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:(adsbygoogle = window.adsbygoogle || []).push({});

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 :)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# From QM to Stat Mech

Loading...

Similar Threads for Stat Mech |
---|

I Doubt about partition functions in QFT and in stat Mechanics |

**Physics Forums | Science Articles, Homework Help, Discussion**