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Simple 2 level paramagnet system- How to find the number of states

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data
    We have [itex]N[/itex] non interacting particles in external field [itex]\vec{H}[/itex]. The hamiltonian is given as [itex]H = -h\sum^{N}_{i=1}{\sigma_{i}}[/itex] with [itex]\sigma =\pm 1 [/itex] and [itex]h = - \mu |\vec{H}|[/itex]. Calculate the number of states with given energy [itex]E[/itex] with help of this relation:

    [itex]\Omega = \sum^{}_{\sigma}{\delta{(E-H)}}[/itex]

    where [itex]\sum^{}_{\sigma}[/itex] represent summing over all configurations
    Hint: Use Integral representation of delta function. Integrand can be written in form: [itex]exp[N f(k)][/itex]. Make a Taylor approximation of second order with respect to [itex]k[/itex] and solve integral.




    2. Relevant equations
    So, integral representation of delta function looks like that:
    [itex]\delta(x) = \frac{1}{2\pi}\int^{\infty}_{-\infty}{exp(-ikx)}dx[/itex]

    I'm asking for some hints, or advices, my main problem is that i'm not really sure how to handle that summation over all configurations and also, do we really need to play here with this delta function? We know that following is true:

    [itex]E = (N_{up} - N_{down})(- \mu_{B}|\vec{H}|)[/itex]

    and also [itex]N_{up} + N_{down} = N[/itex]

    So if [itex]E[/itex] and [itex]N[/itex] are given, we can find out [itex]N_{up}[/itex]
    and [itex]N_{down}[/itex] and from there we know that the number of all (to energy [itex]E[/itex]) corresponding configurations equals:

    [itex]\Omega = \frac{N!}{N_{up}!N_{down}!}[/itex]

    I'm sure i must be missing something, can someone be so kind and enlighten me:D
     
  2. jcsd
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