# 2nd_Report_Many_body-Gross_Pitaevskii (2014)

Trabajo InglésUniversidad | Universidad de Barcelona (UB) |

Grado | Física - 4º curso |

Asignatura | Mecànica Quàntica d'N-cossos i Sistemes Ultrafreds |

Año del apunte | 2014 |

Páginas | 5 |

Fecha de subida | 04/08/2014 |

Descargas | 0 |

Subido por | xoc |

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Many Body Quantum Mechanics and Ultracold Systems
2nd Assignment
——–
1
The Gross-Pitaevskii equation: introduction
Let us consider N identical bosons trapped by an external potential Vext. (r ), in equilibrium at
a temperature of T
0 K. If there is no interaction between them, they will be in the ground
state of the external potential, but if the bosons interact one-to-one via a potential V (ri − rj ),
then this interaction will substantially modify the structure of the condensate. The problem
consists in the ground state computation of the hamiltonian
N
2
−
H=
i=1
2m
∇2i + Vext. (r ) +
1
2
V (ri − rj ).

i
(1.1)
i=j
Generally, it is very difficult to solve the problem in an exact fashion. A first attempt is try
to find an approximate solution for the ground state. To do so, we proceed with the variational
method. Since we are dealing with bosons, a good trial state is the totally symmetric wave
function Ψ(r1 , r2 , . . . , rN ) = ψ(r1 )ψ(r2 ) . . . ψ(rN ) = i ψ(ri ), in which all bosons are in the
same single-particle state. The idea is to find the function ψ which minimizes the functional
E[ψ] = Ψ|H|Ψ , with the condition Ψ|Ψ = 1.

Once minimized, one finds that ψ satisfies the equation:
2
−
2m
∇2 ψ(r ) + Vext. (r )ψ(r ) + (N − 1)
d3 r V (r − r )|ψ(r )|2 ψ(r ) = µψ(r )
(1.2)
(µ is the chemical potential). The equation above is a non-linear Schr¨odinger equation (Hartreelike type), where every boson feels an external potential besides an average mean field potential
created by the other N − 1 bosons.

If we consider a pseudopotential of the form
V (r − r ) =
4π
2a
S
m
δ(r − r )
with the normalization condition ψ|ψ =
Pitaevskii (G-P) equation
2
−
2m
(aS is the scattering length),
(1.3)
d3 r|ψ(r )|2 = 1, equation (1.2) becomes the Gross-
∇2 ψ(r ) + Vext. (r )ψ(r ) +
4π
2a
S
m
N |ψ(r )|2 ψ(r ) = µψ(r ).

(1.4)
Notice that aS > 0, and the potential V (r) therefore induces a repulsive interaction (instead
the system is bound! This is thanks to the confinement potential). The energy functional per
particle (from which the equation above is derived), is written as
e[ψ] =
E[ψ]
=
N
d3 r
2m
|∇ψ(r )|2 + Vext. (r )|ψ(r )|2 +
1
2π
2a
S
m
N |ψ(r )|4 .

(1.5)
Actually, the theoretical description of the experiments on Bose-Einstein condensation (BEC)
is based on the G-P equation. The bosons are confined by a magnetic field, typically modeled
by a harmonic oscillator potential (h.o.) ∼ (1/2)mω 2 r2 . It is often helpful to write all the expressions in h.o. units: r ≡ ar1 , E ≡ ωE1 with a =
/(mω). Then, the functional (1.5) reads:
e1 [ψ1 ] =
E1 [ψ1 ]
=
N
d3 r1
1
1
|∇1 ψ1 (r1 )|2 + r12 |ψ(r1 )|2 + 2π¯
aS N |ψ1 (r1 )|4 .

2
2
(1.6)
Once we minimize this with respect to ψ1∗ , we recover the G-P equation, now scaled in units
of the h.o.,
1
1
aS N |ψ1 (r1 )|2 ψ1 (r1 ) = µ
¯ψ1 (r1 ).

(1.7)
− ∇21 + r12 + 4π¯
2
2
The various contributions to the energy are:
ekin. =
1
d3 r1 |∇1 ψ1 (r1 )|2 ,
2
eh.o. =
1
d3 r1 r12 |ψ(r1 )|2 ,
2
d3 r1 |ψ1 (r1 )|4 , (1.8)
eint. = 2π¯
aS
with the constraint imposed by (1.7): ekin. + eh.o. + 2eint. = µ
¯. Moreover, analogously to problem
8 of the set of problems (solved in class), we can find the virial theorem for this system. One
finds: 2(ekin. − eh.o. ) + 3eint. = 0.

2
Results and Conclusions
In this exercise we study some of the physical quantities of a condensate of weakly interacting
and confined bosons at low density. In particular, we consider such a condensate made of 87 Rb.

To do this, given a set of parameters, such as the number of particles, the scattering length,
etc., we numerically solve equation (1.7) using the imaginary time method.

Due to the spherical symmetry of the problem, and making use of the fact that all the atoms
are in the ground state ( = 0), we move from a 3-dimensional problem to a 1-dimensional one
via the ansatz (ignoring the subindex 1) ψ(r) = (R(r)/r)Y00 . The equation we want to solve
finally becomes:
1 2
R(r) 2
1 d2
+
r
+
a
¯
N
R(r) = µ
¯R(r).

(2.1)
−
S
2 dr2 2
r
a) Using the fortran’s program, and introducing the same input’s there on virtual campus,
we obtained the following values:
N
ekin.

eh.o.

eint.

102
103
104
105
106
0.65650
0.43752
0.24036
0.12367
0.06122
0.8599
1.3670
2.9770
7.2380
18.0600
0.1356
0.6199
1.8240
4.7430
12.0000
ei
1.6520
2.4247
5.0415
12.1040
30.1200
µ
¯
1.7875
3.0445
6.8659
16.8470
42.1190
Let us analyze the data. Clearly, the kinetic energy decreases as N increases. Indeed, as
there are more atoms confined in the same region, these have less phasic space to move in.

2
On the other hand, if N increases, the interaction term associated to the average mean
field potential is higher, and since this is a repulsive interaction, the atoms explore greater
distances. This fact has as a consequence an increase in the energy associated to the
confinement potential (the h.o. potential). We have already commented on the relation
µ
¯ = ei + eint. , which, according to what we have argued so far, implies a higher chemical
potential for a larger number of atoms. Intuitively, the higher N is, the more strongly
bound the system becomes and thus, the more energy is required to extract a particle.

b) The Thomas-Fermi (T-F) approximation consists in ignoring the kinetic term, i.e. in
removing the second derivative in (2.1). Of course, in this limit we would hope for a null
kinetic contribution to the total energy, as is observed:
N
ekin.

eh.o.

eint.

102
103
104
105
106
0
0
0
0
0
0.4722
1.1470
2.8620
7.1800
18.0300
0.2863
0.7498
1.9010
4.7840
12.0200
ei
0.7585
1.8969
4.7631
11.9640
30.0520
µ
¯
1.0448
2.6468
6.6640
16.7480
42.0710
Contrasting with a), the obtained results simply manifest that the higher the number
of particles is, the more successful the approximation is. In a system with higher N , the
interaction terms, both the confinement term and the term associated to the average mean
field created by the rest of the atoms, (according to the table in the first section and to the
reasonings followed above) become dominant, such that the kinetic term can be neglected.

c) At the end of the report one finds two graphics where the function ρ(r) = |R(r)/r|2 is
represented for N = 103 and N = 105 , hence considering both the kinetic term (G-P line)
and the Thomas-Fermi approximation (T-F line). As can be seen, the agreement between
the two lines is better for N = 105 , in accordance with the previous section. The reason
for which the density profile becomes extended for higher N values has been addressed
previously (the repulsion is higher).

Using this approximation we can find an analytic expression for the density ρ(r). Disregarding the second derivative:
√
θ( 2¯
µ − r)
1
1 2
r +a
¯S N ρ(r) = µ
¯ =⇒ ρ(r) =
µ
¯ − r2 .

(2.2)
a
¯S N
2
2
The step function which appears in the analytic expression becomes manifest in the fast
decay presented by the T-F lines for large r. We can calculate the maximum radius Rmax. =
√
2¯
µ in each case, which is effectively where the densities become zero, as being 2.30 and
5.78, respectively. Using the normalization condition r2 drρ(r) = 1, we immediately
see that µ
¯ = (1/2) (15¯
aS N )2/5 . Moreover, by means of the thermodynamic relation µ
¯=
∂E1 /∂N , we get: e1 = ei = (5/7)¯
µ. We thus elaborate the following table which does
not need to be commented on (it speaks for itself) (¯
aS = 0.00433),
3
N
µ
¯
G-P
√
2¯
µ
103
3.0445
16.8470
2.4676
5.8047
105
µ
¯
T-F/Analytic
√
2¯
µ
2.6468/2.6546
16.7480/16.7495
2.3008/2.3042
5.7876/5.7878
ei
2.4247
12.1040
ei
1.8969/1.8962
11.9640/11.9639
d) Let us see whether the obtained values in the first section verify the virial theorem. To do
so, we compute the combination 2(ekin. − eh.o. ) + 3eint. , which, the virial theorem assures
us is zero. For the different values of N we have considered, we have:
N
102
103
104
105
106
2(ekin. − eh.o. ) + 3eint.

−6 · 10−5
−3 · 10−6
−3 · 10−8
−3 · 10−11
−9 · 10−6
As we see, the table above shows that the virial theorem is perfectly satisfied.

4
N = 103
G-P
T-F
0.6
ρ(r) (a−3 )
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
r (a)
N = 105
0.04
G-P
T-F
0.035
ρ(r) (a−3 )
0.03
0.025
0.02
0.015
0.01
0.005
0
0
1
2
3
4
5
r (a)
5
6
7
8
9
...