lim P(t a,b) = Differentiate (1) and use the definition of the probability current, j = i (

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1 PHYS851 Quntum Mechnics I, Fll 2009 HOMEWORK ASSIGNMENT 7 1. The continuity eqution: The probbility tht prticle of mss m lies on the intervl [,b] t time t is Pt,b b x ψx,t 2 1 Differentite 1 n use the efinition of the probbility current, j i 2m ψ x ψ ψ x ψ, to show tht Pt,b j,t jb,t. 2 t Next, tke the limit s b 0 of both 1 n 2, n combine the results to erive the continuity eqution: x jx,t t ρx,t. We strt by ifferentiting Eq. 1, with tψx,t i2m t Pt,b b 2 ψx,t V x x 2 b i Pt,b t 2M Integrting by prts gives t Pt,b z x ψ x,t t ψx,t + ψx,t t ψ x,t ψx,t this gives x ψ x,t 2 2 x2ψx,t ψx,t x 2 ψ x,t [ i ψ x,t 2M x ψx,t ψx,t b x ψ x,t + i 2M b b b x x ψ x,t x ψx,t x x ψ x,t x ψx,t ψ x,t ψx,t ψx,t x x ψ x,t j,t jb,t 5 Tking the limit s b 0, we cn then write the integrl re s simply the height times the with: lim Pt,b b 0 ψx,t 2 b ρx,tb 6 where x is tken to be the point onto which n b converge. This gives iviing both sies by b gives ρx,tb j,t jb,t. 7 t jb,t j,t ρx,t lim jx,t 8 t b 0 b x b ] 3 4 1

2 2. Boun-sttes of elt-well: The inverte elt-potentil is given by V x g δx, 9 where g > 0. For prticle of mss m, this potentil supports single boun-stte for E E b < 0. Bse on imensionl nlysis, estimte the energy, E b, using the only vilble prmeters,, m, n g. The only energy scle we cn form from g, n M is E mg2 2 the scttering length. Thus we shoul expect the nswer to be 2 M 2, where 2 mg is E b 2 M 2 10 b Assume solution of the form: ψ b x ce x λ, 11 n use the elt-function bounry conitions t x 0 to etermine λ, s well s the the energy, E b. You cn then use normliztion to etermine c. Wht is X 2 for this boun-stte? Clling x < 0 region 1, n x > 0 region 2, we hve The first bounry conition is ψ 1 x ce x λ 12 ψ 2 x ce x λ 13 ψ 2 0 ψ 1 0, 14 which is stisfie by construction. Integrting the energy eigenvlue eqution from x ǫ to x ǫ gives ǫ ǫ ǫ xeψx 2 x 2 2M ǫ x 2ψx x g δxψx ǫ ǫ Tking the limit ǫ 0 then gives 2ǫEψ0 2 2M With 12 n 13 the bounry conitions give ψ 2 0 ψ 1 0 gψ0 15 ψ 2 0 ψ ψ0 16 c λ c λ 2 c 17 The solution is λ. For x 0, the energy is purely Kinetic, so we cn use E b ψx 2 x 2 ψx 2M 2 to fin 2 E b 2M

3 which is consistent with our estimtion. For normliztion, we nee c x ψx 2 1 xe 2x 1 c so tht c 1 20 As X 0 by symmetry, the vrince will be given by x X 2, where X xx 2 e x 2 uu 2 e u 0 21 We cn solve this integrl by integrtion by prts: X which woul le to x 2. 0 ue u e u 3

4 3. Inverte elt scttering: Consier prticle of mss m, subject to the inverte elt-potentil, V x g δx, with g > 0. Only this time, consier n incoming prticle with energy E > 0. Wht re the trnsmission n reflection probbilities, T, n R? Treting this like ny other scttering problem, we efine region 1 s x < 0 n region 2 s x > 0, then we let The first bounry conition is which gives ψ 1 x e ikx + re ikx 23 ψ 2 x te ikx 24 ψ 1 0 ψ r t 26 The secon bounry conition is the sme s tht in the previous problem: ψ 2 0 ψ ψ0 27 which gives ik1 r ikt r 28 with t 1 + r, this becomes solving for r then gives 1 r 1 + 2i 1 + r 29 k r 1 1 ik Tking g correspons to 0, which gives r 1, which mkes sense. Tking g 0 + gives, which gives r 0. So the nswer seems resonble. The reflection probbility is then 1 R 1 + k 2 31 so tht T 1 R which is exctly the sme s the non-inverte elt sctterer. 30 k2 1 + k

5 4. Combintion of elt n step: Consier prticle of mss m, whose potentil energy is where ux is the unit step function n V 0 > 0. V x V 0 ux + gδx, 33 Wht re the two bounry conitions t x 0 tht ψx must stisfy? The first bounry conition is continuity of ψ, ψ 2 0 ψ To fin the secon bounry conition, we integrte the energy eigenvlue eqution from ǫ to ǫ n tke the limit ǫ 0, which gives ψ 20 ψ ψ0 35 b For n incient wve of the form e ikx, use the plug n chug pproch to fin the reflection n trnsmission mplitues, r n t. With the nstz where K k 2 k 2 0, with k2 0 2MV 0/, so tht K < k. Putting these into the bounry conition equtions gives: putting t 1 + r then gives which gives n then with t 1 + r we fin ψ 1 x e ikx + re ikx 36 ψ 2 x te ikx r t ik1 r ikt r 38 1 r K k 2i 1 + r 39 k 2 ik K r 2 + ik + K t 2ik 2 + ik + K in the limit K k we shoul recover the solution to the previous problem, which clerly oes work

6 c Compute the reflection probbility, R, n the trnsmission probbility, T. Wht is the reltionship between T n t 2? n R r k K k + K T t 2K k 4kK k + K Lstly, compute the trnsfer mtrix for this potentil t the iscontinuity point, x 0. To compute the trnsfer mtrix, we strt from putting this in the bounry conitions gives ψ 1 x Ae ikx + Be ikx 44 ψ 2 x Ce ikx + De ikx 45 A + B C + D 46 ika B ikc D 2 C + D 47 In mtrix form, this gives 1 1 ik ik A B ik 2 ik C D 48 Solving for C,D T gives C D A 2 + ik 2 ik ik ik B i 2 K A 2 2K K 1 ik ik B 1 2i + k + K 2i + K k A 2K 2i K k 2i + k + K B 49 so we fin M step+δ K,k 1 2K 2i + k + K 2i + K k 2i K k 2i + k + K For the cse g 0 we hve, which oes give M step k,k. For the cse K k, we cn see tht we will recover M δ k. 50 6

7 e Compre your nswer to the mtrices n where K With n we fin k 2 2mV 0 2 M δ,step M step K,kM δ k, 51 M step,δ M δ KM step K,k, 52 n 2 /Mg. Comment on your result. 1 i M δ k i k k i k 1 + i k M step K,k 1 K k K k 2K K k K + k M δ,step M step,δ M step+δ 55 This shows tht putting the elt to the left of the step n tking the limit s the seprtion goes to zero, n putting the elt to the right of the step n tking the seprtion to zero, both give the sme result s putting the elt n the step t the sme point. 7

8 5. Delt function Fbry Perot Resontor: Consier trnsmission of prticles of mss m through two elt-function brriers, escribe by the potentil where g > 0 n L > 0. V x gδx + gδx L, 56 First, compute the llowe k-vlues for n infinite squre well of length L, where k 2mE/. For n infinite squre well, the llowe k-vlues re k n nπ/l where n 1,2,3,... b Next, use the trnsfer-mtrix pproch to compute the full trnsfer mtrix of the resontor. M M δ km f klm δ k 2 e iθ/2 + i 2 e iθ/2 2i cosθ/2 + sinθ/2 2i cosθ/2 + sinθ/2 2 e iθ/2 i 2 e iθ/2 c Use the full trnsfer-mtrix to compute the trnsmission probbility, T, in terms of the imensionless prmeters θ 2kL n 1/k, where 2 /mg. 57 First we compute R M 21 /M 22 2, which gives R T 1 R 4 2 cosθ/2 + sinθ/ cos θ sin θ cos θ sin θ Mke plots of T versus θ for 1, 2, n 4. Compre the loction of the trnsmission resonnces on ech plot to the loctions of the llowe k-vlues from prt. Here is the plot of T versus θ 2kL. We see tht s increses, the resonnces nrrow, n pproch the llowe k vlues from prt. 8

9 6. Consier prticle of mss m incient on squre potentil brrier of height V 0 > 0, n with W. Consier the cse where the incient energy, E, is smller thn V 0. Compute the probbility to tunnel through the brrier, T, s function of the incient wvevector, k. We cn efine the stte just before the first step to be 1,r T, n the stte just fter the step to be A,B T. Then just before the secon step C,D T, n just fter the secon step t,0 T. The full trnsfer mtrix for the step is then M M step k,km f iγwm step K,k coshγw + i k 2 γ 2 2kγ sinhγw i k2 +γ 2 2kγ sinhγw i k2 +γ 2 2kγ sinhγw coshγw i k2 γ 2 2kγ sinhγw 60 We then fin T etm/m 2,2 2 4k 2 γ 2 4k 2 γ 2 cosh 2 γw + k 2 γ 2 2 sinh 2 γw 61 where γ 2MV 0 E/, n k 2ME/ b Write out the full form of the wvefunction of the prticle in the tunneling region. To write the full form of the wvefunction, we first efine x 0 to be the loction of the first step. Then we hve ψ 1 x e ikx + re ikx 62 ψ 2 x Ae γx + Be γx 63 ψ 3 x te ikx W 64 where to fin t we use r t etm/m 2,2 k 2 + γ 2 sinhγw 2ikγ coshγw + k 2 γ 2 sinhγw 2ikγ 2ikγ coshγw + k 2 γ 2 sinhγw to fin A n B we use to fin n A A B 1 M step iγ,k r e γw kk + iγ 2ikγ coshγw + k 2 γ 2 sinhγw e γw kk iγ B 2ikγ coshγw + k 2 γ 2 sinhγw

10 c Tke limit s W 0 n V 0, while holing V 0 W constnt, n show tht your nswer grees with the result for elt-function potentil, V x gδx, with g V 0 W. For comprison with the elt function, we only nee to look t the full trnsfer mtrix. For V 0 W g, we fin tht γw 2MV 0 W 0. Expning the cosh n sinh functions to leing orer then gives noting tht we hve M 1 i γ2 W 2k i γ2 W 2k γ 2 W 2k MV 0W 2 k 1 i M which recovers the elt-function result. i k k i k 1 + i k i γ2 W 2k 1 + i γ2 W 2k Mg 2 k 1 k M δ k 72 10