1. Lecture 7. Systems with a “Limited” Energy Spectrum The definition of T in statistical mechanics is consistent with our intuitive idea of the temperature (the more energy we deliver to a system, the higher its temperature) for many, but not all systems.

2. “Unlimited” Energy Spectrum Einstein solids, ideal gases T > 0 E the multiplicity increase monotonically with U :   U f N/2 U S C T UU S U T U C U Pr. 1.55 ideal gas in thermal equilibrium self-gravitating ideal gas (not in thermal equilibrium) T > 0 Pr. 3.29. Sketch a graph of the entropy of H 2 0 as a function of T at P = const, assuming that C P is almost const at high T.  At T  0, the graph goes to 0 with zero slope. The rate of the S increase slows down (C P  const). When solid melts, there is a large  S at T = const, another jump – at liquid–gas phase transformation. S T ice water vapor

3. “Limited” Energy Spectrum: two-level systems e.g., a system of non-interacting spin-1/2 particles in external magnetic field. No “quadratic” degrees of freedom (unlike in an ideal gas, where the kinetic energies of molecules are unlimited), the energy spectrum of the particles is confined within a finite interval of E (just two allowed energy levels). S U T > 0T < 0 E the multiplicity and entropy decrease for some range of U in this regime, the system is described by a negative T S U T U 2B2B Systems with T 0.

4. ½ Spins in Magnetic Field The total energy of the system:  N  - the number of “up” spins N  - the number of “down” spins Our plan: to arrive at the equation of state for a two-state paramagnet U=U (N,T,B) using the multiplicity as our starting point.  - the magnetic moment of an individual spin E E 1 = -  B E 2 = +  B 0 an arbitrary choice of zero energy The magnetization:  (N,N  )  S (N,N  ) = k B ln  (N,N  )   U =U (N,T,B)

5. From Multiplicity – to S(N  ) and S(U) The multiplicity of any macrostate with a specified N  : U S 0 NBNB- N  B N ln2 Max. S at N  = N  (N  = N/2): S=Nk B ln2

6. From S(U,N) – to T(U,N) Energy E1E1 E2 E2 The same in terms of N  and N  :

7. The Temperature of a Two-State Paramagnet 0 E1E1 E2 E2 E1E1 E2 E2 Systems with neg. T are “warmer” than the systems with pos. T: in a thermal contact, the energy will flow from the system with neg. T to the systems with pos. T. Energy E1E1 E2 E2 T E1E1 E2 E2 - N  B N  B U E1E1 E2 E2 T = +  and T = -  are (physically) identical – they correspond to the same values of thermodynamic parameters.

8. The Temperature of a Spin Gas The system of spins in an external magnetic field. The internal energy in this model is entirely potential, in contrast to the ideal gas model, where the energy is entirely kinetic. B E6E6 E5E5 E4E4 E3E3 E2E2 E1E1  B B At fixed T, the number of spins n i of energy E i decreases exponentially as energy increases. spin 5/2 (six levels) - lnn i EiEi EiEi EiEi T =  - lnn i EiEi T = 0 For a two-state system, one can always introduce T - one can always fit an exponential to two points. For a multi-state system with random population, the system is out of equilibrium, and we cannot introduce T. Boltzmann distribution no T the slope  T

9. The Energy of a Two-State Paramagnet U approaches the lower limit (-N  B) as T decreases or, alternatively, B increases (the effective “gap” gets bigger). U  B/k B T N BN B - N  B U T 1 The equation of state of a two-state paramagnet:  (N,N  )  S (N,N  ) = k B ln  (N,N  )   U =U (N,T,B)

10. S(T/B) for a Two-State Paramagnet S U 0 N  B- N  B Problem 3.23 Express the entropy of a two-state paramagnet as a function of T/B. Nk B ln2

11. S/Nk B x =  B/k B T T/B  , S = Nk B ln2 T/B  0, S = 0 ln2  0.693 S(T/B) for a Two-State Paramagnet (cont.) S/Nk B ln2  0.693 k B T/  B = x -1 high-T (low-B) limit low-T (high-B) limit

12. Low-T limit Which x can be considered large (small)? S/Nk B x =  B/k B T ln2  0.693

13. The Magnetization, Curie’s Law  B/k B T NN In the case of electrons,  =  B = eħ/2m e ~ 9.3·10 -24 J/T = 5.8 ·10 -5 eV/T. In the external magnetic field B = 1T,  B = 5.8 ·10 -5 eV = k B · 0.67K. Thus, the spin gas is very disordered at room temperature (N  ~ N  ) - the EPR technique must have a very high resolution. To neglect the “disordering” effect of a finite temperature, the experiments must be performed at T << 1K. In the case of protons,  =  N = eħ/2m P ~ 5·10 -27 J/T (m P ~ 2000 m e ), and the typical energies are a factor of 10 3 smaller than in the case of electron spins. For this reason, experiments on ordering the nuclear spins require mK temperatures. M 1 The high-T behavior for all paramagnets (Curie’s Law)  B/k B T NN - N  The magnetization:

14. The Heat Capacity of a Paramagnet The low -T behavior: the heat capacity is small because when k B T << 2  B, the thermal fluctuations which flip spins are rare and it is hard for the system to absorb thermal energy (the degrees of freedom are frozen out). This behavior is universal for systems with energy quantization. The high -T behavior: N  ~ N  and again, it is hard for the system to absorb thermal energy. This behavior is not universal, it occurs if the system’s energy spectrum occupies a finite interval of energies. kBTkBT 2 B2 B kBTkBT 2 B2 B compare with Einstein solid  B/k B T C 1 Nk B /2 E per particle C equipartition theorem (works for quadratic degrees of freedom only!) Nk B /2

15. Negative T in a nuclear spin system Fist observation – E. Purcell and R. Pound (1951) The spins equilibrate among themselves on a timescale  2 ~ 10 -2 s (spin/spin relaxation time). This time is much shorter than the timescale  1 ~ hours at mK (spin/lattice relaxation time), over which the spins will cool off (or warm up) to match its surroundings. This allows introducing the spin T (which is different from the temperatures for the lattice and conduction electrons). By doing some tricks, sometimes it is possible to create a metastable non-equilibrium state with the population of the top (excited) level greater than that for the bottom (ground) level - population inversion. Note that one cannot produce a population inversion by just increasing the system’s temperature. The state of population inversion in a two-level system can be characterized with negative temperatures - as more energy is added to the system,  and S actually decrease.

16. Cooling Rhodium Nuclei to – 0.75 nK A.S. Oja and O.V. Lounasmaa, Rev. Mod. Phys. 69, 1 (1997) Initially, both spins and lattice are cooled down to 0.2 mK by dilution refrigerator + adiabatic demagnetization. Then the spins in rhodium are cooled down into the nano-K range by adiabatic demagnetization - the spins are thermally isolated by the slow spin- lattice relaxation (  1 = 14 h) from the conduction electrons which are anchored to 200 µK. the spins are polarized in a magnetic field B = 4·10 -4 T= 4G. the 4G magnetic field is flipped in about 1 ms – within  2 ~ 10 s, the negative temperature is achieved in the system of nuclear spins. the system starts to lose its negative polarization at a time scale  1, crossing in a few hours, via infinity, from negative to positive temperatures.

17. Metastable Systems without Temperature (Lasers) For a system with more than two energy levels, for an arbitrary population of the levels we cannot introduce T at all - that's because you can't curve-fit an exponential to three arbitrary values of #, e.g. if occ. # = f (  E) is not monotonic (population inversion). The latter case – an optically active medium in lasers. E1E1 E2 E2 E3 E3 E4 E4 Population inversion between E 2 and E 1 Sometimes, different temperatures can be introduced for different parts of the spectrum.

18. Problem A two-state paramagnet consists of 1x10 22 spin-1/2 electrons. The component of the electron’s magnetic moment along B is   B =  9.3x10 -24 J/T. The paramagnet is placed in an external magnetic field B = 1T which points up. (a)Using Boltzmann distribution, calculate the temperature at which N  = N  /e. (b)Calculate the entropy of the paramagnet at this temperature. (c)What is the maximum entropy possible for the paramagnet? Explain your reasoning. (a) kBTkBT E 1 = -  B B E 2 = +  B B B spin 1/2 (two levels) B B -  B

19. Problem (cont.) If your calculator cannot handle cosh’s and sinh’s: k B T/  B S/Nk B 0.09

20. (b) the maximum entropy corresponds to the limit of T   (N  =N  ): S/Nk B  ln2 For example, at T=300K: E1E1 E2 E2 kBTkBT k B T/  B S/Nk B T   S/Nk B  ln2 T  0 S/Nk B  0 ln2 Problem (cont.)