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Homework Help: Reif Ch7, Decomposition of partition function

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    For a system A consists of two parts A' and A'' which interact only weakly with each other, if the states of A' and A'' are labeled respectively by r and s, then a state of A can be specified by the pair of numbers r,s and its corresponding energy [itex]E_{rs}[/itex] is simply additive, i.e.,
    [itex]E_{rs}[/itex] = [itex]E^{'}_{r}[/itex] + [itex]E^{''}_{s}[/itex]

    The partition function Z for the total system A is a sum over all states labeled by rs, i.e.,


    Z=[itex]\sum_{r,s}e^{-\beta(E^{'}_{r}+E^{''}_{s})}[/itex] = [itex]\sum_{r,s}e^{-\beta E^{'}_{r}} \ e^{E^{''}_{s}}[/itex] = ([itex]\sum_{r}e^{-\beta E^{'}_{r}}[/itex])([itex]\sum_{r}e^{-\beta E^{''}_{s}}[/itex]) = [itex]Z^{'}Z^{''}[/itex]


    My question is: how the sum of product [itex]\sum ()()[/itex] is converted to product of sum ([itex]\sum[/itex])([itex]\sum[/itex]), they are not generally equal

    Thanks
     
    Last edited: May 5, 2012
  2. jcsd
  3. May 5, 2012 #2

    vela

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    Do you understand why you can do this?
    $$\sum_r \big[\sum_s e^{-\beta E_r}e^{-\beta E_s}\big] = \sum_r \big[ e^{-\beta E_r}\sum_s e^{-\beta E_s}\big] $$
     
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