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Boltzmann Factor in QM Language

  1. May 29, 2008 #1

    I'm stuck on "the summit of statistical mechanics" (as Feynman calls it): the definition of the Boltzmann Factor.

    The probability of measuring the system with energy E is P(E) = 1/Z * e^-E/kT.

    I've taken courses in QM and can't understand why P(E) does not depend on the ket of the system |psi>.

    From QM, I want to write down P(Ei) = <psi|Ei><Ei|psi>. So why doesn't 1/Z * e^-E/kT depend on |psi>? Is it a hidden assumption about the nature of the system that all |psi> in the Hilbert Space have P(Ei) equal?

    Thanks in advance,
  2. jcsd
  3. May 29, 2008 #2
    The issue lies in the difference between a pure state, which represents an ensemble of identical systems, and a mixed state, which represents an ensemble of systems in different states. So you might want to consider an ensemble of systems in different energy eigenstates, represented by a thermal density matrix. The diagonal components are then the Boltzmann factors.
  4. May 30, 2008 #3
    What if I consider a system in a single energy eigenstate?

    If the system is prepared in an energy eigenstate, say E1, then surely a measurement of E1 has probability 1 and not 1/Z * e^-E1/kT ?

    Or can I not prepare a system that has a definite energy?
  5. May 30, 2008 #4


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    The "problem" is that you are at non-zero temperature (otrhwise the Boltzmann factor would be 1), meaning you are implicitly assuming that your system is actually interacting with a heat bath of some sort.
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