(1) Correspondence of the density matrix to traditional method

Transcription

1 (1) Correspondence of the density matrix to traditional method New method (with the density matrix) Traditional method (from thermal physics courses) ZZ = TTTT ρρ = EE ρρ EE = dddd xx ρρ xx ii FF = UU TTTT The partition function ZZ FF = kkkk ln ZZ SS = TTTT ρρ ln TTTTTT ρρ TTTTTT [derive from the thermodynamic relations] HH = SS zz TTTT HHHH TTTT ρρ = TTTT SS zzρρ TTTT ρρ pp = TT,NN cc vv = [HH ] UU = FF + TTTT or UU = kktt cc vv = SS = or S = NN,VV (entropy) ln ZZ cc vv TT TT = TT FF TT (internal energy) MM = TT (magnetization) pp = TT,NN [if there is work involved] dddd = ββββ NN,VV [a system ability to store additional energy with each T increase] ΔHH ΔEE = kktt cc vv (energy fluctuations)

2 () Free massive particle [ideal gas] with the density matrix KK xx, tt, xx, xx ρρ xx = TTTT ρρ = ρρ xx, xx dddd = VV mm ππππħ mm ππππħ / ee mm ββħ xx xx For a free particle it is better to use the pp basis set. The propagator is pp UU pp = ee iipp mm tt δδ(pp pp ). With the replacement iiii ββ, the density matrix in this basis set is a diagonal matrix with ee EE ii kkkk = ee pp mmmmmm on the diagonal. Also, the off-diagonal elements are zero, which makes it easy to work with logarithms of this matrix (just take the logarithms of the diagonal elements) and traces of products of matrices. 4ππpp Then, TTTT ρρ = VV ee pp 4ππππ h mmmmmm dddd = mmmmmm ππ h [Mathematica], equal to the expression obtained above (traces of representations are independent of the basis set chosen) Internal energy UU = HH = TTTT HH ρρ Entropy SS = TTTT ρρ TTTTTT ln TTTT ρρ ln ρρ = 4ππππ pp h Then, SS = ππ VV ππ mmmmmm VV mmkkkk ππ = 1 1 TTTTTT ZZ h ρρ TTTT ρρ ln ρρ = TTTTTT TTTTTT ee pp mmmmmm pp + ln VV mmkkkk ππ mmmmmm 4ππππ mm pp4 ee pp mmmmmm dddd = kkkk + ln TTTT(ρρ) with TTTT(ρρ) from above and dddd = 4ππππ = + ln VV + ππ kkkkkk 5 h = ππ mmmmmm ln mmmmmm ππ Once UU and SS are known, everything else can be calculated as FF = UU TTTT, cc vv = [Mathematica] UU = kkkk VV mmmmmm (πππ) VV, etc.

3 () With the traditional method We can apply classical statistics to the free particle, provided the density is not too high ZZ = TTTT ρρ = pp ee EE kkkk pp dddd where ρρ = 4ππpp VV h ZZ = 4ππππ ee pp mmmmmmpp dddd h = VV mmmmmm ππħ FF = kkkkkkkk ZZ = kkkk ln VV + SS = = kk ln VV + ln NN,VV UU = FF + TTTT = kkkk pp = = kkkk TT,NN VV cc vv = kk ΔHH = kktt cc vv = kkkk (ideal gas law) dddd dddd = 4ππpp VV h (NN = 1 fixed) / = VVnnQQ with nn QQ mmmmmm ππħ mmmmmm ln ππħ mmmmmm + kk ππħ For instance, the Maxwell-Boltzmann distribution gives EE EE = kkkk Note: kk TT dddd SS [does not work out because there is no ideal gas at very low temperatures TT] Note: classical and quantum free particle models are the same (the only difference is that the classical model small volume ττ h ) The entropy increase indefinitely with VV and TT for an ideal gas because that entropy comes from the particle motion in the ideal gas (new regions of xx phase space open up at higher volume VV, and new regions of the pp phase space open up at higher TT because of higher velocities). UU = FF + TTTT = kkkk All results are consistent in the two methods /

4 () Free massive particle plots Does not satisfy Nernst theorem because the ideal gas neglects all phase transitions (not a good approximation possible in practice)

5 Spin ½ in a constant magnetic field [classical statistics, no state overlap, no exchange energy] Density matrix method (example of using EEEE ρρ EEE ) ρρ = 1 ee βββββ/ [N = 1 fixed]. ZZ βββββ/ ee ωω HH ZZ = TTTT ee ff ωω ee ff HH = ωω tanh ff Z SS zz = TTTT 1 1 ee ff ee ff SS zz = 1 tanh ff [comparing to MM, we get the Lande g-factor gg = ee mmmm ] Traditional method ZZ = ee βββββ/ + ee βββββ/ = cosh ff where ωω = eeee and ff = ββββ. mmmm FF = kkkk ln( cosh ff) [for NN > 1 we get FF = kkkkkkkkkk cosh ff μμ = kkkk ln cosh ff λλ = ee μμ kkkk = 1 independent on level cancels out (Gibbs factor reduces to the Boltzmann factor)] SS = = kk ln ZZ ωω tanh ff = kk ln cosh ff ff tanh ff NN,VV TT Alternatively, cc vv TT () The two-state paramagnet ff dddd ff mmmmmm S = ^TT_mm dddd = kk = kk = kk ff tanh ff ln cosh ff, where TT cosh ff TT cosh ff ff ff mmmmmm is the upper limit For the upper limit = we get SS = kk ln [the finite limit for entropy] (each spin has multiplicity WW = pointing up and down) The entropy saturates to ln because the multiplicity is saturated for complete random spin arrangement [the paramagnet temperature still increases indefinitely]. That S saturates can be seen in thermodynamics from the 1 st law: at the highest temperatures MM cccc. BBBBBB. Therefore, dddd = TTTTTT = BBBBBB, too, and SS cccc. (both sources of entropy for an ideal gas are neglected here: the volume VV cccccccccc., the kinetic energy and velocities ) ff dddd cosh ff

6 () The two-state paramagnet [cont d] UU = FF + TTTT = ωω tanh ff < because HH iiiiii MM BB < MM = = ee tanh ff [the Brillouin function for SS = 1 ] TT mmmm Results for UU and MM are consistent in the two methods We get UU = MMMM = eeee mmmm reverse of internal energy) cc vv cc BB = BB = tanh ff = ωω tanh ff (which is just the BB = ωω 4kkTT 1 cosh ff = kk ff cosh ff cc BB peaks when cc BB = ff tanh ff = 1 ff 1. cc BB.44 kk cc vv peak in the middle because cc vv at TT, The specific heat is small near TT = and at TT because the entropy and dddd = TTTTTT are changing very slowly in these regions. Energy fluctuations from thermodynamics ΔHH = kktt cc vv One standard procedure to simplify a magnetism problem is to replace the internal energy operator UU by the average of its eigenvalues [MFT-type models]. This replaces an operator with a function and neglects the fluctuations about the average values.

7 () Statistics of the two state PM We can also get SS by calculating the multiplicity WW Take 4 spins. For UU = 4 or, we get WW = 1 For UU = or, we get WW = 4 For UU = or, we get WW = 4 = 6 Take 6 spins. For UU = 6 or, we get WW = 1 For UU = 4 or, we get WW = 6 For UU = or, we get WW = 6 5 = 15 For UU = or, we get WW = = In general, for NN spins WW = NN! = NN! and NN!NN! NN! NN NN! NN! SS = kk ln NN! NN NN! The energy is UU = MMMM NN NN = MMMM NN NN Both expressions for SS and UU depend on one variable NN, which can be eliminated to find SS(UU) in the limit of many spins The plots of entropy/spin for 4 and 6 spins already show the main features For the very high temperatures we have WW = NN SS = kkkk ln One of the intuitive ways to find MM TT is to calculate SS(UU), then find TT = and apply the 1st law TTTTTT = BBBBBB to find MM(TT)