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Ising model without spin-spin interaction

  1. Sep 29, 2014 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hey guys!
    I'm not convinced by what I obtained so far, please tell me whether I'm in the right or wrong direction.
    The Hamiltonian has the form ##H=-h\sum _i s_i## where ##s_i=\pm 1## and ##h=kT##.
    1)Calculate the partition function and the Gibbs free energy.
    2)Calculate the magnetization as ##M=\langle \mu \sum s_i \rangle## and ##M=-\frac{\partial G}{\partial B}##. Plot M in function of h and show that the system is paramagnetic.

    2. Relevant equations

    Partition function, Gibbs free energy, etc.
    3. The attempt at a solution
    1)##Z=\sum _s \exp (\beta h \sum _i s_i )=\exp (\beta h)+\exp (-\beta h ) =2\cosh (\beta h)##. I am not sure at all this is right, I feel like a robot who doesn't understand anything and just plugs and chugs. I can't make any sense of what I did with the 2 sums... nevertheless I "summed over all energy states" as I should have, I guess.
    So then ##G(T)=-kT \ln \left [ 2 \cosh \left ( \frac{h}{kT} \right ) \right ]##.
    2)##M=-\left ( \frac{\partial G}{\partial B} \right )=...=-\mu \tanh \left ( \frac{h}{kT} \right ) =-\frac{h}{B} \tanh \left ( \frac{h}{kT} \right )##. The fact that B appears in this expression raises a red flag to me...
    So I would like to know if all is ok so far (I guess around 0.1% chances to be true) or if I goofed somewhere, which I guess is when I tried to calculate the partition function.

    Thanks guys.
     
  2. jcsd
  3. Sep 30, 2014 #2

    Orodruin

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    Is this the problem exactly as stated? It seems strange to me to have an external field proportional to the temperature and you have not used it in that way. You also did not introduce ##B##.

    Your partition function is the partition function for a single spin, but this should not matter much since your spins are not interacting.
     
  4. Sep 30, 2014 #3

    fluidistic

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    My bad... I just realize that I typed h=kT instead of ##h=\mu B##...!
    About my partition function, so I should multiply it by N?
    Thank you.
     
  5. Sep 30, 2014 #4

    Orodruin

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    Well, for independent spins (without interactions), the total partition function for ##N## spins would just be ##Z_N = Z_1^N##, where ##Z_1## is the partition function for one spin. This gives ##\ln Z_N = N \ln Z_1## so this just multiplies the magnetization by a factor ##N##. (Not very surprising ...)

    So your expression looks fine to me apart from an additional minus sign (the minus from ##G = -kT \ln Z## and that from ##M = -(\partial G/\partial B)## cancel). Since ##h## is proportional to ##B##, you do not really have the problem of the ##B## in the denominator, but everything is multiplied by the magnetic moment ##\mu## (hardly surprising either). In the end, you have a total magnetization which is ##N## multiplied by the magnetic moment of a single spin, multiplied by a function (##\tanh(\beta h)##) which describes how the spins are oriented at a given temperature and external magnetic field. The magnetization approaches ##N\mu## when ##\beta h \gg 1## corresponding to low temperature or strong magnetic fields (or rather, a strong magnetic field in comparison to the temperature) and goes to zero in the absence of a magnetic field - so nothing surprising here either.
     
  6. Sep 30, 2014 #5

    fluidistic

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    Thank you for the explanations.
    I have still to think how to reach that ##Z_N=Z_1^N## using the expression ##
    Z=\sum _s \exp (\beta h \sum _i s_i )
    ## because so far I haven't been able to write the first terms of the series. But I'll try to do it alone and if I need help I'll post here.

    So I've corrected my results and I've showed that the system is paramagnetic (partial derivative of M with respect to B yields a positive expression).

    Now I'm trying to get M via the expression
    ##M=\langle \mu \sum s_i \rangle## but I do need to know how to write up the series terms by terms.
     
  7. Sep 30, 2014 #6

    Orodruin

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    My hint for your new question: The expectation value has certain properties that you can use ...
     
  8. Sep 30, 2014 #7

    fluidistic

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    Ok that interests me. What I was tempted to perform is the calculation of ##\frac{\sum _s \mu \sum _i \exp (\beta h \sum _i s_i)}{Z_N}## but if there's another way, I'm all ears in finding it.

    Meanwhile I can't make sense of the given Hamiltonian. On my draft I have ##H=-h (s_1+s_2+...+s_N)## where N is the number of particles. Now I don't even know how to interpret the fact that ##s_i=\pm 1##. Does this mean that for a fixed i, s_i is equal to either 1 or -1? And about half of the s_i are worth 1 and the other half -1? Or not necessarily...?
    Then I need to sum over s the exponent of beta times the Hamiltonian, I don't even see "s" in the Hamiltonian and I am summing over that index...

    EDIT: nevermind my question about the Hamiltonian and sums! I figured it out...
    Still interested in knowing what's your hint is about! :)
     
    Last edited: Sep 30, 2014
  9. Sep 30, 2014 #8
    One thing about averages is you can sometimes get them by just taking derivatives of the logarithm of the partition function.
    Your Hamiltonian is non interacting which tends to make things easier. You can play around with manipulating sums and factoring and such. I would not qualify as an expert, however I think alternatively you should just be able to approximate the partition function or perhaps even solve for it exactly once you know the multiplicity of states as a function of energy, ##\Omega(E)##.

    I don't know if this will be helpful.
     
  10. Oct 1, 2014 #9

    Orodruin

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    The expectation value is linear so, since ##\mu## is a constant, you could rewrite it
    $$
    \langle \mu \sum_i s_i\rangle = \mu \sum_i \langle s_i \rangle.
    $$
    For each of the ##s_i##, the expectation value is independent of the other spins (no interaction) so we only need to consider the one spin in order to take this expectation value and therefore only consider the partition function ##Z_1##. The expectation value becomes
    $$
    \langle s_i\rangle = \frac 1{Z_1} \sum_{s_i = \pm 1} s_i e^{s_i \beta h} = \frac{e^{\beta h}-e^{-\beta h}}{e^{\beta h} + e^{-\beta h}} = \tanh(\beta h).
    $$
    It thus follows
    $$
    M = \mu \sum_i \tanh(\beta h) = \mu N \tanh(\beta h).
    $$

    Generally, the spins will be quite randomly oriented unless you have a magnetic field that is strong enough to align most of them in its direction. For a given temperature and magnetic field, the number of + spins will be binomially distributed with ##p = e^{\beta h}/Z_1##.
     
  11. Oct 1, 2014 #10

    fluidistic

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    Thanks a lot guys!
    I am in a rush right now but I wonder when can you say that you've aligned a fraction p of spins into the direction of the magnetic field z? I ask this because in my mind most spins will have a z component but generally they'll make an angle "theta" with respect to the z-axis and so in the end the number of spins that are exactly aligned with the applied field is almost 0? Or I'm wrong and spins can only be either aligned or anti-aligned with the applied field?
     
  12. Oct 1, 2014 #11

    Orodruin

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    In the Ising model you either align or anti-align with the z axis, there is no middle ground. Essentially it describes a decoherent quantum state of spin 1/2 particles.

    Of course, you could do statistical mechanics also with a completely classical model where the spins could have any direction, but that will likely be more involved. For example, your partition function will go from being a sum over the two states to being an integral over the different spin directions, etc.
     
  13. Oct 2, 2014 #12
    I don't usually post such direct clues, but I can see you have tried. The problem stems from the hamiltonian expression. If you take ## \beta h = \frac{kT}{kT} = 1## so that can't be right.

    The energy of one site is ## E_i = \pm \mu B##. (spin up or down)

    Energy for one site is then ## Z_1 = exp(\beta \mu B) + exp(-\beta \mu B) = 2 cosh (\beta \mu B)##

    Since sites are non-interacting, the overall partition function for ##N## sites is simply ## Z_1^N##

    Secondly, the expression you gave for Gibb's energy is wrong. That expression is for Helmholtz energy: ## F = -kT ln (Z)##
     
  14. Oct 2, 2014 #13

    fluidistic

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    Right now I'm in a rush I have to go to class in a few.
    I did not really understand the part "the problem stems from the Hamiltonian expression".
    Yesterday I've learned why Gibbs free energy is just ##-kT\ln (Z)## for this particular problem (and not Helmholz's). The reason is that the true Hamiltonian of the system would be the sum of all spin-spin interactions only, while if one also adds the spin-external magnetic field interaction like in this problem, one does not get the Hamiltonian of the system only. My prof. called that Hamiltonian H'.
    After some math (variational principle), he reached that if ##Z=-kT\ln (H')## then ##-kT\ln (Z)## is worth Gibbs potential instead of Helmholtz.
    So according to his logic if we had taken the Hamiltonian as only the spin-sping interactions and we would have performed ##-kT\ln (Z)##, then we would have found Helmholtz's potential.
    He said that several books were wrong on this topic but nevertheless they followed by working on ##\frac{\partial G}{\partial B}## using the expression for "A" they claimed to had found instead of "G". Now I didn't have the chance whether this is true or not.
     
  15. Oct 19, 2014 #14
    I recommend reading "Concepts in Thermal Physics" by Blundell. It will clear it up.
     
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