Consider a Bose gas confined in an isotropic harmonic

Chapter 7, Problem 74P

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Problem 74P

Consider a Bose gas confined in an isotropic harmonic trap, as in the previous problem. For this system, because the energy level structure is much simpler than that of a three-dimensional box, it is feasible to carry out the sum in equation numerically, without approximating it as an integral.

(a) Write equation for this system as a sum over energy levels, taking degeneracy into account. Replace T and μ with the dimensionless variables t = kT/ hf and c = µ/hf.

(b) Program a computer to calculate this sum for any given values of t and c. Show that, for N = 2000, equation is satisfied at t = 15 provided that c = −10.534. (Hint: You’ll need to include approximately the first 200 energy levels in the sum.)

(c) For the same parameters as in part (b), plot the number of particles in each energy level as a function of energy.

(d) Now reduce t to 14, and adjust the value of c until the sum again equals 2000. Plot the number of particles as a function of energy.

(e) Repeat part (d) for t = 13, 12, 11, and 10. You should find that the required value of c increases toward zero but never quite reaches it. Discuss the results in some detail.

Equation: