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Lecture 7. Systems with a “Limited” Energy Spectrum

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. S. T. S. U. water. ice. vapor.

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Lecture 7. Systems with a “Limited” Energy Spectrum

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  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. S T S U water ice vapor T U E Einstein solids, ideal gases “Unlimited” Energy Spectrum the multiplicity increase monotonically with U :   U f N/2 self-gravitating ideal gas (not in thermal equilibrium) Pr. 3.29. Sketch a graph of the entropy of H20 as a function of T at P = const, assuming that CP is almost const at high T. ideal gas in thermal equilibrium S Pr. 1.55  U At T0, the graph goes to 0 with zero slope. The rate of the S increase slows down (CP  const). When solid melts, there is a large S at T = const, another jump – at liquid–gas phase transformation. U T T > 0 T > 0 U C C U

  3. E “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 ofE (just two allowed energy levels). 2B S the multiplicity and entropy decrease for some range of U T > 0 T < 0 S U U in this regime, the system is described by a negative T T U Systems with T < 0 are “hotter” than the systems at any positive temperature - when such systems interact, the energy flows from a system with T < 0 to the one with T > 0 .

  4. ½ Spins in Magnetic Field N - the number of “up” spins N - the number of “down” spins  The magnetization: E E2 = + B an arbitrary choice of zero energy 0 The total energy of the system: E1 = - B  - the magnetic moment of an individual spin 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.  (N,N)  S (N,N)= kB ln  (N,N)   U=U (N,T,B)

  5. S N ln2 0 - NB NB From Multiplicity – to S(N)and S(U) The multiplicity of any macrostate with a specified N: U Max. S at N = N  (N= N/2): S=NkBln2

  6. Energy E2 E1 From S(U,N) – to T(U,N) The same in terms of N andN :

  7. Energy E2 E1 Energy E2 E2 E1 E1 The Temperature of a Two-State Paramagnet T= +  and T= -  are (physically) identical – they correspond to the same values of thermodynamic parameters. E2 T E1 E2 E1 N B 0 U - N B 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.

  8. E6 E5 E4 E3 E2  B E1 T = 0 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. Boltzmann distribution At fixed T, the number of spins niof energy Eidecreases exponentially as energy increases. B spin 5/2 (six levels) Ei Ei Ei Ei the slope  T T =  no T - lnni - lnni - lnni - lnni 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.

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

  10. S NkBln2 S(T/B) for a Two-State Paramagnet U Problem 3.23 Express the entropy of a two-state paramagnet as a function of T/B . 0 - N B N B

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

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

  13. N  B/kBT - N The Magnetization, Curie’s Law The magnetization: M N The high-T behavior for all paramagnets (Curie’s Law) 1  B/kBT In the case of electrons,  = B = eħ/2me~ 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 = kB· 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/2mP ~ 5·10-27 J/T (mP ~ 2000 me), and the typical energies are a factor of 103 smaller than in the case of electron spins. For this reason, experiments on ordering the nuclear spins require mK temperatures.

  14. C NkB/2 1  B/kBT The Heat Capacity of a Paramagnet The low-T behavior: the heat capacity is small because when kBT << 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. 2 B kBT 2 B kBT C NkB/2 compare with Einstein solid equipartition theorem (works for quadraticdegrees of freedom only!) E per particle

  15. Negative T in a nuclear spin system 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. 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).

  16. Cooling Rhodium Nuclei to – 0.75 nK 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 . A.S. Oja and O.V. Lounasmaa, Rev. Mod. Phys. 69, 1 (1997) • the spins are polarized in a magnetic field B = 4·10-4T= 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. E4 Population inversion between E2 and E1 E3 E2 E1 Sometimes, different temperatures can be introduced for different parts of the spectrum.

  18. Problem • A two-state paramagnet consists of 1x1022 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. • Using Boltzmann distribution, calculate the temperature at which N= N/e. • Calculate the entropy of the paramagnet at this temperature. • What is the maximum entropy possible for the paramagnet? Explain your reasoning. spin 1/2 (two levels) (a) B B - B E2 = + BB kBT E1= - BB

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

  20. Problem (cont.) the maximum entropy corresponds to the limit of T   (N=N): S/NkB  ln2 (b) For example, at T=300K: E2 kBT E1 ln2 S/NkB T   S/NkB  ln2 T  0 S/NkB  0 kBT/ B

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