Two indistinguishable molecules are absorbed on a simple two-dimensional square

Question

Two indistinguishable molecules are absorbed on a simple two-dimensional square crystalline surface, each at one particular lattice site (the two molecules cannot be absorbed on one and the same lattice site). There are lattice sites in total. If the two molecules happen to be absorbed on the adjacent sites, attractive interaction between them results in a negative potential -? . Otherwise, molecules do not interact. Consider the volume is fixed and neglect all other energies, find the multiplicity ? for E= 0 and E= -? .

Explanation / Answer

For this problem, we must think about what microstates are possible in the system. Some possibilities for a system with M = 16 are shown in Fig. We notice that each lattice site has a total of four neighbors, with the exception of the edge sites. However, the number of edge sites will be much less than the total number of sites M in a macroscopic system M = 10^23 ; therefore, we will neglect this subtle effect for now.

The number of possible E = - microstates is simply the total number of ways we can place the first molecule times the number of possible neighbors to it,

    (E = - ) = M x 4

On the other hand, all remaining possible configurations have an energy of zero. The total number of configurations is the M number of spots to put the first molecule, times the (M-1) number of spots to then place the second. Therefore (E = 0) = M(M-1) - 4M

What if the two molecules were indistinguishable? That is, what if the fourth and fifth microstates above looked exactly the same? In that case, there are fewer total distinct microstates and we have overcounted by the number of ways of swapping the places of different particles. For indistinguishable molecules, we would obtain,

    (E = - ) = (M x 4)/2

(E = 0) = [M(M-1) - 4M]/2

We will find out later on that indistinguishability has an actual effect on the properties of a system, and is due to the rules of quantum mechanics.

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