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Fermi's Golden Rule

  1. Dec 25, 2015 #1
    Consider a system with countable quantum states. One can define Jij as the rate of transition of probability from i-th to j-th quantum state. In H-theorem, if one assumes both $$ H:=\Sigma_{i} p_{i}log(p_{i})$$ $$J_{ij}=J_{ji}$$ then they can prove the H always decrease. The latter is Fermi's Golden Rule states that the transition rate's matrix is symmetric.

    I have seen in Federick Reif's book Fundamentals of Statistical and Thermal Physics, he has proven Fermi's rule. Briefly, consider a quantum system which obeys Schrödinger's equation:$$i\hbar (d/dt)(\psi)=H\psi$$
    where H is Hermitian. Then one can use these relations to _prove_ Fermi's Golden Rule in this specific case: (I show i-th eigenvector with Ψi.)
    $$J_{ij}\alpha |\langle{\psi_{j},H\psi_{i}}\rangle|^2=\langle{\psi_{j},H\psi_{i}} \rangle\overline{\langle{\psi_{j},H\psi_{i}}\rangle}$$ and H is Hermitian, so: $$J_{ij}\alpha |\langle{\psi_{j},H\psi_{i}}\rangle|^2=\overline{\langle{H\psi_{j},\psi_{i}} \rangle}\langle{H\psi_{j},\psi_{i}}\rangle=\langle{\psi_{i},H\psi_{j}} \rangle\overline{\langle{\psi_{i},H\psi_{j}}\rangle}$$ Hence:
    $$J_{ij}=J_{ji}$$

    As a result, we can proof entropy for an isolated system always increase at least for some special cases with these assumptions:

    I. If our quantum states are countable.

    II. If our system can be described with a Hamiltonian that is Hermitian.

    I have a question: do you have an example of a system does not obey these two assumptions? If so, is Fermi's Golden Rule a principle? How can we prove it using quantum mechanics? Do you know some articles about it?
     
  2. jcsd
  3. Dec 25, 2015 #2

    vanhees71

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    2016 Award

    The H-theorem is valid even for exact S-matrix elements not only for first-order Born matrix elements. You only need the unitarity of the S-matrix, no more (particularly you don't need time-reversal, parity or other discrete symmetries), and the validity of the Boltzmann Stoßzahlansatz (hypothesis of molecular chaos), which is violated for long-range forces like gravity (that's the reason why there is structure in the universe due to small initial fluctuations).

    For details, see my lecture notes on relativistic transport theory (of course, it doesn't matter that I discuss relativistic transport theory; the theorem holds of course also for non-relativistic transport theory, and the proof is identical):

    http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf
     
  4. Dec 28, 2015 #3

    Jano L.

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    The Fermi Golden Rule refers to simple picture of state change, where the system can only be in a state that corresponds to some eigenvalue of the system Hamiltonian. It states that probability that

    system initially at known state ##i## of energy ##E_i## is, after time interval ##t##, in any other state of energy ##E_f## (there may be lots of such states)

    is given approximately by

    $$
    \frac{2\pi}{\hbar} |H'_{if}|^2\rho(E_f) t
    $$
    where ##H'_{if}## is matrix of the perturbation Hamiltonian in the basis of system Hamiltonian eigenfunctions and ##\rho(E_f)## is density of eigenfunctions on the energy line.

    This means according to the golden rule, the probability of the indicated change increases linearly with time. However, this rule can be derived only for transitions that are from one state to a band of states or (similar rule) for transitions from a band of states to one state. As far as I know it is not possible to derive it for a probability of transition coming from a single state to another single state (probabilities of such transitions behave initially as ##t^2## and tend to oscillate in long times).

    The description by the golden rule thus cannot be, without further analysis, be taken as a derivation of simple transition kinetic model, where one can express any transition rate as

    $$
    rate~of~jumps~(j\rightarrow i) = \Gamma_{ij}p_j
    $$
    with ##\Gamma_{ij}## independent of ##p##'s and ##t##.
     
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