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Microcanonical partition function - Dirac delta of operators

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Why is it that the microcanonical partition function is ##W = Tr\{\delta(E - \hat{H})\}##? As in, for example, Mattis page 62?
    Moreover, what's the meaning of taking the Dirac delta of an operator like ##\hat{H}##?

    2. Relevant equations
    The density of states at fixed energy is the microcanonical partition function (using a truth-valued Kronecker delta):

    ##W(E, V, N) = \sum_{\text{all microstates i at V, N}} \delta_{E_i = E}##

    The Dirac delta of an n-dimensional vector ##\vec{x}## (maybe it can be generalized if we see an operator as a matrix?):

    ##\delta (\vec{x}) = \delta(x_1) ... \delta(x_n)##

    3. The attempt at a solution

    Maybe going in reverse to make sense of the result, using all microstates ##| i \rangle## (if the ##|i\rangle##'s are normalized we obtain the original W):

    ##Tr\{\delta(E - \hat{H})\} = \sum_i \langle i | \delta(E - \hat{H})\ | i \rangle = \sum_i \langle i | \delta(E - E_i)\ | i \rangle##

    I don't even know if the last step is legitimate because I don't know what this Dirac delta of an operator is. If we see it as a big product of deltas of each matrix entry the action on a ket is lost (besides: in what basis? I suppose it doesn't matter since we take the trace after, but the formal expression even outside of trace shouldn't change meaning with change of basis to be well-defined, so it seems ill-defined this way).

    Actually I'm not even sure I did something meaningful since we're interested in a large amount of degenerate states with respect to the Hamiltonian: should its trace be taken to entries with same or different energies? Maybe I should've used the degenerate energy eigenkets ##| n \rangle## instead? But then I don't get the desired result.

    Simply put I'm completely confused.
     
  2. jcsd
  3. Sep 24, 2015 #2
    Let's suppose a parameterizes a complete set of energy eingenstates.
    $$H \mid a \rangle = \varepsilon(a)\mid a \rangle$$
    Then we have $$\delta(E - H)\mid a \rangle = \delta(E - \varepsilon(a))\mid a \rangle$$
    You are right we have to think about degeneracy to understand this. Do you know how to do a change of variables when you have a Dirac delta of a function, such as ##\delta(f(x))##? It's going to turn into a sum or integral over all the zeros of the function. Additionally, there will be a factor of ##1/\left| f'(x)\right|## multiplying the new integrand. If ##a_0 ## are the solutions to ##E - \varepsilon(a)=0##
    $$\int da g(a) \delta(E - \varepsilon(a)) = \int da g(a) \sum_{a_0} \frac{1}{\left|- \varepsilon'(a_0)\right|} \delta(a - a_0)$$
    Let me know if that helps.
     
  4. Sep 24, 2015 #3
    I found this definition which was what I needed: for a complete basis of eigenkets of A with eigenvalues ##a_i##

    ##f(A) = \sum_i f(a_i) |a_i\rangle \langle a_i|##

    so

    ##\delta(a - A) = \sum_i \delta(a - a_i) |a_i\rangle \langle a_i|##

    which justifies the formula since you can find an orthonormal basis of degenerate eigenkets of the Hamiltonian (there's more than one but you can choose one) with which to take the trace. Thus the sum is actually counting the degeneracy of the chosen eigenvalue E which is the fixed energy.
     
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