Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji


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      1 (i) For a negative value of βμ with a modulus large compared to 1 (i.e. for μ ≪ —kBT, which corresponds to the right-hand side of the figure), the exponential in the denominator of (25) is always much larger than 1 (whatever the energy e), and the distribution reduces to the classical Boltzmann distribution (27). Bosons and fermions have practically the same distribution; the gas is said to be “non-degenerate”.

      2 (ii) For a fermion system, the chemical potential has no upper boundary, but the population of an individual state can never exceed 1. If μ is positive, with μ ≫ kBT:– for low values of the energy, the factor 1 is much larger than the exponential term; the population of each individual state is almost equal to 1, its maximum value.– if the energy ei increases to values of the order μ the population decreases and when ei ≫ μ it becomes practically equal to the value predicted by the Boltzmann exponential (27).

      Most of the particles occupy, however, the individual states having an energy less or comparable to μ, whose population is close to 1. The fermion system is said to be “degenerate”.

      1 (iii) For a boson system, the chemical potential cannot be larger than the lowest e0 individual energy value, which we assumed to be zero. As μ tends towards zero through negative values and —kBT ≪ μ < 0, the distribution function denominator becomes very small leading to very large populations of the corresponding states. The boson gas is then said to be “degenerate”. On the other hand, for energies of the order or larger than μ, and as was the case for fermions, the boson distribution becomes practically equal to the Boltzmann distribution.

      2 (iv) Finally, for situations intermediate between the extreme cases described above, the gas is said to be “partially degenerate”.

      For a two-particle symmetric operator image we must use formula (C-16) of Chapter XV, which yields:

      (29)image

      As the exponential operator in the trace is diagonal in the Fock basis states |n1, .., ni,.., nj,..〉, this trace will be non-zero on the double condition that the states i and j associated with the creation operator be exactly the same as the states k and l associated with the annihilation operators, whatever the order. In other words, to get a non-zero trace, we must have either i = l and j = k, or i = k and j = l, or both.

      and the case i = k and j = l yields:

      (32)image

      (33)image

      The first term on the right-hand side is called the direct term. The second one is the exchange term, and has a minus sign, as expected for fermions.

      For bosons, the operators a commute with each other.

       α. Average value calculation

      If ij, a calculation, similar to the one we just did, yields:

      If i = j,


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