Homework 1: BEC in a trap

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Lynette White
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1 Homework : BEC in a trap Homework is due Tuesday September 9th. As a reminder, collaboration between students is encouraged, but everyone must turn in their own solutions. This problem is based on NJP 0, (008). In statistical physics class you learned about Bose-Einstein Condensation (BEC) in free space. However, in ultracold atom experiments condensation typically takes place in a harmonic trap as opposed to free space. In this problem we will investigate the difference between condensation in a trap and in free space. Part A: condensation in free space Bose-Einstein condensation occurs when the lowest energy level becomes macroscopically occupied, that is it contains a finite fraction of the total number of atoms. Suppose that the atom density of a d- dimensional system is!. () Show that the atom density that can be accommodated by excited states,! x, is! x = $d k ( ) d ( )(* k+,) + where * k = ħ k m. () For a Bose system,, must be smaller than the ground state energy, why? (3) Find maximum! x as a function of ) = in two and three dimensions. What is the difference k B T between the two answers? (4) What is the critical temperature for BEC to occur in 3D? Part B: condensation in a trap Now let us consider the situation in a d-dimensional trap, in which the single-particle Hamiltonian is: H D = p + m 0 m x x + 0 y y H 3 D = p + m 0 m x x + 0 y y + 0 z z () For the case of a cylindrically symmetric trap 0 x = 0 y = 0 0 (spherically symmetric in 3D), find the spectrum of the single particle states. What are the allowed energies? What is the degeneracy of each energy level? () Assume that k B T >> ħ 0 0. Assuming Bose-Einstein distribution, compute N x the maximum number of atoms that can fit into excited states (that is all states but the ground state) at a given temperature. Hint: use the assumption to transform the sum into an integral. Unlike the free space case, the total number of atoms in a trap must be finite. We will label a system as condensed if all atoms cannot fit into excited states. Show that both in D and 3D there is a critical temperature below which condensation occurs. (3) Bonus part: Relate the critical temperature you found in the previous part to the atom density in the trap. Show that in the 3D case as one relaxes the trap the critical density at the trap center

2 HW.nb trap. Show that in the 3D case as one relaxes the trap the critical density at the trap center approaches the free space result. Show that in the D case that there is a Logarithm in the relation between critical density and trap frequency, this logarithm results in the divergence of the critical density as at fixed temperature. [Part 3 can be done numerically for full bonus credit, however it is possible to obtain asymptotic relations analytically as well. Hint: Only wave functions with zero angular momentum contribute to density at r 3 0.] Solutions: Part A () To obtain the answer, we plug the expression for the energy of the k-th state into the Bose-Einstein distribution n = ( )(*+,) +. () More specifically, we are not allowed to have any state such that *-,=0, as the Bose-Einstein distribution function will diverge, i.e. the state will be occupied by an infinite number of Bosons, which is quite unphysical. (3) In D, the integral becomes: i = 0 Exp ) k m + k ( ) $k which is clearly divergent as can be observed from a power series expansion around k 3 0. Exp ) k m + k m ( ) ) k Hence all atoms can be accommodated by excited states. Series m π β k # Exp β k m $ k 4 π + O[k] π k, {k, 0, } ( π) In 3D, however, the integral is convergent and is related to the Zeta function: id = Simplify Integrate Exp β ħ k m $ #+ d β ħ m #d* Gamma d PolyLog d, π 4 π k d$, {k, 0, }, {β > 0, d >, m > 0, ħ > 0} 3 ( π)

3 HW.nb 3 i3 = id.. d / 3 Zeta 3 π 3* β ħ m 3* The critical temperature can be obtained by equating the integral to the atom density and solving for the temperature. Solve ρ i3.. β / kb T π ρ *3 ħ T. kb m Zeta 3 *3, T [[]] N[%] T ρ*3 ħ kb m Part B () The Hamiltonian is separable, and hence corresponds to (or 3 in 3D) copies of the Harmonic oscillator. Each oscillator has the energy spectrum * i = ħ 0 0 i +. Hence in D the spectrum becomes * i,j = ħ 0 0 (i + j + ), where i and j range from 0 to. In 3D * i,j,k = ħ 0 0 i + j + k + 3. Hence the allowed energy levels are D: * m = ħ 0 0 (m + ) with a multiplicity (m + ) m = 0,,,... 3D: * m = ħ 0 0 m + 3 with a multiplicity m + 3 m + m = 0,,,... () For the D case, the smallest possible value of, is,=. Plugging this value of,, we find that the maximum number of atoms in the excited states is m+ n x = m +) 00 m+ Log +( Exp[) ħ 0 0 m]+ $m = + ħ Exp[) ħ 0 0 m]+ ) 00 ħ Log[) 00 ħ] ħ ) 00 ħ ) + PolyLog,(+) 00 ħ ) 00 ħ For the 3D case, the smallest possible value of, is,=3/. Plugging this value of,, we find that the maximum number of atoms in the excited states is n x = m m + 3 m+ Exp[) ħ 0 0 m]+ m + 3 m+ Exp[) ħ 0 0 m]+ +) 00 Log +( $m = +3 ħ + 5 ) 00 ħ +) 00 ħ PolyLog,( +) 00 ħ + PolyLog 3,( Zeta[3] ) 00 ħ ) ħ ħ 3 ) 3 Plot of critical particle number as a function of temperature

4 4 HW.nb n x D 3D k B T ħ 0 0 (3) Relate the critical temperature you found in the previous part to the atom density in the trap. Show that for 3D, the relationship between T c and! is similar for trap and free space. 3D case We note that only the n=0 trap eigenfunctions contribute to the density at the trap center. Hence, we will find these eigenfunctions and compute the density. Solving for wave functions in the trap (we want n=0 only, as all others will have a node at r 3 0) DSolve $ ψ[r]. m ψ''[r] + r ψ'[r] + m ω r ψ[r] En ψ[r], ψ[r], r # m r ω C[] HermiteH En#ω ω, m r ω r # m r ω C[] HypergeometricF # En#ω,, m 4 ω r ω r We want the wave function to be well behaved at the origin, = n [0] 3 const. This boundary condition implies that the HermiteH functions are the only allowed ones. Moreover, En+0 = n +. Hence 0 E n = 0 n + 3, where n = 0,,,... + Next, we must normalize our wave functions 0 =n [r] 4 r $r = ( +m 0 r 0 HermiteH n +, m 0 r 4 r $r r = 4 m 0 0 ( +u HermiteH[ n +, u] $u = m 0 n+ ( n + )! = n [0] = + +n (m 0) 3?4?4 Gamma + +n Gamma[+ n]

5 HW.nb 5 after some work, we find that = n [0] n (m 0)3? = (for large n). Actually the approximation is decent even for small n. The approximation is plotted with solid line and the exact values of the wave functions at the trap center with the points...0 n [0] (m ) -3/ n Using this expression, we can easily find the density at the trap center 3! D n (m 0) x = 3? n 3? n (m 0) ( ) 0 n + 0 This exactly matches the free space result! D case = m ) 3? Zeta 3 ( ) 0 n + 3? Solve for wave functions in trap (again we want l=0 only, as all others will have a node at r 3 0) Simplify DSolve $ ψ[r]. m r ω m ψ''[r] + r ψ'[r] + m ω r ψ[r] En ψ[r], ψ[r], r, r > 0 C[] HypergeometricU En + ω ω,, #m r ω + C[] LaguerreL # En + ω ω, #m r ω We want the wave function to be well behaved at the origin, = n [0] 3 const. This boundary condition implies that the LaguerreL functions are the only allowed ones. Moreover, the wave function must be well behaved at r 3, lim r3 =[r] 3 0. Hence En+0 0 = n +, and E n = 0( n + ), where n = 0,,,... After playing with LaguerreL polynomials, we find that we can express the wave function as = n [r] = ( + m r 0 LaguerreL n, m r 0, where as before E n = 0( n + ) and n = 0,,,... Next, we must normalize our wave function r r r

6 6 HW.nb 0 =n [r] r $r = 0 ( + m r 0 LaguerreL n, m r 0 r $r = 4 m 0 0 u ( +u LaguerreL n, u $u = m 0 = n [r] = m 0 ( + m r 0 LaguerreL n, m r 0 Hence, = n [0] = m 0 (LaguerreL[n, 0]) = m 0 Using this expression, we find the density at the trap center! x D = n m 0 m 0 ( ) 0 n + = m 0 + ( ) 0 n + Log ++( ) 0 ) 0 + m Log[ ) 0] ) Hence, as 0 3 0, we find that! x D 3 0 as well. Which is consistent with the absence of Bose-Einstein condensation in free space.

The non-interacting Bose gas

Chapter The non-interacting Bose gas Learning goals What is a Bose-Einstein condensate and why does it form? What determines the critical temperature and the condensate fraction? What changes for trapped