Statistical Physics (PHYS 328) HW 7 Due Monday 12/4/17

100 pts

1. (40 pts) Bose condensation for Rubidium: Consider a collection of 10,000 atoms of rubidium-87, confined inside a box of volume (10−5 m)3

(a) Calculate 0, the energy of the (in eV ).

3h2  = = 7.1 × 10−14eV 0 8mL2 where we used m = 87×1.66×10−27kg and L = 10−5m together with the standard value for h and the J/eV conversion.

(b) Calculate the condensation and compare kTc to 0. The condensation temperature is given by

h2 T = (0.527) N 2/3 = (0.224)N 2/3 2πmkL2 0 where the numerical coefficient comes from using our answer for 0 above. For 10,000 atoms kTc is only larger than 0 by a factor of about 100. This isn’t really what we want in the formal large volume limit (where 0 should vanish for fixed Tc, but it is close enough so we can use the formulas derived in class. Plugging in our result for 0 this yields

−8 Tc = 8.6 × 10 K.

This is in good agreement with experiment.

(c) Suppose that T = 0.9Tc. How many atoms are in the ground state? How close is the to the ground-state energy? How many atoms are in each of the (three-fold degenerate) first excited states? At T = 0.9Tc the number of atoms in the ground state is

"  3/2# T 3/2 N0 = 1 − N = [1 − (0.9) ] = 0.146 N = 1460. Tc

From N0 we can extract the chemical potential. We know that 1 kT N0 = β( −µ) ≈ . e 0 − 1 0 − µ

1 As discussed in class, for N0 to be an order 1 fraction of N we need eβ(0−µ) to be very close to 1, and so we can Taylor expand the exponential. That’s what we have done in the last step. Solving for  − µ we find

kT −15 0 − µ = = 4.6 × 10 eV. N0 With this information we can calculate the occupation of the first excited state. Its energy is given by

h2 h2  = (22 + 12 + 12) = 6 = 3 1 8mL2 8mL2 0 so the expected number of particles in any one of these states is 1 N1 = = 87. eβ(1−µ) − 1 The total number of atoms in the 3 degenerate states is

N1stexcited = 3N1 = 260.

This is less than the number of particles in the groundstate by a factor of 5.6. That is, while much fewer particles are in this state it is still a macroscopic occupation. (d) Repeat parts (b) and (c) for the case of 106 atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state. If instead there are 106 atoms, we get

−10 kTc = 1.6 × 10 eV

N0 = (0.146)N = 146, 000 −16 0 − µ = 9.8 × 10 eV

N1 = 1990,N1stexcited = 5960

While N0 grew by a factor of 100 together with N, N1 only went up by a factor of 20 or so. As we increase N, the occupancy of any given excited state becomes less and less important and only the groundstate retains a macroscopic occupation number.

2. (60 pts) 2d :

2 (a) Show that a 2D perfect Bose gas in a box (i.e. confined to a square shaped area) does not undergo Bose-Einstein condensation. The only thing that√ changes in 2d is the density of states. Instead of g() ∼  we have a constant density of states π g() = g0 = 4E0 with h2 E = . 0 8mL2 This is just 1/2 of what we derived for the 2d electron gas, accounting for the fact that a spin 0 boson has no factor of 2 spin-degeneracy. Following the same steps we took in 3d, let us first set up an integral for the net number of bosons in any of the excited states Z ∞ d Nexcited = g0 . 0 β(−µ) − 1 To look for a possible critical temperature Tc we want to look for a place where µ = 0. At this point, the integrand is (eβ − 1)−1. For small epsilon we can expand: eβ − 1 ≈ β

R −1 and so the lower end of the integration goes as 0  d, which diverges (the integral of 1/ is log(), which evaluates to −∞ if we plug in  =√ 0. We can not reach zero µ! We see that the extra factor of  in the 3D case was absolutely√ essential. In this case the integrand√ only diverges as 1/  at small . But√ the integral of 1/  is well defined (since the integral is 2 , which vanishes at  = 0. The same divergence happens for any positive value of µ, since we will always end up integrating over a pole. The only way to keep the integral finite is to have negative µ, so that eβ(−µ) is always bigger than 1. (b) Now suppose that the 2D Bose gas instead of being in a box is con- fined towards the center of the plane by a smooth potential of the form φ(r) = αr2, where r is the radial distance in the plane. Show that Bose-Einstein condensation now is possible, and find the critical temperature at which it occurs. Replacing the box with a simple harmonic oscillator changes the energy spectrum from

2 2  = E0(nx + ny)

3 to  = hf(nx + ny) with nx, ny = 0, 1, 2, 3,.... This will change the density of states in energy space. Of course the density of states in ~n space is still g(~n) = 1. To find the density of states in  we need to count how many different nx and ny values give the same . It is no longer true that  only depends on the magnitude of ~n. So this is a different counting problem. In fact, it’s even easier. For a given energy E = mhf with integer m, there are m + 1 possible states (pick nx to be any integer between 0 and m, there is exactly one ny so that nx + ny = m. So we have m + 1 choices of nx, ny pairs.) When we replace sums by integrals we are implicitly talking about large m and so the ”+1” is negligible. We find 1 · m m  g() = = = . d/dm hf (hf)2 Now we are in business, since our density of states again vanishes as  → 0. We can write the integral Z ∞  d Nexcited = . 0 (hf)2 β(−µ) − 1 This time we indeed can find the critical temperature by setting µ = 0 and demanding that at this critical point all particles are in excited states (Nexcited = N, N0 = 0): Z ∞ 2 2 2  d π k Tc N = 2  = 2 . 0 (hf)  kTc − 1 6(hf) This integral was easily done using Mathematica. We can solve for the critical temperature: s 6(hf)2N T = . c π2k2 While you weren’t asked to do this, it is also interesting to find the condensate fraction for T < Tc. As in 3d, we use the same formula for Nexcited at T < Tc, with µ ≈ 0: Z ∞  d π2k2T 2 T 2 Nexcited = 2  = 2 = N 2 0 (hf)  kT − 1 6(hf) Tc and so "  T 2# N0 = N − Nexcited = N 1 − . Tc

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