. Short Answer. PHY4605 Introduction to Quntum Mechnics II Spring 005 Finl exm SOLUTIONS April, 005 () Write the expression ψ ψ = s n explicit integrl eqution in three dimensions, ssuming tht ψ represents wve function ψ( r). Suppose you hve ψ = n c n n where the n re complete set of orthonorml sttes. Wht conditions does the bove eqution impose on the c n? ψ ψ = d 3 r ψ(r) = n c n = (b) How mny degenerte levels re there for hydrogen tom with principl quntum number n? Are ny of these degenercies lifted by the spin-orbit interction? Justify your nswer. degenercy = n (l + ) = n, l=0 nd if we count spin the totl degenercy would be n. Spin orbit coupling interction is proportionl to L S = (J S L ) where J = L + S. So given nl level is split into different j s. By the rules of ddition of ngulr moment, since s = /, j = l ± /, j 0. So for exmple in the n = 3 level, the possible j vlues re +/(l = 0), /, 3/(l = ), nd 3/, 5/(l = ). The degenercy of ech spin-orbit split level is j +, so the totl degenercy is + + 4 + 4 + 6 = 8 = 3 OK (c) Wht is the big difference between prticle with spin quntum number s = / nd one with s =? One behves s fermion, one s boson, respectively. This mens in the presence of nd prticle of identicl type in ech cse, the two s = / prticles hve to hve wve function which is ntisymmetric under exchnge of ll prticle coordintes, while two s = prticles would hve to hve symmetric wve function.
(d) Suppose tht the opertor corresponding to some observble is clled Q. List properties of this opertor nd/or of its eigenfunctions n. The ltter stisfy Q n = q n n. Suppose further tht the quntum-mechnicl stte of system is given by ψ = n c n n with severl of the c n 0. If you were to mke single mesurement of the observble Q, wht would you get s result? Q must be hermitin, which mens its eigenvlues must be rel. Eigenfunctions corresponding to distinct eigenvlues must be orthogonl. If ψ is mixture of different Q eigensttes, mesurement of Q yields q n with probbility c n. (e) Two quntum mechnicl prticles hve orbitl ngulr momentum l = nd spin ngulr momentum s = 0. Suppose tht there is some coupling of the two prticles. List the vlues tht the totl ngulr momentum j of the two-prticle system my tke on. For ech j, stte wht re the possible vlues for the z component. l + s j l + s so j = l, nd J z hs eigenvlues hm j, where m j cn tke on ny vlues between l nd l seprted by.. Skewed squre well. Consider n infinite well for which the bottom is not flt, s sketched here. If the slope is smll, the potentil V = ɛ x / my be considered s perturbtion on the squre-well potentil over / x /. () Clculte the ground-stte energy, correct to first order in perturbtion theory. Ground stte of box of size : ψ (0) πx 0 = cos. Ground stte energy: E (0) 0 = h k0 = h π. m m st order correction independent of! δe () 0 = ψ (0) 0 ɛ x ψ(0) 0 = ɛ / / = ɛ π 4 4π dx x cos πx
(b) Clculte the energy of the first excited stte, correct to first order in perturbtion theory. st excited stte of box of size : ψ (0) πx 0 = sin. Ground stte energy: E (0) = h k = h π. m m st order correction δe () = ψ (0) ɛ x ψ(0) = = ɛ 4 ɛ / / dx x sin πx (c) Clculte the wve function in the ground stte, correct to first order in perturbtion theory. (You will probbly not be ble to get n nswer in closed form, but tke the clcultion s fr s you cn do not evlute integrls you encounter here). ψ 0 = ψ (0) 0 + δψ 0, δψ 0 = m ɛ x 0 m E m>0 0 E m Note if m is odd, ψ m (0) is odd prity m x 0 = 0. so only m s contribute, m = / cos (m + )πx/ / m ɛ x 4ɛ 0 = dxx cos πx (m + )πx cos 0 ( (m + m + ) ( + m) cos mπ ) = 4ɛ m (m + ) π (I didn t expect you to do these integrls, but on the other hnd no one sked for Grdsteyn-Rhyzhik either.) Denomintor: E 0 E m = h (k 0 k m)/m = ( h /m) [ (π/) ((n + )π/) ] All terms in infinite series for st order correction re now specified, nd they cn be summed up to some order with error which is esy to estimte. (d) At wht vlue of ɛ does perturbtion theory brek down? Justify your nswer. Difference in energy of st excited nd ground sttes in zeroth order is 3 h π /(m ). If the energy shifts in perturbtion theory become comprble, one will get level crossings nd must distrust perturbtion theory. Compring, you find this hppens when ɛ h /(m ). 3
3. p-n scttering The scttering mplitude for p n scttering is modelled by f(θ) = χ f (A + Bσ p σ n )χ i () where χ f nd χ i re the finl nd initil spin sttes of the n p system. Possibilities re: χ i, χ f =,,, () where the first spin refers to the proton nd the second to the neutron. Note the wve functions do not hve to be ntisymmetrized since p, n distinguishble. Using σ p σ n = σ p zσ n z + (σ p +σ n + σ n +σ p ), (3) where σ + = σ x + iσ y = 0 ; σ = σ x iσ y 0 0 = 0 0 (4) 0 Clculte nd tbulte ll 6 scttering mplitudes. Mke tble of results. We ll need the single prticle spinor equtions σ + χ = σ χ = 0 (5) σ + χ = χ, σ χ = χ (6) So for spin sttes m p m n χ p m p χ n m n similrly, σ p σ n = ( σ pz σ nz + [ σ p+ σ n + σ p σ n+ ]) = (7) σ p σ n = + (8) σ p σ n = + (9) σ p σ n = (0) written s χ mp in text of problem sme thing 4
Define ˆV = A + Bσ p σ n ˆV = A + B () ˆV = (A B) + B () ˆV = (A B) + B (3) ˆV = (A + B) (4) So scttering mplitutde f is tensor in spin spce f (mp m n finl of f: A + B 0 0 0 0 A B B 0 0 B A B 0 0 0 0 A + B ;m p m n initil ). Allowed vlues For tble of cross-sections, tke bsolute vlue to the squre of ll elements. σ = A B (5) 4. Selection rules (0 pts.) Recll tht the selection rule for electric dipole trnsitions in hydrogenic toms, i.e.those with nonzero mtrix elements φ nlm d φ n l m, (d = er) re tht l must be l ± nd tht m must be m ± (or m). () A hydrogen tom is initilly in n n = 4,l = excited stte. It decys to the ground stte by sequence of llowed dipole trnsitions. How mny steps re required? Write possible sequence or two. Since l =, there re two steps required, e.g. 4m m 00, 4m 3m 00. (b) (b) A hydrogen tom initilly in n n = 4, l = stte decys directly to its ground stte. Show tht s dipole trnsition this is forbidden, but it is possible s n electric qudrupole trnsition, with perturbtion proportionl to Q ij = e(3r i r j δ ij r ). The probbility of n electric dipole trnsition directly to the ground stte from 4m is zero becuse 4m r 00 = 0, formlly. The r comes from the perturbtion to the energy of n tom in the presence of n electric field in the dipole pproximtion, V qɛ r, where ɛ is unit vector describing the polriztion of the photon. Clerly for photon with finite 5
wvelength coming in t some ngle, the electric field cn distort the tom in other wys, nd one cn expnd the distortion in sphericl hrmonics, of which the dipole pprox. is only the lowest. See, e.g., J.D. Jckson, Electrodynmics, for discussion of the multipole expnsion. For now, you don t need to worry bout this; the question is only whether the opertor given cn cuse trnsitions from 4m nd 00, i.e., cn 4m Q ij 00 0 for some i, j? The r term doesn t help, since the ngulr prt of the integrl will involve YmY 0 dω, which is zero just by the orthonormlity of the sphericl hrmonics. Let s tke the simplest cse of i, j = z, z nd consider the first term (with m=0) 40 z 00 dω Y0 cos θ Y 0 dω π 0 = 4 5, sin θ dθ ( 3 cos θ ) cos θ so yes, the qudrupole opertor cn cuse trnsitions between the two sttes in question.. 5. Exchnge Symmetry (0 pts.). Suppose you hve three prticles, one in stte ψ, one in ψ b, nd one in ψ c. Construct wve function for the three prticles if they re () distinguishble If they re distinguishble we don t hve to worry bout spin-sttistics theorem, so we could write ψ(r, r, r 3 ) = ψ (r )ψ b (r )ψ c (r 3 )χ χ χ 3, where the χ s re the spin sttes of the 3 prticles. prticulr symmetry under interchnge of, etc. Note there is no (b) fermions For fermions we hve to ntisymmetrize: ψ(r, r, r 3 ) = ψ (r )ψ b (r )ψ c (r 3 )χ χ χ 3 + ψ b (r )ψ c (r )ψ (r 3 )χ χ χ 3 +ψ c (r )ψ (r )ψ b (r 3 )χ χ χ 3 ψ b (r )ψ (r )ψ c (r 3 )χ χ χ 3 ψ (r )ψ c (r )ψ b (r 3 )χ χ χ 3 ψ c (r )ψ b (r )ψ (r 3 )χ χ χ 3, where I ve ssumed no spin-orbit coupling. 6
(c) bosons For bosons we hve to symmetrize: ψ(r, r, r 3 ) = ψ (r )ψ b (r )ψ c (r 3 )χ χ χ 3 + ψ b (r )ψ c (r )ψ (r 3 )χ χ χ 3 +ψ c (r )ψ (r )ψ b (r 3 )χ χ χ 3 + ψ b (r )ψ (r )ψ c (r 3 )χ χ χ 3 + ψ (r )ψ c (r )ψ b (r 3 )χ χ χ 3 + ψ c (r )ψ b (r )ψ (r 3 )χ χ χ 3, 6. Vritionl principle (5 pts.)use the vritionl principle to estimte the ground stte energy of the nhrmonic oscilltor, H = ˆp /m + λx 4, nd compre with exct result E 0 =.06λ /3 ( h /m) /3. Let s try n Gussin tril wve function, φ = A exp( λ x ), with λ s vritionl prmeter. To normlize, we tke φ φ = A dx e η x = A π η = so A = η / (/π) /4. The expecttion vlue of the Hmiltonin is then ( ) φ H φ = A e η x h d m dx + λx4 e η x dx ( ) = A h m (η4 x η ) + λx 4 e η x dx [( h ) ( = A η 4 π m (η ) 3 η [ = A h ] π π m η + 3λ 6η 5 = h m η + 3λ 6η 4 ) π + λ 3 η 4 π (η ) 5 ] So minimizing d H /dη = 0, we find η = 3λm/4 h 6, nd vlue of H t minimum is ( ) 4/3 3 H min = λ /3 ( h /m) /3 = 0.68λ /3 ( h /m) /3, 4 which is (oops!) smller thn the result I quoted for the exct result, which is of course impossible. Looks like I got the wrong exct result somewhere. 7. Ahronov-Bohm phse shift (5 pts.) A single electron psses through lrge cpcitor of chrge Q nd cpcitnce C.Its velocity is v (ssume unltered by cpcitor s field) nd the cpcitor hs length L. Estimte the chnge of phse of the electron s wve function. 7
From notes, the result of vrying potentil will be vrying phse of electronic wve function: ψ(x, t) = ψ 0 (x, t)e is/ h, S = t 0 V (t )dt (6) where ψ 0 is wve fctn. in bsence of bttery. If the electron psses through lrge cpcitor with smll gp, we cn neglect the fringing fields, nd pproximte the potentil difference between the cpcitor pltes s usul s V = Q/C, then the electron experiences potentil which is zero except for time t = L/v, when it is Q/C. The phse shift is then S = LQ/(vC). 8. Stern-Gerlch (0 pts.). A Stern-Gerlch pprtus is ligned long the z direction, nd second one is ligned t 45 with respect to this in the z x plne. A neutrl spin-/ prticle is prepred with spin ẑ nd then pssed through both in succession. () Wht is the probbility detector locted fter the two SG setups finds the prticle to hve spin up ( ẑ)? The prticle psses through the st detector with wve function ( 0) intct. The second detector projects onto the eigenfunctions of the opertor S x / + S z /, i.e. rotted by 45 from z. This opertor is represented in spin spce by ( ) S = (S x + S z ) = h which hs normlized eigenvectors ψ = ( ). The stte ( 0) cn be decomposed ( ) 0 (+ ) ( ) + ( ) ( + = ψ ) + ψ 4 nd ψ = so the prticle will pss through the second detector with probbility ( + )/4 =85%. The stte will then be ψ, nd we wnt to mesure with third detector long z. From the definition of this stte, we see tht the probbility of mesuring ( 0 probbility is therefore 73%. ) is ( + / ( + )) =85%. Totl (b) Wht is the probbility detector mesures the spin to be long the ŷ- direction? 8
The up eigenvector of S y in the originl coordinte system is ( ) i /. Therefore if we mesure fter it mkes it through the nd detector in stte ψ we tke the inner product of the two sttes squred to find (4 + )/(4( + )) = 50%. The initil spin stte is now chnged to definitely ˆx. (c) Redo question ) By symmetry must be the sme. 9