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I Boltzmann distribution for spin-1/2 dipole: high T limit

  1. May 5, 2016 #1
    The analysis of the distribution of spins for a paramagnetic solid in a B field shows that the probability of a dipole being aligned/anti-aligned with the B field ##\to 0.5## as ##T \to \infty##.

    The intuitive justifications that I've read say that this is "expected" as thermal motion tends to cause disorder (e.g. this is from Mandl). However, what's wrong with this intuitive argument:

    1. It costs energy E, say, to align a dipole with the B field.

    2. A dipole will align if it receives energy E from the thermal motion of its heat bath.

    3. As ##T \to 0##, the probability that a dipole-heat bath interaction exchanges energy E ## \to 0##, since the proportion of particles in the heat bath with energy at least E ##\to 0## as ##T \to 0## (so we have agreement with the Boltzmann distribution prediction here)

    4. As ##T \to \infty##, the probability that a dipole-heat bath interaction exchanges energy E ## \to \infty##, since the proportion of particles in the heat bath with energy at least E ##\to 1## as ##T \to \infty##

    5. Hence, we would expect that, on average, in the high T limit, that each dipole spends an increasingly large proportion of its time in the aligned state, so p(alignment) ## \to \infty## as ##T \to \infty##
  2. jcsd
  3. May 6, 2016 #2


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    You need to remember that all the interactions are reversible, i.e. the probability of the dipole absorbing energy=the probability of emitting energy.

    This is why you can never have more than 50% of the population in the excited state in a simple two-level system. Note that this is a general result which hold not only for interactions with a heat bath but also with other "baths", it is also true for e.g. an atom interacting with an EM field.

    Hence, in your statement 4 the probability that your dipole emits energy to the heat bath also goes to infinity.
  4. May 9, 2016 #3
    Could you expand on this, please? You seem to be stating it as some fundamental rule, but I'm not sure on what this is based - is it a result from QFT, or am I forgetting some simple principle?
  5. May 9, 2016 #4

    Charles Link

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    C.P Slichter's Principles of Magnetic Resonance discusses this scenario in detail in pp.4-10 of his book, including the probabilities of the transitions and the transition rates and the Boltzmann factor. I think you would find this section very good reading.
    Last edited: May 9, 2016
  6. May 9, 2016 #5
    You can also realize that as T goes to infinity, the interaction energy with the magnetic field becomes negligible. The system behaves pretty much like at zero field.
    The orientation of the dipoles is due to the random thermal "kicks" from the temperature bath.
  7. May 9, 2016 #6
    I've managed to take a look at this. It looks very helpful, but at first sight I'm confused. Slichter initially discusses the case of spin transitions due to a driving B field, and quotes a result from time dependent perturbation theory to show that the "up" and "down" transitions are equal, but then goes on to discuss the heat bath case without an external field i.e the one that I'm interested in, and shows that the transition probabilities differ. That seems to contradict what f95toli said in post #2.

    Anyway, I need to think about this in some detail. Thanks. That's a useful reference anyway.
    Last edited: May 9, 2016
  8. May 9, 2016 #7

    Charles Link

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    The ratios of the two populations is ## exp-(\Delta E/(kT)) ## which is unity in the high temperature limit. When the energy difference (## \Delta E ## which is caused by the magnetic field) between the two states is very much smaller than ## kT ##, then the field is incapable of aligning the spins, and the system behaves almost like it didn't exist. Except at very high temperatures, the field will still have a weak effect, and that's why it is paramagnetism. The susceptibility ## \chi ## is weak, (## M=\chi H ##), but it is non-zero. Slichter's book emphasizes the case of nuclear spins, and some of his discussion in this section gets slightly more detailed, apparently because the nuclear spins don't always have a strong coupling to the surrounding lattice. With electron spins, the coupling is much stronger and thermal equilibrium will automatically be achieved. His driving (sinusoidal ) field of the first section will always be present for the electron spin case because of IR and r-f fields of the thermal (blackbody) background of the thermal reservoir. Much of his discussion is still applicable to the case of electron spins, and hopefully it proves helpful. ...editing... reading more carefully, yes, I see C.P. Slichter first discusses a case where the transition probabilities are equal and then goes on to show that in the case of a thermal reservoir, these transition probabilities are not equal... Makes it somewhat complicated, but I think, some very good reading...
    Last edited: May 9, 2016
  9. May 10, 2016 #8

    Charles Link

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    In addition to my comments in post #7, a couple of additional comments: The Boltzmann factor ## exp(-E/(kT)) ## gets applied to a number of systems in statistical physics. Besides the application to the paramagnetic system, it also gets applied to systems of bosons and fermions. In the case of the bosons and fermions, the mean number of occupied states is more complicated than the simple product of the number of states times the Boltzmann factor. The paramagnetic system, though non-trivial, is somewhat simpler than the boson and fermion cases. Reif's book Statistical Physics does a very good treatment of the Boltzmann factor.
  10. May 10, 2016 #9
    Having thought a little more, I guess this is referring to the principle of detailed balance, rather than some kind of QM transition probability argument?
  11. May 10, 2016 #10

    Charles Link

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    I think it is both. Slichter mentions this result and then, although I need to study it further before I fully understand it, he provides a solution to what he calls a "paradox".
  12. May 10, 2016 #11
    I think that you're right. Eqn 1.24 seems to combine detailed balance and the transition probabilities.

    I don't really understand the argument though. Slichter is limiting his analysis to a two-state heat bath, which has to be in the complementary state to the nucleus to allow a transition, and that won't apply in general.

    Is this stuff incredibly subtle, or am I a dimwit? (I'm assuming the former at the moment).
  13. May 10, 2016 #12

    Charles Link

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    I think incredibly subtle. I think Slichter probably spent several weeks or more to figure out his resolution of the paradox. If I spend a couple hours on it, I might have his basic equations memorized, but I most likely wouldn't know the complete "why" behind it. A similar thing applies to the boson and fermion case-I think that one is even more difficult: The mean occupation number of bosons in a state of energy ## E_s ## is given by ## n_s=1/(exp((E_s-\mu)/kT)-1) ## where ## \mu ## is the chemical potential. For fermions, the "-1" in the denominator becomes a "+1". These formulas are very well known and in widespread use, (and they originate from the Boltzmann factor), but I would have to believe that only a handful fully understand the complete details of the derivations. The first goal should be to get a working knowledge of the equations. Any of the finer details you can pick up is icing on the cake.
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