1549380323-Statistical Mechanics Theory and Molecular Simulation

410 Quantum ideal gases

wherexiis the combined coordinate and spin labelxi= (ri,si). TheN-particle func-
tion Φ(x 1 ,....,xN) is the solution to eqn. (11.2.2) before any symmetry conditions are
imposed. Since eqn. (11.2.2) is completely separable in theN-particle coordinate/spin
labelsx 1 ,...,xN, the Hamiltonian can be written as a sum of single-particle Hamilto-
nians:

Hˆ=

∑N

i=1

hˆi

ˆhi=pˆ

2

i

2 m

. (11.2.3)

Moreover, sinceHˆ is independent of spin, the eigenfunctions must also be eigenfunc-
tions ofSˆ^2 andSˆz. Therefore, the unsymmetrized solution to eqn. (11.2.2) can be
written as a product:

Φα 1 m 1 ,...,αNmN(x 1 ,...,xN) =

∏N

i=1

φαimi(xi), (11.2.4)

whereφαimi(xi) is a single-particle wave function characterized by a set of spatial
quantum numbersαiandSzeigenvaluesmi. The spatial quantum numbersαiare
chosen to characterize the spatial part of the eigenfunctions according to a set of ob-
servables that commute with the Hamiltonian. Each single-particle functionφαimi(xi)
can be further decomposed into a product of a spatial functionψαi(ri) and a spin
eigenfunctionχmi(si). The spin eigenfunctions are defined via components of the eigen-
vectors ofSˆzgiven in eqn. (9.4.3):

χm(s) =〈s|χm〉=δms. (11.2.5)

Thus,χ ̄h/ 2 ( ̄h/2) = 1,χh/ ̄ 2 (− ̄h/2) = 0, and so forth. Substituting this ansatz into the
wave equation yields asingle-particle wave equation:

−

̄h^2

2 m

∇^2 iψαi(ri) =εαiψαi(ri). (11.2.6)

Here,εαiis a single-particle energy eigenvalue, and theN-particle eigenvalues are just
sums of these:

Eα 1 ,...,αN=

∑N

i=1

εαi. (11.2.7)

Note that the single-particle wave equation is completely separable inx,y, andz.
If we impose periodic boundary conditions in all three directions, then the solution
of the wave equation is simply a product of one-dimensional wave functions of the
form given in eqn. (9.3.14). The one-dimensional wave functions arecharacterized by
integersnx,i,ny,i, andnz,ithat arise from the quantization of momentum due to the