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Microcanonical ensemble, density operator

  1. Jun 13, 2010 #1
    hi,

    usually the density operator for the microcanonical ensemble is given by

    [tex]\rho = \sum_n p_n|n><n|[/tex]

    where |n> are energy eigenstates and p_n is the probability that our system is in this state.
    p_n = const. if the energy corresponding to |n> is in the energy inteval (E,E+∆E), otherwise p_n =0.

    I.e. we assume, our system is composed of energy-eigenstates. Why is this allowed? Why don't we have to assume, that our system is composed of general quantum states? (with the same conditions for p_n, that is, p_n should be const. if the energy (or much better the expectation value for the energy, since we have no energy eigenstates anymore) corresponding to the general quantum state is within the energy interval)


    --derivator
     
  2. jcsd
  3. Jun 16, 2010 #2
    One assumes that the sytem is completely isolated, so energy is conserved. This makes it possible to consider such a system in a consistent way.

    But, of course, you can consider using different stes that are not energy eigenstates. However, due to the invariance of the trace under a change of basis, that doesn't matter as long as the quantum states the system can be in are such that a measurement of the energy cannot yield values outside of E and E + Delta E.
     
  4. Jun 21, 2010 #3
    Hi,

    I don't see, why we can justify this with the invariance od the trace und a change of basis.

    Could you explain it?

    --derivator
     
  5. Jun 21, 2010 #4
    The expectation value of some observable A is given by

    <A> =Tr[rho A] =

    sum over n of <n|rho A|n>


    Obviously, due to invariance of the trace you could sum over some other set of basis states, we don't have to prove that. You want to show that you could have defined rho as:

    rho = 1/N sum over s of |s><s|

    where N is the number of different energy eigenstates with energy in the range between E and E +Delta E. The states |s> span some arbitrary orthogonal basis in the space of the energy eigenstates with the energies in that range.

    So, we have:

    <A> = 1/N sum over n and s of <n|s><s|A|n>


    Clearly, the sum over |s><s| acts as the identity operator when applied to the bra vector <n|, just like a sum over |n'><n'| with the n' running over all the energy eigenstates with energy eigenvalues in the interval would have been.
     
  6. Jul 2, 2010 #5
    No, they don't have to be necessarily orthogonal nor they have to span a basis.
     
  7. Jul 2, 2010 #6
    The Liouville equation for the time evolution of the statistical operator is:

    [tex]
    i \, \hbar \, \frac{\partial \, \rho}{\partial t} = [H, \rho]
    [/tex]

    where [itex][A, B] = A B - B A[/itex] stands for the commutator of the two operators.

    In equilibrium, the density operator is, by definition, time-independent. But, that means it must commute with the total Hamiltonian of the system.

    The Hamiltonian, like any operator, is diagonal in the basis of its eigenstates. But, the eigenstates of the Hamiltonian are the stationary states.

    If a matrix X commutes with a matrix A and the matrix A is diagonal, then the matrix X is also diagonal. Therefore, the matrix of the equilibrium statistical operator must be diagonal in the basis of stationary states. This ensures the expansion you wrote in your post.
     
  8. Jul 6, 2010 #7
    nitpick: take X to be a non-diagonal matrix, A to be the identity.
    then [X,A] = 0, A is diagonal but X is not!

    you want something more like: for commuting operators we can always
    find a basis that simultaneously diagonalizes them.
     
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