Chapter 13 Classical Quantum Statistics 13 1 Boltzmann

Chapter 13 Classical & Quantum Statistics 13. 1 Boltzmann Statistics • It deals distinguishable, non-interacting particles. • There are two constrains NJ = N N J · εJ = U where NJ is the number of particles with single-particle energy εJ.

• The goal: find the occupation number of each energy level when thermodynamic probability is a maximum (i. e. the configuration of an equilibrium state)! • The number of ways of selecting NJ particles from a total of N to be placed in the j level is • For level 1: • For level 2, there are only (N-N 1) particles left, thus

• We consider that each energy level may contain more then one quantum state (degeneracy, g ≥ 1 )! • Using gj to represent the number of quantum state on energy level j. • Therefore, there are g 1 quantum state on level 1, where each of the N 1 particles would have g 1 choices. Therefore, the total possibility would be g 1 N 1. After considering the arrangement of these N 1 particles W 1 becomes For the second energy level, W 2 becomes

The thermodynamic probability for a system with n energy levels is The above equation is subjected to two constraints listed at the beginning of this section

13. 2 Lagrange multiplier • Suppose that there are only two energy levels in a system, thermodynamic probability, W, can be expressed based on the number of particles on each level W = W (N 1, N 2) • The arrangement of N 1 and N 2, which gives the largest value of W can be found via differentiating the above equation against N 1 and N 2, respectively. at the maximum dw = 0

If N 1 and N 2 are independent variables, one has and However, N 1 and N 2 are connected through N = N 1 + N 2 For N = N(N 1, N 2), one has

Thus, one gets = Let the ratio be a constant =α Therefore, –α =0 • For a system with n energy levels, there will be n differential equations. • Note that when taking the derivative against N 1 , N 2 , N 3 , … Nn are kept constant!

When there are two constrains, two parameters are needed! For example: N = NJ U = NJ · E J therefore –α –β =0 • α and β are the Lagrange multipliers.

13. 3 Boltzmann Distribution Since ln W is a monotonically increasing function of W, maximizing ln W is equivalent to maximizing W. Obviously, the logarithm is much easy to work with! W = N! ) ln W = ln. N! + ln( ) --- using ln (x/y) = lnx + lny ln W = ln. N! + ln · g. JNJ – ln NJ! --- using ln (x/y) = lnx – lny ln W =ln. N! + ln · g. JNJ – ln(NJ!)--- using, again, ln (x/y) = lnx - lny Using stirlings application to the last term… ln W =ln. N! + (NJlng. J) – ( NJln. NJ – NJ )

Using the method of Lagrange multiplier +α +β· =0 Therefore, lng. J – ln. NJ + α + βEJ = 0 (see chalkboard) ln = 0 + α + βEJ = e α + βE J

The ratio of NJ/g. J is called the Boltzmann distribution, which indicates the number of particles per quantum state. Now we relate α and β to some physical properties! Since lng. J – ln. NJ + α + βEJ = 0, NJ·lng. J – NJ·ln. NJ + αNJ + βNJεJ = 0

Sum the above eqn over all energy levels NJ·lng. J – NJ·ln. NJ + αNJ + βNJεJ = 0 NJ·lng. J – NJ·ln. NJ + α·N + β·U = 0 The first two term can be replaced by ln. W – ln. N! –N + α·N + β·U = 0 (see inclass derivation) ln. W = ln. N! + N - α·N - β·U

Since S = k·ln. W S = k(ln. N + N - αN – βU) s = kln. N - αk. N -βk. U But kln. N + αk. N = so Therefore, s = so - βk. U Because T·ds = d. U + P·d. V = ds = + ·d. V = ·d. U +… therefore, =

Since s = so + k βU, = -kβ thus, = -k·β -β = = e-α·e e-α = ·e NJ = g. J e-α · e NJ = g. J·e-α · e e-α = = ·e is called partition function!