Class 27 M Nov 2 chapter 5 recall H atom solutions, particularly dimensionless coordinates http://www.physics.csbsju.edu/QM/H.02.html http://www.physics.csbsju.edu/QM/H.03.html http://www.physics.csbsju.edu/QM/H.04.html http://www.physics.csbsju.edu/QM/H.05.html read about spin-statistics and the periodic table: http://www.physics.csbsju.edu/QM/H.06.html http://www.physics.csbsju.edu/QM/H.07.html http://www.physics.csbsju.edu/QM/H.08.html http://www.physics.csbsju.edu/QM/H.09.html HW: 5.7, old exam: 346t208.pdf #5 ---- Class 28 W Nov 4 chapter 5 read: delta.04.html thru delta.09.html http://www.physics.csbsju.edu/QM/delta.04.html HW: http://physics.nist.gov/PhysRefData/Handbook/Tables/titaniumtable5.htm Titanium has an outer electron configuration 3d^2 4s^2. Either using table 11 in the handout or directly by adding the two orbital angular momentum and two spin angular momentum find the possible (for fermions) total angular momentum configurations. Record the ground state according to Hund's Rules. Go to the above nist.gov web site and record the energy levels that correspond to all the possible angular momentum combinations you found above. Using those values sketch an energy level diagram of those states. The nist site records many actual states that do not match any option you might get by the above process. Explain what those states are and give an example. Do all of the above for Nickel http://physics.nist.gov/PhysRefData/Handbook/Tables/nickeltable5.htm Note: in order to find the high-up 1S0 sate you need to use the full database: http://physics.nist.gov/PhysRefData/ASD/levels_form.html ---- Class 29 F Nov 6 finish chapter 5 (Section 5.4 and beyond will be part of StatMech next semester; give them only the briefest skim.) HeNe_laser.pdf HW text problem: 5.16 ----- class 30 M Nov 9 perturbation theory chapter 6 HW problems 6.2,6.3, 6.4b ----- class 31 W Nov 11 degenerate perturbation theory stark effect for n=4 old test: 346t313.pdf #6 ----- class 32 F Nov 13 variational (Rayleigh-Ritz) method ch 7 cosh2.pdf #1 & #2 ----- class 33 M Nov 16 variational (Rayleigh-Ritz) method start WKB read: http://www.physics.csbsju.edu/QM/fall.01.html http://www.physics.csbsju.edu/QM/fall.02.html http://www.physics.csbsju.edu/QM/fall.03.html http://www.physics.csbsju.edu/QM/fall.04.html http://www.physics.csbsju.edu/QM/fall.05.html http://www.physics.csbsju.edu/QM/fall.06.html http://www.physics.csbsju.edu/QM/fall.07.html i.e., read the first and at the bottom click "next"; continue HW: http://www.physics.csbsju.edu/QM/fall.15.html #11 in class (not HW): see #19 http://www.physics.csbsju.edu/QM/H.13.html e=(-2*a^11*b^2 + a^12*b^2 - 16*a^10*b^3 + 8*a^11*b^3 - 56*a^9*b^4 + 29*a^10*b^4 - 204*a^8*b^5 + 64*a^9*b^5 - 426*a^7*b^6 + 226*a^8*b^6 - 426*a^6*b^7 + 368*a^7*b^7 - 204*a^5*b^8 + 226*a^6*b^8 - 56*a^4*b^9 + 64*a^5*b^9 - 16*a^3*b^10 + 29*a^4*b^10 - 2*a^2*b^11 + 8*a^3*b^11 + a^2*b^12 - 6*a^11*b*c + 3*a^12*b*c - 48*a^10*b^2*c + 24*a^11*b^2*c - 170*a^9*b^3*c + 85*a^10*b^3*c - 360*a^8*b^4*c + 176*a^9*b^4*c - 1232*a^7*b^5*c + 165*a^8*b^5*c - 2000*a^6*b^6*c + 1147*a^7*b^6*c - 1232*a^5*b^7*c + 1147*a^6*b^7*c - 360*a^4*b^8*c + 165*a^5*b^8*c - 170*a^3*b^9*c + 176*a^4*b^9*c - 48*a^2*b^10*c + 85*a^3*b^10*c - 6*a*b^11*c + 24*a^2*b^11*c + 3*a*b^12*c - 6*a^11*c^2 + 3*a^12*c^2 - 48*a^10*b*c^2 + 24*a^11*b*c^2 - 171*a^9*b^2*c^2 + 86*a^10*b^2*c^2 - 364*a^8*b^3*c^2 + 184*a^9*b^3*c^2 - 536*a^7*b^4*c^2 + 269*a^8*b^4*c^2 - 2363*a^6*b^5*c^2 + 48*a^7*b^5*c^2 - 2363*a^5*b^6*c^2 + 2868*a^6*b^6*c^2 - 536*a^4*b^7*c^2 + 48*a^5*b^7*c^2 - 364*a^3*b^8*c^2 + 269*a^4*b^8*c^2 - 171*a^2*b^9*c^2 + 184*a^3*b^9*c^2 - 48*a*b^10*c^2 + 86*a^2*b^10*c^2 - 6*b^11*c^2 + 24*a*b^11*c^2 + 3*b^12*c^2)/ (2*(a^10*b^2 + 8*a^9*b^3 + 28*a^8*b^4 + 120*a^7*b^5 + 198*a^6*b^6 + 120*a^5*b^7 + 28*a^4*b^8 + 8*a^3*b^9 + a^2*b^10 + 3*a^10*b*c + 24*a^9*b^2*c + 86*a^8*b^3*c + 184*a^7*b^4*c + 823*a^6*b^5*c + 823*a^5*b^6*c + 184*a^4*b^7*c + 86*a^3*b^8*c + 24*a^2*b^9*c + 3*a*b^10*c + 3*a^10*c^2 + 24*a^9*b*c^2 + 87*a^8*b^2*c^2 + 192*a^7*b^3*c^2 + 294*a^6*b^4*c^2 + 1872*a^5*b^5*c^2 + 294*a^4*b^6*c^2 + 192*a^3*b^7*c^2 + 87*a^2*b^8*c^2 + 24*a*b^9*c^2 + 3*b^10*c^2)) g = Table[ ContourPlot[e, {a, .15, 1.25}, {b, .15, 1.25}, Contours -> {-.52, -.5, -.48, -.46, -.44, -.42, -.4, -.38, -.36, -.34, -.32}, ColorFunction -> Hue, DisplayFunction -> Identity], {c, 0, 1, .05}] ListAnimate[g] FindMinimum[e,{a,1.1},{b,.5},{c,.35}] Out[2]= {-0.525919, {a -> 1.07487, b -> 0.477447, c -> 0.312549}} with c=0... e /. c->0 FindMinimum[%,{a,1.1},{b,.5}] Out[4]= {-0.513303, {a -> 1.03923, b -> 0.283221}} ----- class 34 W Nov 18 WKB text: chapter 8 HW old exam 346t313.pdf #4 & #5 cosh2.pdf #5 ----- class 35 W Nov 20 return to ch 5: relativity & H-atom ...PT ----- class 36 M Nov 23 return to ch 5: spin orbit & H-atom...PT take home test due Wednesday Nov 25 --- class 37 M Nov 30 chapter 9: TDPT HW 9.1 & 9.2 9.1: the easy way to do this is grab the Mathematica expressions for the H-atom phi and psic from previous problems Done directly 9.2 is rather a mess. In order to get aid from Mathematica it helps to get to dimensionless quantities. Lets measure time with x=omega0 t Then Hab/(hbar omega0) = h is a "small" dimensionless quantity that rates the strength of the perturbation compared to hbar omega0 = Delta E Note: this h has nothing to do with Planck's constant Note: if you see the trick there is an easy way to prove the desired result: |ca|^2+|cb|^2=1 without ever solving the differential equation. I'll give you some extra credit if you can find it, but I wanted you to see the solution and, at least graphically, understand the result. Copy & paste the below into mathematica and see that the result is one. Now (more importantly) line-by-line comment why/how this is solving 9.2 DSolve[{ca'[x] == -I h Exp[- I x] cb[x], cb'[x] == -I h Exp[ I x] ca[x], ca[0]==1, cb[0]==0}, {ca,cb},x] ca[x_]=First[ ca /. %][x] cb[x_]=First[ cb /. %%][x] ca2[t_]=Simplify[ComplexExpand[ca[t] Conjugate[ca[t]] ], Element[h,Reals]] cb2[t_]=Simplify[ComplexExpand[cb[t] Conjugate[cb[t]] ], Element[h,Reals]] Simplify[ca2[t]+cb2[t], Element[h,Reals]] OK now view the following plots and write (words on a sheet of paper, no need to print out this or any other mathematica output) what these plots show: Plot[Evaluate[{ca2[t],cb2[t]} /. h->.1],{t,0,10}] Plot[Evaluate[{ca2[t],cb2[t]} /. h->.2],{t,0,10}] Plot[Evaluate[{ca2[t],cb2[t]} /. h->.4],{t,0,10}] Plot[Evaluate[{ca2[t],cb2[t]} /. h->.8],{t,0,10}] Plot[Evaluate[{ca2[t],cb2[t]} /. h->1.6],{t,0,10}] Plot[Evaluate[{ca2[t],cb2[t]} /. h->3.2],{t,0,10}] Remark: Hab need not be real, but wolog we can take it as real by changing the |a> and/or |b> states by a phase factor. Remark2: Very commonly omega0 would be >1GHz, so these oscillations would be fast compared to human scale. -- class 38 W Dec 2 chapter 9: more TDPT Note: Sarah scheduled a 15 minute quiz on TDPT...next Monday! HW 9.11, tdpt.pdf #1a-c, -- class 39 F Dec 4 chapter 11: scattering Read: bessel functions... http://www.physics.csbsju.edu/QM/square.14.html http://www.physics.csbsju.edu/QM/square.15.html http://www.physics.csbsju.edu/QM/square.16.html http://www.physics.csbsju.edu/QM/square.17.html HW: in folder scatter see file: scatter.problem.txt -- class 40 M Dec 7 in folder scatter: partial_waves.pdf Note: this is mostly a plug and chug problem, best done on Mathematica where you can write things like: f[theta_]:=1/(2 k I) (Sum[(Exp[2 I d2[[l+1]] ]-1) (2l+1) LegendreP[l, Cos[theta]],{l,0,2}]) d2={32.5,8.6,.4} Pi/180 So the hard part is converting a KE = 5 MeV, to a k value, and converting m^2 to barns -- class 41 W Dec 9 chapter 11: scattering math of partial waves ---- class 42 F Dec 11 chapter 11: scattering born approximation HW: In scatter/square_U=18thru63.pdf find 11 expanded resonances (for U=18,27,30,33,39,42,45,48,57,60,59). For each resonance find (by reading the graphs) the maximum cross section, the k location of that maximum, and the l value of the resonance. Make a spreadsheet table of (U, k location, maximum, l, maximum/(2l+1)) Fit (k,maximum/(2l+1)) data to a power law and make a log-log plot of the results. What is the expected value of B? What is the fit value of B? What you are finding is called the unitary limit HW: In that handout, report the U that show a negative scattering length. Imagine making a plot of the cross section @k=0 as a function of U (the depth of the potential)...this zero energy cross section seems to rise and fall. Note that the actual radius of the potetnial is "a" so in our dimensionless units, the physical cross section is pi. Thus the cross section can be hundreds of times the physical extent of the potential; it can also be much smaller than the physical area. U=59 is particularly small and U=61.5 is particularly big so huge changes in zero energy cross section can happen with small changes in U. Explain what is going on. FYI: my guess is that somewhere between U=21 and U=24 there is a U with a huge low energy cross section. ---- class 43 M Dec 14 HW in folder scatter: scatter_problem2.pdf (except part f) --- final: W Dec 16 3:30pm help: T Dec 15 3pm