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Statistical operator of hydrogen atom

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Individual hydrogen atoms have been prepared in the energy state n = 2. However, nothing is known about the remaining quantum numbers. Fine structure and all corrections can be ignored.

    What is the micro-canonical statistical operator.

    2. Relevant equations

    [itex] \hat{\rho_{mc}} = \frac{1}{\Omega (E)}\delta(E-\hat{H})[/itex]

    [itex] \hat{H} = -\frac{\hbar^{2}}{2m_{p}}\Delta_{p} -\frac{\hbar^{2}}{2m_{e}}\Delta_{e} - \frac{e^{2}}{4\pi \epsilon_{0}}\frac{1}{\left|r_{p} - r_{e}\right|}[/itex]

    [itex] \Omega (E) = Tr( \delta(E-\hat{H}) )[/itex]
    3. The attempt at a solution

    I don't understand this. If the atoms are all in state n = 2 then with have a system in a pure state: therefore the probability [itex]p_{i} = \frac{1}{\Omega (E)} = 1[/itex] and [itex]\rho = |2\rangle \langle 2|[/itex].

    is that so?
     
    Last edited: Apr 22, 2013
  2. jcsd
  3. Apr 22, 2013 #2

    TSny

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    All of the atoms have the same energy determined by the quantum number n = 2. But that doesn't mean that all the atoms are in the same quantum state. Consider the "remaining quantum numbers" and list all the possible quantum states that have this energy.
     
  4. Apr 23, 2013 #3
    n = 2
    l = 0, 1
    ml = -1, 0, 1
    ms = -1/2, 1/2

    so if the generalised state vector is [itex]|n,l,m_{l},m_{s}\rangle [/itex] then

    [itex]\hat{\rho} = \frac{1}{P} \cdot \sum\limits_{i = 0}^{1} \left(\sum\limits_{j = 0}^{2} \left(\sum\limits_{k = 0}^{1}|2,i,j-1,k - \frac{1}{2}\rangle\langle k - \frac{1}{2}, j-1,i,2| \right)\right)[/itex]

    where [itex]\frac{1}{P}[/itex] is the probability associated with each state, which I assume to be all equal, e.g. normal distribution (I think 1/12 but not sure )
     
    Last edited: Apr 23, 2013
  5. Apr 23, 2013 #4

    TSny

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    When ##l = 0## what are the possible values of ##m_l##?

    Otherwise, I think you're on the right track.
     
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