is given by A. nd to get h for. 0 A 3 + 4 + 0.8535533906 4 + 0.464466094 i + To find the probbility for + h in the z-direction for the up stte of S n express the stte in the eigenspinors to S z. χˆn+ i + i The probbility is given by the squre of the coefficient: + 0 i + 0.8535533906 + 0 + 5. Let the commuttor ct on wve function Ψy nd p y i h d dy [y, p y]ψy h y Ψy dy + h Ψy + 4y h dψy dy [y, p y] + h + 4i hyp y + h + 4 h y d dy. d y Ψy h y Ψy y Ψy 4y dψy Ψy dy dy dy dy + h + i4 hyp y Ψy concluding for the commuttor: b The energy levels for hydrogen like system re given by: E n 3.6 Z [ev], here we hve n Z 4 : E Es Es E E 3.54 6 6 3.54 6 3 6.48 ev 4 c The ngulr prt of the wve function cn be written s sphericl hrmonic: 3 cos θ Y 0 Which gives l och m 0. The prt depending on r r / µe r/3µ principl quntum number n 3 och l consistent with Y 0. corresponding to the 6