Lecture 31 Highlights Quantum statistical mechanics, continued… We found the statistical weight W of the arrangement (n1, n2, n3, n4, …ns, …): , ∏∞==121,...),...,,(sssPnnnWfor three different types of statistics. This weight is proportional to the probability of finding this particular distribution of occupation numbers. 1) Distinguishable classical particles ∏∞==121!!,...),...,,(ssnssDistngNnnnWs (1) 2) Indistinguishable identical Fermions ∏∞=−=121)!(!!,...),...,,(ssssssFermionsngngnnnW(2) 3) Indistinguishable identical Bosons ∏∞=−−+=121)!1(!)!1(,...),...,,(ssssssBosonsgngnnnnW (3) The next step is to maximize by varying all of the occupation numbers, subject to the number and total energy constraints: and ∑. We will include the constraints using the method of Lagrange multipliers. This method allows one to perform a constrained maximization. We will form a new function to maximize, namely; ,...),...,,(21 snnnW∑∞==1iiNn∞==1iiiEEn ⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+=∑∑∞=∞= 112121,...),...,,(),,...,,...,,(iiiiissEnEnNnnnWnnnGβαβαTo maximize this function we must enforce these conditions: snGs∀=∂∂0 and 0=∂∂=∂∂βαGG. The form of G already satisfies the last two conditions. Because of the products appearing in Eqs. (1)-(3), it is easier to maximize the logarithm of W. This will yield the same result since W and lnW have maxima at the same values of their arguments. The newly defined G for distinguishable particles now is: ⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛=∑∑∏∞=∞=∞=11121!!ln),,...,,...,,(iiiiissnssDistEnEnNngNnnnGsβαβα ()⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−+−+=∑∑∑∞=∞=∞= 111!lnln!lniiiiissssEnEnNngnNβα To take the derivative of G with respect to we must now decide what to do with the logarithm of . One approach is to employ Stirling’s approximation: , good for . With this approximation, Gsn!snxxxx −≅ ln!ln 1>>xDist becomes: 1()⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛−++−+≅∑∑∑∞=∞=∞= 11121lnln!ln),,...,,...,,(iiiiisssssssDistEnEnNnnngnNnnnGβαβαTaking the derivative of G with respect to some particular (called in the lecture) and setting it equal to zero (to find the maximum), yields; sntn Distinguishable particles )(sEssegnβα+−=For the other cases one gets 1)(+=++sEssegnβα Identical Fermions 1)(−=++sEssegnβα Identical Bosons What are the Lagrange multipliers αandβ? They are determined by the number and energy constraints ∑ and . . The challenge is to determine the energies and degeneracies of all of the single-particle states of the system - this is the hardest part of quantum statistical mechanics. Calculating the total energy of an ideal gas, which is a relatively easy case, Griffiths (pp. 239-240) finds that∞==1iiNn∑∞==1iiiEEnTkB/1=β, where Tis the absolute temperature of the gas. The other parameter αis re-defined in terms of the chemical potential μasμβα−≡. The chemical potential is a measure of how much energy is required to change the particle number of the system from to . The three distribution functions can now be written as: N 1+N Distinguishable particles TkEssBsegn/)(μ−−= 1/)(+=−+ TkEssBsegnμ Identical Fermions 1/)(−=−+ TkEssBsegnμ Identical Bosons
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