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I Quantum thermodynamics of single particle

  1. Oct 20, 2016 #1
    Hello everybody,

    I have two questions:

    1) Is it possible to define a temperature for single particle (or atom or molecule)? If so, how?

    2) How to model with quantum Hamiltonian an exchange of energy between a single atom (or molecule) and a reservoir at given temperature ##T## ?

    Thank you everybody.

    Konte
     
  2. jcsd
  3. Oct 20, 2016 #2

    Bystander

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    What are your thoughts?
     
  4. Oct 20, 2016 #3
    I don't understand your answer.
     
  5. Oct 21, 2016 #4

    Demystifier

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    Yes, by taking
    $$\rho=e^{-H/kT}$$
    where ##H## is the single-particle Hamiltonian.
     
  6. Oct 22, 2016 #5
    Thank you for your answer.
    I still have question:
    - in this post https://www.physicsforums.com/threa...echanics-and-temperature.426455/#post-2869963, the forumer xerma mentioned a ##\rho = \frac{e^{-H/kT}}{Z}##. What is the difference between those two definitions of ##\rho## (that I suppose both density matrix operator) ?

    Thank you very much.
     
  7. Oct 23, 2016 #6

    vanhees71

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    The latter formula is correct. The statistical operator must be of trace ##1##. Thus the statistical operator of the canonical ensemble is
    $$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}), \quad \beta=\frac{1}{k_{\text{B}} T}.$$
     
  8. Oct 23, 2016 #7
    Thank you for your answer.
    How to demonstrate this expression of ##\hat{\rho}##?
    Because, after searching, I always have another alternative form ##\hat{\rho}= \sum_i p_i | \psi_i \rangle \langle \psi_i|##

    Thanks.

    Konte
     
  9. Oct 23, 2016 #8

    vanhees71

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    This expression you get from the maximum entropy principle. If you look for all statistical operators that lead to a given expectation value ##U=\mathrm{Tr} (\hat \rho \hat{H})## of the energy and minimize the entropy,
    $$S=-k_{\text{B}} \mathrm{Tr} \hat{\rho} \ln \hat{\rho},$$
    you get to the canonical statistical operator (of course you need the normalization ##\mathrm{Tr} \hat{\rho}=1## as another constraint).

    The "alternative form" is just the expansion of the statistical operator in terms of its eigenvectors. Note that there can be also generalized eigenvectors if the operator has a continuous spectrum.
     
  10. Oct 23, 2016 #9
    @vanhees71
    Ok, thank you for this interesting answer. Could you indicate me some lectures that can help to understand and make it clearer for a novice as me, please?
    I suppose, all of those concepts are valid and applicable for the case of a single system (like single atom or single molecule) ?

    Thanks.
    Konte
     
  11. Oct 23, 2016 #10

    vanhees71

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    Well, these are very general concepts. My favorite book, using the information-theoretical approach, is

    A. Katz, Principles of Statistical Mechanics, W. H. Freeman and Company, San Francisco and London, 1967.

    Very good is also Landau&Lifshitz, vol. 5 or Reif.
     
  12. Oct 23, 2016 #11
    Thanks a lot.

    Konte
     
  13. Oct 24, 2016 #12
    Hello,

    I'm back, just because a little doubt persist on my understanding:
    Is the operator ##\hat{\rho} = \frac{e^{-\beta \hat{H}}}{Z}## still meaningful even for describing a pure state?
    Thanks

    Konte
     
  14. Oct 24, 2016 #13

    vanhees71

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    Of course not. An equilibrium state is only a pure state for ##T \rightarrow 0##. Note that using the equilibrium (canonical) distribution means that you look at a single particle within a heat bath at temperature ##T##!
     
  15. Oct 24, 2016 #14
    Thank you,

    So even for a single particle within a heat bath at ##T##°, the concept of mixed states is meaningful?

    In other words, single particle is describable as a canonical ensemble once it is "surrounded" by a heat bath at fixed ##T##° (equilibrium)?

    Konte.
     
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