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From QM to Stat Mech

  1. Dec 7, 2014 #1
    Hey guys, I'd like to ignore Hamiltonian mechanics and use a specific example of the many particle Schrodinger equation in Griffith's chapter 5.4 to try to get to statistical mechanics, and see what happens, I need help making sense of it:

    Given 3 non-interacting particles in a 1-d potential well of length a we solve the Schrodinger equation [itex]\frac{-\hbar^2}{2m}\nabla^2\psi = E \psi[/itex]
    to find the solution
    [itex]\psi_{n_1,n_2,n_3}(x_1,x_2,x_3) = (\sqrt{\tfrac{2}{a}})^3\sin(\tfrac{n_1 \pi}{a}x_1)\sin(\tfrac{n_2 \pi}{a}x_2)\sin(\tfrac{n_3 \pi}{a}x_3)[/itex]
    such that
    [itex]E_{n_1n_2n_3}=\tfrac{\hbar^2 \pi^2}{2ma^2}(n_1^2+n_2^2+n_3^2)[/itex]

    If the total energy was [itex]E = 243 \cdot \tfrac{\hbar^2 \pi^2}{2ma^2}[/itex] then there would be 10 ways to satisfy [itex] n_1^2+n_2^2+n_3^2 = 243[/itex]:
    [itex](9,9,9), \\ (3,3,15), (3,15,3), (15,3,3), \\ (5,7,13), (5,13,7), (7,5,13), (7,13,5), (13,5,7), (13,7,5)[/itex]

    Assuming the particles are distinguishable we'd have 10 distinct wave function solutions, e.g. [itex]\psi_{9,9,9}(x_1,x_2,x_3), ...[/itex]. This is all many-particle quantum mechanics. Now how do I get to statistical mechanics from this?

    Reading Griffith's, he says we invoke 'the fundamental assumption of statistical mechanics' which says in 'thermal equilibrium' every distinct state with total energy [itex]E[/itex] is equally likely. What does this mean with regard to the wave function?

    Is it saying all the wave functions are 'equal' to one another
    [itex]\psi_{9,9,9}(x_1,x_2,x_3) = \psi_{3,3,15}(x_1,x_2,x_3) = \psi_{3,15,3}(x_1,x_2,x_3) = ...[/itex]
    in the sense that we can choose any one of the wave functions and use it to describe the system of particles as a whole?

    Is it saying only some of the wave functions are equal to one another
    [itex]\psi_{9,9,9}(x_1,x_2,x_3) \\
    \psi_{3,3,15}(x_1,x_2,x_3) = \psi_{3,15,3}(x_1,x_2,x_3) = \psi_{15,3,3}(x_1,x_2,x_3) \\
    \psi_{5,7,13}(x_1,x_2,x_3) = \psi_{5,13,7}(x_1,x_2,x_3) = \psi_{7,5,13}(x_1,x_2,x_3) = \psi_{7,13,5}(x_1,x_2,x_3) = ...[/itex]
    and they're all equally likely, it's just that (any one of the) last wave function(s) is the statistical mechanical wave function of the system since it can be achieved in the most number of ways? This is what the Boltzmann distribution seems to imply, is this what it says?

    Is it saying we have to completely dispose of the wave function? I fear this is what most people would think, but it doesn't make any sense to me to throw the wave function away. We should be allowed to use all or some of them. Also, since you can express statistical mechanics in terms of a partition function (which is a Feynman path integral which is a solution of the Schrodinger equation) there should be a direct relationship to these explicit wave functions.

    Appreciate any input :)
     
  2. jcsd
  3. Dec 7, 2014 #2

    atyy

    User Avatar
    Science Advisor

    The usual interpretation is that there is an ensemble of many systems, each in a particular energy eigenstate. For extreme simplicity, let's take energies in a narrow band, such that there are only 3 energy eigenstates A,B and C. Then in thermal equilibrium 1/3 of systems will be in state A, 1/3 in state B and 1/3 in state C.

    In this case, since all the systems in the ensemble are in different pure states, we have to represent the ensemble by the density operator, which allows us to describe the relative frequency of the various pure states in the ensemble. Thus the usual view is that a thermal system is not in a pure state, but a mixed state.

    As an aside, one should remember that statistical ensembles are not necessarily real, and just a way to calculate thermodynamic properties. So it is possible to have a different picture in which the system is in a thermal pure state: http://arxiv.org/abs/1112.0740.
     
    Last edited: Dec 7, 2014
  4. Dec 8, 2014 #3
    Ensemble's are defined in a Hamiltonian mechanics phase space, I'd like to ignore that. Furthermore the density operator is just Schrodinger's equation for a density matrix but the density matrix is diagonal so it reduces to my Schrodinger equation. There's nothing suspect about my set-up, I'm just trying to determine how to get to stat mech from an explicit solution of Schrodinger. I do not understand the logic of statistical mechanics in this concrete situation unfortunately. I think this example would be a good way to see the difference between many body quantum mechanics, equilibrium stat mech & non-equilibrium stat mech if we flesh it out :)

    I've since been led to believe that yes we can choose any of the wave functions [itex]\psi_{999} = \psi_{3,3,15}=...[/itex] as a means to describe the whole system, so if that's the case what is the point of finding the Maxwell-Boltzmann distribution? I know the point is to find the distribution of states which maximize [itex]\Pi_s \tfrac{n!g_s^{n_s}}{n_s!}[/itex], which is another way of saying we want to find the set of allowable eigenvalues so that the number of eigenstates [itex]\psi(x_1,...,x_n) = \phi_1(x_1)\cdot...\cdot\phi_s(x_n)[/itex] is maximal (which leads us to say one of [itex]\psi_{5,7,13} = \psi_{7,5,13} = ...[/itex] wave functions will describe the system), but who cares if all wave functions are equally allowable? Surely you at least partition the set of wave functions into 3 sets like I did above, so that finding the state that gives the largest number of eigenstates has some meaning, right?

    Once we have one of the six [itex]\psi_{5,7,13} = \psi_{7,5,13} = ...[/itex], what do we do with it? I have a feeling if you used it to do anything you'd get wrong answers. But if we can choose any of these wave functions they must all be related to the partition function, i.e. the path integral, i.e. the formal operator solution of the Schrodinger equation.
     
  5. Dec 8, 2014 #4

    atyy

    User Avatar
    Science Advisor


    No, if you want a single system in a pure state to produce the thermal statistics, then it cannot be in an energy eigenstate. The equal probability of all states within a small energy band is the microcanonical ensemble.

    If you'd like to have a single system in a pure state produce thermal statistics you can look at:

    http://arxiv.org/abs/cond-mat/0511091
    Canonical Typicality
    Sheldon Goldstein, Joel L. Lebowitz, Roderich Tumulka, Nino Zanghi

    http://arxiv.org/abs/quant-ph/0511225
    The foundations of statistical mechanics from entanglement: Individual states vs. averages
    Sandu Popescu, Anthony J. Short, Andreas Winter

    http://arxiv.org/abs/1112.0740
    Thermal Pure Quantum States at Finite Temperature
    Sho Sugiura, Akira Shimizu


    http://arxiv.org/abs/1007.3957
    Strong and weak thermalization of infinite non-integrable quantum systems
    Mari Carmen Bañuls, J. Ignacio Cirac, Matthew B. Hastings

    http://arxiv.org/abs/1003.5424
    The approach to thermal equilibrium and "thermodynamic normality" --- An observation based on the works by Goldstein, Lebowitz, Mastrodonato, Tumulka, and Zanghi in 2009, and by von Neumann in 1929
    Hal Tasaki
     
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