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I'd like to know how to get to statistical mechanics from the many-particle Schrodinger equation using a specific example, without using any Hamiltonian mechanics, phase spaces or ensembles, as a concrete way to understand what's going on. I do not understand the meaning of assuming thermal equilibrium as it relates to the wave function (in choosing a state) or what the Maxwell-Boltzmann distribution means in this context. My attempt is given below.

Given 3 non-interacting particles in a 1-d potential well of length a we solve the Schrodinger equation $\frac{-\hbar^2}{2m}\nabla^2\psi = E \psi$ to find the solution

$$\psi_{n_1,n_2,n_3}(x_1,x_2,x_3) = (\sqrt{\tfrac{2}{a}})^3\sin(\tfrac{n_1 \pi}{a}x_1)\sin(\tfrac{n_2 \pi}{a}x_2)\sin(\tfrac{n_3 \pi}{a}x_3)$$ such that $$E_{n_1n_2n_3}=\tfrac{\hbar^2 \pi^2}{2ma^2}(n_1^2+n_2^2+n_3^2)$$ If the total energy was $$E = 243 \cdot \tfrac{\hbar^2 \pi^2}{2ma^2}$$

then there would be 10 ways to satisfy $ n_1^2+n_2^2+n_3^2 = 243$: $$(9,9,9), \\ (3,3,15), (3,15,3), (15,3,3), \\ (5,7,13), (5,13,7), (7,5,13), (7,13,5), (13,5,7), (13,7,5)$$

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

Reading Griffith's QM, section 5.4, he says we invoke 'the fundamental assumption of statistical mechanics' which says in 'thermal equilibrium' every distinct state with total energy $E$ 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

$$\psi_{9,9,9}(x_1,x_2,x_3) = \psi_{3,3,15}(x_1,x_2,x_3) = \psi_{3,15,3}(x_1,x_2,x_3) = ...$$

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

$$\psi_{9,9,9}(x_1,x_2,x_3) \\ \psi_{3,3,15}(x_1,x_2,x_3) = \psi_{3,15,3}(x_1,x_2,x_3) = \psi_{15,3,3}(x_1,x_2,x_3) \\ \psi_{5,7,13}(x_1,x_2,x_3) = \psi_{5,13,7}(x_1,x_2,x_3) = \psi_{7,5,13}(x_1,x_2,x_3) = \psi_{7,13,5}(x_1,x_2,x_3) = ...$$

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 :)

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  • $\begingroup$ Are you familiar with classical statistical mechanics? The issues described in your question are identical (each microstate is given the same probability etc.). The 10 states you have listed are to be equally likely, not equal. $\endgroup$ – alarge Dec 7 '14 at 22:40
  • $\begingroup$ This calculation would basically derive (an example of) classical statistical mechanics from quantum mechanics without any Hamiltonian mechanics (as you assume to derive classical statistical mechanics). You can't assume the things you're trying to derive. $\endgroup$ – bolbteppa Dec 7 '14 at 22:42
  • $\begingroup$ If what you're after is the distinction between Maxwell-Boltzmann / Fermi-Dirac / Bose-Einstein, there's little quantum mechanics anywhere to be seen (in the sense of the Schrödinger equation): One is about distinguishable particles, the other about indistinguishable and non-overlapping ones, and in the last one the particles are both indistinguishable and can overlap. Wavefunctions need not be introduced, just energy states (you can just think about classical states if you think this is easier, and make each of the three assumptions above about the character of the particles). $\endgroup$ – alarge Dec 7 '14 at 22:56
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I can offer a few comments on your project, although without solving your specific problem.

I believe my outlook is rather different from yours. Historically, statistical mechanics arose from macroscopic thermodynamics; you cannot divorce the two. We can take Gibbs and Boltzmann as the creators of statistical mechanics, with a hat tip to Maxwell.

For simplicity, I focus first on Gibbs, who was primarily a macroscopic, classical thermodynamicist. In his day, most thermodynamicists were adamant that their science was purely macroscopic, and had nothing to say concerning microscopic reality (presumably atomic and/or molecular.) Gibbs set out to create a mathematical structure which would start from the Hamiltonian mechanics of molecules, and end up with a thermodynamics whose variables (energy, entropy, etc.) were all expressible in terms of molecular parameters. The success of his effort is manifest in the fact that he worked before the discovery of quantum mechanics, and yet his conceptual structure is still what we use today.

Gibbs' work is still in print, but difficult to read. I recommend Tolman (Statistical Mechanics, now available as a Dover reprint) and Landau (Statistical Physics, Vol. 1.) You will have to get beyond Griffiths if you wish to make progress here; and I would be wary of trying to build your logical structure upon his statement of fundamental principles: for example, two states of equal energy will have different statistical weights if one is degenerate and the other not.

Boltzmann's aim was similar to that of Gibbs: i.e. to connect macroscopic thermodynamics with microscopic molecular mechanics. He was roundly attacked for this, and it is likely these attacks contributed to his suicide. Unfortunately, he did not see Einstein's early papers on Brownian motion (which had already appeared) and which provided the first definitive verification --i.e. with respect to a specific physical system-- that statistical mechanics correctly describes thermodynamics.

The things you wish to avoid --Hamiltonian mechanics, ensembles, phase spaces, etc.-- are not easy to avoid, and attempting to avoid them will likely lead you into much frustration. The Schroedinger equation is not a necessary starting point: if you begin with a Hamiltonian, you get a statistical mechanics which is readily adaptable to either classical or quantum-mechanical usage.

I know my answer is not what you are looking for; but I hope it is some small help in spite of that.

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  • $\begingroup$ Yes they are literally the best two books I've found, Tolman & Landau, Fowler is also good. I'm really just asking where many-body qm ends and where stat mech begins, and what happens to the wave function when stat mech begins, for a specific example to understandd what's going on, Griffith's example is good for the specificity in the numbers and explicitness of computation. Any thoughts on the question? $\endgroup$ – bolbteppa Dec 11 '14 at 20:24
  • $\begingroup$ I don't usually think of where QM ends and SM begins. To me, QM is about calculating dynamics, and SM about calculating thermodynamics. They are different tools that do different jobs, and they don't obviously (to me at least) merge one into the other, and vice versa. The only place where wave functions enter SM (in my world at least) is when I need to calculate a density matrix, whose basis elements are outer products of pairs of wave-functions. But even then, the density matrix isn't intrinsically SM. OK you can calculate the density matrix at thermal equilibrium; then you're doing SM. $\endgroup$ – hyperpolarizer Dec 13 '14 at 7:34
  • $\begingroup$ Continuing in this vein: the thing I usually take from QM to use in SM is the eigenvalues for energy. Give me those and I can calculate the partition function; give me that and I can calculate all the thermodynamic quantities I want (e.g. energy, entropy, etc.) Hope I am not too far off topic. BTW enjoy Landau and Tolman; Fowler I know only by (its immense) reputation. $\endgroup$ – hyperpolarizer Dec 13 '14 at 7:39
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It means that every state (wavefunction) that has the same total energy has also the same probability, that is, if you have the system prepared in the same way many times, and measure the state, the probability of measuring a specific state is not a function of the energy.

Is it saying all the wave functions are 'equal' to one another 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?

yes

Is it saying only some of the wave functions are equal to one another?

He wasnt talking about that, but it is true anyways. You must also assume that the particles are indistinguishable, so there are actually only 3 different states, (9,9,9), (3,3,15), and (5,7,13) (changing the order of the numbers do not change the state). If you do not assume indistinguishable particles you obtain the wrong result, which is known as Gibbs paradox

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