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How to deal with index in the problem of thermal field?

  1. Jan 25, 2009 #1


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    In the problem of harmonic oscillator, for single mode, that is, the energy

    [tex]H = \hbar\omega(n + 1/2)[/tex]

    It is easy to find the average of energy by considering the density operator

    [tex]<H> = \frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}[/tex]

    But for multimode (assume no polarization), we have to consider all modes specified with [tex]k[/tex]

    [tex]H = \sum_k \hbar\omega_k(n_k + 1/2)[/tex]

    So the average energy would be

    [tex]<H> = \prod_k\frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}
    = \prod_k\frac{\sum_{k_1} \hbar\omega_{k_1}(n_{k_1} + 1/2) \exp[\beta\sum_{k_2}\hbar\omega_{k_2}(n_{k_2}+1/2)] }{\exp[\beta\sum_{k_3}\hbar\omega_{k_3}(n_{k_3}+1/2)]}

    Since each sum and prod should have independent index so I use k, k1, k2, k3. I am very confusing how to deal with the index to get the simplifed expression.

    In some textbook, it first let the partition function be
    [tex]Z=Tr(e^{\beta H})[/tex]

    so the energy average be

    <H> = -\frac{\partial \ln Z}{\partial \beta} = \sum_{k}\dfrac{\hbar\omega_{k}}{\exp\{\beta\hbar\omega_{k}\}-1}

    My question is how to deal with index if I don't use the partion function.
  2. jcsd
  3. Jan 26, 2009 #2


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    Homework Helper

    same way as with the partition function. it's just easier to calculate the partition function and take a derivative.
  4. Jan 26, 2009 #3


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    But there is three different index, how can I make it into one index?
  5. Jan 26, 2009 #4


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    Well, I found one problem. For single mode, which way is the correct one to write the partition function?

    [tex]Z= \prod_k\exp[\beta\sum_{k}\hbar\omega_{k}(n_{k}+1/2)][/tex]


    [tex]Z= \prod_k\exp[\beta\hbar\omega_{k}(n_{k}+1/2)][/tex]


    And what about the hamiltonian in the numerator?
  6. Jan 26, 2009 #5


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    Homework Helper

    Neither is the correct expression for the partition function. This is apparent since the occupation numbers still appear on the LHS...

    But, before going on, note that
    \prod_j e^{f_j}
    =e^{f_1}e^{f_2}\ldots e^{f_N}\ldots=e^{f_1+f_2+\ldots}
    =e^{\sum_j f_j}\;.

    The correct partition function is
    and this equals
    \prod_k \sum_{n_k}e^{-\beta(n_k+1/2)\hbar\omega_k}\;,
    where the sum, for bosons, runs from 0 to infinity.
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