UNIVERSITY OF OSLO FACULTY OF MATHEMATICS AND NATURAL SCIENCES


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1 UNIVERSITY OF OSLO FACULTY OF MATHEMATICS AND NATURAL SCIENCES Solution to exm in: FYS30, Quntum mechnics Dy of exm: Nov Permitted mteril: Approved clcultor, D.J. Griffiths: Introduction to Quntum Mechnics, the printed notes: Time evolution of sttes in quntum mechnics, Symmetry nd degenercy nd WKB connection formule, one hndwritten Asheet pges with your own notes, nd K. Rottmnn: Mtemtisk formelsmling. Two dimensionl hrmonic oscilltor H 0 ψr, φ = Eψr, φ [ e ikφ h r fr k fr + ] m r r r r mω r fr = Ee ikφ fr [ h m r r r r k + ] r mω r fr = Efr Singlevluedness under rottion by π: ψr, φ + π = ψr, φ = e ikπ = = k Z. b Eigenvlues: hk. L z ψr, φ = L z fre ikφ = i hfr φ eikφ = hkfre ikφ = hk ψr, φ c H 0 = x x + + y y + = x x + y y + No further justifictions re needed. The three lowest energy levels, their degenercy nd sttes: Energy deg. sttes 00 0, , 0,
2 d We use x = h mω terms of the y ldder opertors. We get x + x nd hmω px = i x x nd similr for y nd py in L z = xp y yp x = h i [ x + x y y y + y x x ] = i h x y x y Lowest energy level: L z 00 = 0, = k = 0, nd H 0 00 = 00 = E =. Thus 00 is simultneously n eigenket of both L z nd H 0. For sttes in the first excited level we hve L z 0 = i h 0 nd L z 0 = i h 0. Multiplying the second eqution with i nd dding nd subtrcting the equtions we see tht the normlized eigensttes of L z re 0 ± i 0 with eigenvlues k = ± h. Both 0 nd 0 re eigensttes of H 0 with eigenvlue, thus 0 ± i 0 re simultneous eigensttes for both H 0 nd L z. e First order perturbtion theory requires the clcultion of the mtrix elements w mn = m H n where m nd n lbel degenerte unperturbed sttes, E 0 m = E 0 n. If there is no degenercy, m = n. Using the fct tht H 0 nd H re both hermitin which implies rel energies E 0 m the mtrix elements become w mn = m H n = m [H 0, A] n = m H 0 A n m AH 0 n = n A H 0 m m AH 0 n = E 0 m n A m E 0 n m A n = E 0 m m A n E 0 n m A n = E 0 m E 0 n m A n = 0 So ll the w mn s re zero which implies tht the first order energy correction is zero. q.e.d. Tking the hermitin conjugte of the eqution H = [H 0, A] nd using the fct tht H 0 nd H re hermitin we get H = A H 0 H 0 A = [H 0, A ]. So if A = A, we get H = [H 0, A] = [H 0, A] = H which is only true for H = 0. Thus A cnnot be hermitin, it must be ntihermitin A = A. Note tht the commuttor of two hermitin opertors is lwys ntihermitin [x, p] = i h for instnce. f H mω = g h Lets try A p x. This term commutes with p x, p y nd y. However its commuttion with x is nontrivil: [ x, p x ] = x p x p x x = x p x xp x x + xp x x p x x = x[x, p x ] + [x, p x ]x = i hx
3 which implies [ H 0, p x ] = mω [x, p x ] = imω hx = [ H 0, ] mω mω imω h g h p x = g h x = H Thus A = g mω im h p x = ip x g m 3 Observe tht A = A. Another wy of writing this is to use ldder opertors. With hmω p x = i x x we get A = g x x. The first order energy correction to ny stte is therefore 0. The second order correction cn be computed s E00 = n x n y H 00 E 00 E nxn n y xn y 00 The opertor H mω = g x = g h x + x thus H 00 = g 0. Therefore the sum hs just single nonzero term 0, nd we get E 00 = g = g. g Set H = H 0 + H. The tril stte is normlized. Therefore the vritionl principle is ψ H ψ E gs. Compute H ψ first: H cos θ 00 + sin θ 0 = cos θ 00 + sin θ 0 + g cos θ 0 + g sin θ Then ψ H ψ = cos θ + sin θ + g sin θ cos θ = + sin θ + g sin θ Differentiting this w.r.t. θ gives Squring the identity tn θ = sin θ cos θ + g cos θ = 0 = tn θ = g sin θ sin θ we cn solve the bove eqution for sin θ: sin θ g sin θ = = sin θ = g + g & cos θ = + g 3
4 When tking the squre roots we need to choose signs such the extremum condition is stisfied, which mens tht sin θ nd cos θ must hve different signs. To help us select the sign we must lso clculte the second derivtive to ensure we hve minimum. The. derivtive condition for minimum is d ψ H ψ dθ = cosθ g sinθ > 0 which mens tht cos θ > 0 nd sin θ < 0. Therefore sin θ = g + g & cos θ = + g We need to express sin θ in terms of these. This cn be done by observing tht cos θ = cos θ sin θ = sin θ = sin θ = cos θ. Therefore 3 ψ H ψ = 3 cos θ + g sin θ = = 3 g + + g = + g 3 g + g + g g h The electric field cuses n dditionl term in the Hmiltonin H E = qe 0ˆn re t /τ. Initilly, before the pulse rrives t =, the system is in the ground stte 00. First order timedependent perturbtion theory gives then the following probbility mplitude for finding the system in one of the sttes f { 0, 0 } fter the pulse hs pssed t t = f t = dt f H E ψ i e iωt i h = i h qe 0 dte t /τ +iωt f x 00 Since the mtrix element 0 x 00 = 0 there is zero probbility to find the system in the stte 0 fter the pulse. The other mtrix element is 0 x 00 = nd the integrl over time is dte t /τ +iωt = dte t τ iωτ/ e ω τ / = e ω τ / τ h mω due u = e ω τ / τ π
5 The integrl cn lso be found in Rottmnn, Chpter X 3., 5: π dxe x +bx+c c/ = eb setting x = t, = /τ, b = iω nd c = 0 we get the bove nswer. Therefore the trnsition mplitude becomes 0 t = qe 0 i h h πτe ω τ / mω nd the probbility for finding the system in the first excited level fter the pulse is P t = π q E 0 m τ e ω τ / i Two spin/ fermions. Let us denote the sttes n x n y n x n y s s where n xj, n yj re the hrmonic oscilltor quntum numbers of prticle j, nd s j is the zcomponent of the spin of prticle j. Since b the spin interction will only give slight energy chnges so the lowest energy sttes re found by putting s mny fermions s possible in the lowest oscilltor levels. The prticles re identicl fermions so the stte must be ntisymmetric under the interchnge of both orbitl oscilltor nd spin quntum numbers. Ground stte: 0 = The spin interction term cn be written H s = b h S S = b S tot S S b = SS + 3/ The spin prt of the bove stte hs totl spin S = 0 s it is the spin singlet thus it is n energy eigenstte with energy E 0 = 3 b nd it is not degenerte. The next to lowest energy sttes must hve one fermion in n excited level nd the other in the 00 stte. The first excited singleprticle energy level is degenerte so we cn choose ny bsis for it. Here we choose 0 nd 0, but we could s well hve chosen the eigensttes of L z gotten in d. One prticle in 00 nd the other in 0 or 0 gives n 5
6 energy + = 3. The sttes re = = = = = = = = From the spinspin interction we know tht ll spin triplet sttes gets n dditionl energy + b 3b while the spin singlets re lowered in energy by. There re singlet sttes nd 5, thus E = 3 3b E = 3 + b hs degenercy 6. hs degenercy. There re 6 triplet sttes, which mens tht j To chieve the lowest energy the oscilltor levels must be filled from below under the constrint tht the wve function is ntisymmetric under interchnge of prticle lbels. An ntisymmetric stte cn be gotten by forming the Slter determinnt not required for credit, which for the cse n = cn be written det 00, 00. The tble lists the singleprticle sttes tht goes into the Slter determinnt. For degenerte oneprticle sttes the sequence of oneprticle sttes is rbitrry. n Sttes E n [] E n /n [] 00 00, , 00, 0 /3=.33 00, 00, 0, 0 6 3/= , 00, 0, 0, 0 8 8/5= , 00, 0, 0, /6= , 00, 0, 0, 0 0, 0 3 3/7=.86 6
7 .8 E n /n n Figure : Energy per prticle vs. n A plot of the totl energy E n divided by the number of prticles n, right column, is shown in Fig.. We see tht the lowest oscilltor energy level cn tke fermions, the next, nd the third 6, fourth 8 etc. An inert element is n element with its shells completely filled. Therefore n inert element in the d world with hrmonic oscilltor potentil insted of Coulombpotentil will be elements where the singleprticle energy levels re completely filled. This hppens for n =, 6,, 0, 30,,... nd cn be seen in the plot of E n /n s the lst integer before the energy per prticle mkes jump. 7
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