MASSACHUSETTS INSTITUTE OF TECHNOLOGY. Physics Department , Physics of Solids I. Problem set #11

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.3, Physics of Solids I Problem set #. Size of boson condensation in a trap Consider a as of bosonic sodium atoms confined in a quadratic potential well U(r = mω r where m is the mass of the sodium atom. The characteristic lenth of the oscillator potential is r = /mω = 5 3 cm. At zero temperature T =, the bosons form a condensed state. Here we like to study how the size of boson condensate depend on the number of bosons. (a Inore the interaction between the sodium atoms, find the size of the condensed sodium atoms at T =. How does the size of the condensation depends on the number of particles? (b For interactin bosons, the shape of condensation at T = is determined by m r + (U(r µ + ψ(r ] ψ(r = In Thomas-Fermi approximation, we assume the wave function ψ is smooth and drop the r term. In this case the shape of condensation is determined by (U(r µ + ψ(r ]ψ(r = Now, how does the size of the condensation depends on the number of particles? (Hint: µ is determined by requirin that the total number of bosons N = d 3 x ψ. Solution: (a For 3-D quadratic potential well, round state enery 3 ω. Without interaction, all the bosons condensate to the round state. Therefore the condensate size r satisfies that which ives mω r = 3 ω r = 3 = 3r cm mω actually any result that differs from above result by an order one factor is also correct. And the size is independent of particle number N. (b For interactin Bosons, the shape of condensation at T = is determined by m r + (U( r µ + ψ( r ] ψ( r = In Thomas-Fermi approximation, we drop the r term, (U( r µ + ψ( r ] ψ( r =

2 Thus the density of condensate is ψ( r = µ U( r Usin the normalization condition ψ( r d 3 r = N, we now have rmax 4πr dr µ mω r = N That is rmax 4πr dr µ mω r = N Noticin that when r = r max, ψ( r =, which means that µ = mω r max, therefore the size of the condensate is. Bunchin of bosons 4π 5 mω r 5 max = N ( 5N r max = 4πmω Interactin bosons have a special property that all the bosons prefer to occupy the same orbital (for example to have the same wave vector in order to lower the total enery. In other words, bosons prefer to share. In this problem, we will study such a phenomenon. The Bunchin phenomenon is closely related to the superfluidity of the condensed boson as. We have seen that free bosons have a vanishin total enery at T =. For interactin bosons, the enery will be non-zero. In the classical-field-theory (or oscillator description of interactin bosons, a state of bosons is described by a complex field φ(x, where φ(x corresponds to the boson density. The enery for a state described by φ(x is iven by U = /5 dx m xφ + φ 4] Here we consider a boson as in D. Let us assume the as contain N bosons which live on a circle of circumference L (a The round state (the minimal enery state of the D boson as is described by a constant field φ(x = φ. Find the value of φ in order for the field φ(x = φ to describe the boson as of N bosons. Find the total enery of the boson as. (Express the enery in term of the boson number N and the system size L. (b In the round state, all the bosons have a vanishin momentum. Now let us consider a state where all the bosons have the same but non-zero momentum p = k. Such a state is described by the followin field φ(x = φ e ikx. Find the value of the real constant φ in order for the field φ(x = φ e ikx to describe the boson as of N bosons. Find the total enery of the boson as. (Express the enery in term of the boson number N and the system size L. Note that k is quantized: k = π L inteer. (c Now let us consider a state where half bosons have the same but non-zero momentum p = k and half bosons have zero momentum. Such a state is described by the followin field φ(x = φ + φ 3 e ikx. Find the values of the real constants φ and φ 3 in order for the field

3 φ(x = φ + φ 3 e ikx to describe the boson as of N bosons. Find the total enery of the boson as. (Express the enery in term of the boson number N and the system size L. (d Explain why the enery difference between (b and (c in the small k limit is not zero and why (c even has a hiher enery then (b? You may want to sketch the density φ(x in (b and (c.] 3. Coherent lenth and size of vortex core ( pts Consider an interactin bosonic as in D. The Ginzbur-Landau free enery is iven by + ( A = dx m xψ + + U(x ψ + b 4 ψ 4 Where ψ is the amplitude of condensed bosons (the order parameter and a(t = a ( T T c for T near T c. Here a, b and m are constants. The external potential U(x has the followin form U(x x< = +, U(x x> = (a Show that there is a boson condensation for T < T c and find the amplitude of condensed bosons ψ(x for x +. (b Near x =, the amplitude of condensed bosons is suppressed by the potential U(x. To ain a more quantitative understandin of the suppression, we assume ψ(x to have a form ψ(x x< =, ψ(x <x<ξ = x ξ ψ(+, ψ(x ξ<x = ψ(+ We want to adjust ξ to minimize the total free enery for the above form of boson condensation. Calculate the ξ dependence of the free enery. (Hint: there is an infinite term in the free enery which is independent of ξ. Find the value of ξ that minimizes the free enery. (c Show that near T c, ξ diveres as ξ T c T ν. Find the critical exponent ν. (The lenth scale ξ is called the coherent lenth. It is a very important lenth scale in superfluid. For example, the size of the vortex core is iven by ξ. Solution: (a The Ginzbur-Laudau free enery is A = ( dx m xψ + + U(x ψ + b 4 ψ 4 where a(t = a ( T T and b, m are positive constants. ( δa = dx m ( xδψ x ψ + + U(x ψδψ + b ψ ψδψ + c.c. where c.c. means complex conjuate. The first term in δa is Therefore δa = dx m ( xδψ x ψ = m d(δψ x ψ = dx( m xψδψ ( dx m ( xψδψ + + U(x ψδψ + b ψ ψδψ + c.c. 3

4 Minimizin the Ginzbur-Laudau free enery, that is δa =, requires the coefficient of δψ to be zero, which leads to m ( xψ + + U(x ψ + b ψ ψ = At x +, ψ(x keeps constant, thus we can nelect the xψ term, (a(t + b ψ ψ = where we have takin into account that U(+ =. For T < T c, a(t <, therefore ψ has non-zero solution, which means condensation happens, and the amplitude of condensed bosons is ψ = a(t (. b (b Insert the expression of ψ(x into the interation to calculate Ginzbur-Laudau free enery, A = A + A + A 3, where with ψ = ψ(+, A = dx m xψ ξ = ψ m ξ = ψ mξ A = dx = A 3 = ξ dx a(t + U(x ψ x ξ ψ + ξ dx a(t ψ (. = a(t 6 ξ ψ + a(t ψ ( ξ = a(t 3 ξ ψ + a(t ψ (.3 = ξ dx b 4 ψ 4 b x 4 4 ξ 4 ψ 4 + ξ b 4 ψ 4 = b ξ ψ 4 + b 4 ψ 4 ( ξ = b 5 ξ ψ 4 + b 4 ψ 4 (.4 We can find that A and A 3 divere. That is because we have infinite volume and infinite number of bosons. However we can let the ξ-independent terms in free enery A equal an arbitrary constant A, thereafter A = ψ mξ a(t 3 ξ ψ b 5 ξ ψ 4 + A The value of ξ to minimize A can be obtained by lettin A/ ξ =, which ives, mξ = a(t b 3 5 ψ 4

5 Recall Eq. (., we have ξ = 5 4ma(T = 5T c 4ma (T c T (c At critical temperature, ξ diveres as ξ T c T ν, from part (b we can see ν = /. 5

8.04: Quantum Mechanics Professor Allan Adams Massachusetts Institute of Technology. Problem Set 5

8.04: Quantum Mechanics Professor Allan Adams Massachusetts Institute of Technology Tuesday March 5 Problem Set 5 Due Tuesday March 12 at 11.00AM Assigned Reading: E&R 6 9, App-I Li. 7 1 4 Ga. 4 7, 6 1,2