Download lecture 4:Sisyphus cooling, evaporative cooling, and magnetic

Degenerate Quantum Gases 2006
lecture 4:Sisyphus cooling, evaporative cooling, and
magnetic trapping
Jani-Petri Martikainen
[email protected]
http://www.helsinki.fi/˜jamartik
Department of Physical Sciences
University of Helsinki
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.1/43
Sisyphus cooling
Earlier we found the Doppler limit kB TDoppler = ~Γ2 e as
expressed by the excited state decay rate of the
two-level atom.
For sodium atoms this is ~Γe = 480 µK
experiments soon found out that temperatures they
could reach with the same setup were actually lower
than the Doppler limit!
Also, these temperatures were not reached at the
detuning |δ| = Γe /2 as predicted by the two-level theory
for Doppler cooling
Something beyond the simple Doppler picture was
clearly going on.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.2/43
Sisyphus cooling
understanding of these phenomena was quickly
reached and the lowest attainable temperature was
found to scale with the recoil energy
kB T ∼ Er = (~q)2 /2m.
(1)
This is the kinetic energy atom gains after absorbing a
photon.
For sodium in a laser light with λ = 1000 nm this gives
Tmin ∼ 400 nK.
orders of magnitudes lower than the Doppler limit and
rather close to the quantum degenerate regime.
Note ∼-sign. There can be a large numerical coefficient
of 30 in front!
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.3/43
Sisyphus cooling
Doppler cooling ignored two important ingredients
1. The radiation field produced by two counter
propagating lasers is inhomogeneous.
2. Real atoms are not two-level systems:in the absence
of magnetic fields alkali atoms have degenerate
ground states
assume that the beam propagating towards positive z is
linearly polarized along x-axis and the other beam
towards negative z is polarized along y -axis.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.4/43
Electric field polarization
Total electric field is of form
E(z, t) = E(z)e−iωt + E ∗ (z)eiωt
(2)
where the field vector
iqz
E(z) = E0 êx e + êy e
√
=
2E0 eiqz ˆ(z).
−iqz
= E0 e
iqz
êx + êy e
−2iqz
(3)
It is clear that the polarization ˆ(z) of the light field
varies in space.
For z = 0 the field is linearly polarized along
the
axis
45o
√
tilted from the x-axis i.e. ˆ(0) = (êx + êy )/ 2
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.5/43
Electric field polarization
√
For z = λ/4 we have ˆ = (êx − êy )/ 2 and the field is
linearly polarized along −45o tilted axis.
√
For z = λ/8 we get ˆ = (êx − iêy )/ 2 and the field is
circularly polarized in the negative sense (σ− ) about the
z -axis.
For z = 3λ/8 we get circularly polarized field in the
positive sense (σ+ )
The polarization state everywhere can be expressed as
a superposition of the two
√ circular polarization states
with vectors (êx ± iêy )/ 2
one finds that the intensities of the two polarization
components vary as (1 ∓ sin 2qz)/2.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.6/43
Polarization gradient
Superposition of the counter propagating fields gives
rise to the position dependent polarization of the
electric field... polarization gradient.
lin
σ−
lin
σ+
lin
σ−
E0
E0
y
z=0
z=λ/8
z=λ/4
z=3λ/8
x
z=λ/2
z
See movie as well...
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.7/43
Atomic structure
Assume an atom with a doublet ground state Jg = 1/2
Ground state manifold is coupled by the radiation field
to the quadruplet (Je = 3/2) excited state manifold.
−3/2
1
−1/2
1/2
3/2
2/3
2/3
1/3
g−
g+
1
Two degenerate ground
states and four excited
states.
Squares of the
Clebsch-Gordan coefficients
for the Jg = 1/2 → Jg = 3/2
transition are also indicated
in figure.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.8/43
Atom in the light field
The presence of the radiation field will shift the energies
of the two ground states g± .
These shifts have two-components one from each of
the polarization components of the light field.
due to the selection rules each ground state is coupled
with 3 excited states.
For example, σ+ field can couple |1/2, 1/2i ground state
to |3/2, 3/2i excited state, σ− couples it to |3/2, −1/2i
state, while linear polarization field can couple to
|3/2, 1/2i state.
The relative strength of these couplings are given by
the associated Clebsch-Gordan coefficients.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.9/43
Atom in the light field
A transition in a field of some polarization contributes to
the total energy shift with a factor that is a product of
the intensity of the polarization component and the
square of the Clebsch-Gordan coefficient.
polarization gradient implies that transition probabilities
are position dependent
Since the intensities of the polarization states vary in
space, also the energy shifts are position dependent.
In the end we would get Stark shifts for the ground
states
V ± (z) = V0 (−2 ± sin 2qz) ,
(4)
where V0 is a constant that we could calculate if we
wish.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.10/43
Atom in the light field
Where dressed |g+ i state has an energy maxima,
dressed |g− i has an energy minima.
|g± i atoms experience an optical lattice
Energy shifts
−1
−1.2
g+
−1.4
−1.6
−1.8
−2
−2.2
−2.4
−2.6
g−
−2.8
−3
0
0.2
0.4
0.6
0.8
1
z/λ
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.11/43
Optical pumping
We still must understand how atoms are transferred
between the states g+ and g− . Consider a point where
the field is circularly polarized in a positive sense.
1. For such polarization state g+ with m = 1/2 can only
couple to an excited state with m = 3/2. This excited
atom can again only decay back to the g+ state since
all other transitions are forbidden in the dipole
approximation.
2. By contrast g− ground state with m = −1/2 can be
excited to m = 1/2 excited state. As can be seen in
the level diagram this state can decay back to g− OR
to the g+ state.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.12/43
Optical pumping
The related weights for g− transitions are given by the
squares of the Clebsch-Gordan coefficient i.e. 1/3 and
2/3 respectively.
Therefore, the excited state at m = 1/2 is more likely to
spontaneously decay into the g+ state. The net effect is
therefore to pump atoms from g− into g+ which has at this
location lower energy than g− .
The argument works the similarly in locations where the
field is circularly polarized in the negative sense. Atoms
are then pumped from g+ into g− which at this location
has a lower energy than g+ .
with linear polarization there is no net pumping since
CG-coefficient is higher for the transition back to where
we came from...
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.13/43
Atomic cooling
In total, atomic transitions therefore tend to transfer atom
from the top of the hill to the bottom of the hill.
For a moving atom this means cooling.
1. The atom moves up hill and loses kinetic energy in
the process. It reaches the top of the hill and is just
about to start gliding down in order to convert the
gained potential energy back into kinetic energy.
2. Pumping, however, intervenes and suddenly
transfers the atom to the other ground state, into a
state which is at the potential minima.
3. The previously gained potential energy was
irreversibly carried away by the spontaneously
emitted photon.
4. The atom now starts its climb up the next hill and so
and so on and so on....
˜
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/ quantumgas/ – p.14/43
Sisyphus cooling
This mechanism is referred to as Sisyphus cooling since it
reminded Dalibard and Cohen-Tannoudji of the Greek myth
in which Sisyphus was condemned to an eternal
punishment in Tartarus, where he forever had to push a
heavy rock up a steep hill.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.15/43
Limits of Sisyphus cooling
In Sisyphus cooling the lowest attainable temperature is
set by the recoil energy ER = (~q)2 /2m multiplied by a
fairly large numerical factor (30).
This corresponds to that kinetic energy kick which the
spontaneously emitted photon, on its way out, delivered
to the atom.
It is also an energy scale for the center of mass of an
atom localized in a box of size ∼ λ... I.e. cooling stops
when the atom wavefunction is so spread out that “it
cannot distinguish lattice maxima from its minima
anymore”.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.16/43
Beyond the recoil limit
Even this limit can and has been broken by several
orders of magnitude using the so-called
velocity-selective coherent population trapping
(VSCPT) or so-called Raman cooling.
The interested reader is instructed to read some review
article on laser cooling or Metcalf & van der Straaten
book to gain further understanding.
For example, download L. Guidoni and P. Verkerk, PhD
tutorial in J. Opt. B:Quantum Semiclass. Opt. 1
R23-R45 (1999)
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.17/43
Evaporative cooling
Lower the trap depth and
let the fast atoms escape.
Remaining atoms will then
rethermalize through good
elastic collisions.
Collisions are important. For example, evaporative
cooling of identical fermions will become ineffective at
lower temperature since their scattering cross-section is
reduced at lower temperatures (fermionic property)
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.18/43
Evaporative cooling
For bosons, cross-section does not vanish, and bosonic
stimulation (i.e. transition rate increases with the final
state population... opposite to Pauli blocking) ends up
making the evaporative cooling even more effective.
Q: What should we use for trap depth V (t) as a function
of time?
A: Optimization problem that depends on densities,
collisional properties etc...
Reach a low temperature with as much atoms as
possible remaining trapped.
Evaporation takes us to the BEC regime!
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.19/43
Atomic structure
Nuclear spin I (odd multiple of 1/2 for bosons) and
electronic spin J = S + L
For alkali atoms electrons have no orbital angular
momentum so L = 0 and J = S = 1/2.
Coupling between nuclear spin and electronic spin
yields two possibilities for the total spin quantum
number: F = I ± 1/2.
Levels split by hyperfine interaction (A is constant, I and
J are nuclear and electronic angular momentum
operators)
Hhf = AI · J
(5)
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.20/43
Atomic structure
Total angular momentum operator is F = I + J
Square of this gives a way to express I · J in terms of
the quantum numbers I , J , F , determining the squares
of the angular momentum operator
1
I · J = [F (F + 1) − I(I + 1) − J(J + 1)]
2
(6)
alkalis have J = S = 1/2 so the splitting between
F = I + 1/2 and F = I − 1/2 becomes
∆Ehf = E(F = I + 1/2) − E(F = I − 1/2) = (I + 1/2) A
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.21/43
Level structure in a magnetic field
In B = 0 the states are 2F + 1-fold degenerate
Magnetic field lifts this degeneracy
Magnetic field in the z -direction...
Hspin = AI · J + CJz + DIz
(7)
C = gµB B and D = −µB/I
Since nucleus is heavy compared to electron D is
typically much smaller than C and then the g-factor
g≈2
For BEC:s we have often I = 3/3 and S = 1/2, so lets
focus to that
Other possibilities can be dealt similarly
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.22/43
Level structure in a magnetic field
Task: Diagonalize Hspin in a basis of eight states
|mI = 0, ±3/2, mJ = ±1/2i.
We define the raising and lowering operators
I± = Ix ± iIy and J± = Jx ± iJy and use the identity
1
I · J = Iz Jz + [I+ J− + I− J+ ] .
2
These raising and lowering operators operating on
some eigenstates |mI i of nuclear spin I result in
p
I+ = |mI i = (I − mI )(I + mI + 1)|mI + 1i
(8)
(9)
and
p
I− = |mI i = (I + mI )(I − mI + 1)|mI − 1i.
Department of Physical Sciences, University of Helsinki
(10)
http://theory.physics.helsinki.fi/˜quantumgas/ – p.23/43
Level structure in a magnetic field
...similarly for the eigenstates of S (or J ). These
relations make the algebra straightforward.
Since [H, Fz ] = 0 the z -component of the total spin is
conserved
This follows immediately from the standard angular
momentum commutation relations)
Therefore, it only couples states with same value of
mI + mJ , since raising of mJ by 1 must be accompanied
by the lowering of mI by 1.
splits into 2 × 2 blocks (at most) since the state
|mI , −1/2i is only coupled to |mI + 1, 1/2i
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.24/43
Level structure in a magnetic field
The states |3/2, 1/2i and | − 3/2, −1/2i do not mix (or
“couple”) with any other states and therefore their
energies can be calculated easily.
The result is
3 1
3
1
3
E( , ) = h3/2, 1/2|Hspin |3/2, 1/2i = A + C + D
2 2
4
2
2
3 1
3
1
3
E(− , − ) = A − C − D
2 2
4
2
2
linear in the magnetic field
(11)
6 more to solve...
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.25/43
Level structure in a magnetic field
If mI + mJ = 1 the relevant coupled states are
|3/2, −1/2i and |1/2, 1/2i.
The matrix elements of the Hamiltonian for these states
!
√
3
− 34 A −√12 C + 34 D
2 A
.
(12)
3
1
1
1
2 A
4A + 2C + 2D
Eigenvalues for this matrix are given by
r
3 2 1
A
A + (A + C − D)2 .
E =− +D±
4
4
4
(13)
4 solutions to left...
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.26/43
Level structure in a magnetic field
If mI + mJ = −1 the relevant states are | − 3/2, 1/2i and
| − 1/2, −1/2i, the matrix can be formed from the earlier
result by the mapping C → −C , D → −D.
2 solutions to left...
remaining block is for states with mI + mJ = 0 i.e. states
|1/2, −1/2i and | − 1/2, 1/2i.
The matrix is given by
− 14 A + 12
(C − D)
A
A
− 14 A − 12 (C − D)
Department of Physical Sciences, University of Helsinki
!
.
(14)
http://theory.physics.helsinki.fi/˜quantumgas/ – p.27/43
Level structure in a magnetic field
... eigenvalues
A
E=− ±
4
r
A2
1
+ (C − D)2 .
4
(15)
All 8 energy levels are now solved!
In the limit B → 0 we find (again) the 5-fold degeneracy
of the F = 2 states and the 3-fold degeneracy of the
F = 1 states.
Since |D| |C| let us look at limit D = 0 and g = 2.
define the dimensionless magnetic field through
(2I + 1) µB B
C
=
b=
A
∆Ehf
Department of Physical Sciences, University of Helsinki
(16)
http://theory.physics.helsinki.fi/˜quantumgas/ – p.28/43
Level structure in a magnetic field
energy levels are
3 1
E( , ) = A
2 2
3 1
E(− , − ) = A
2 2
3 b
+
4 2
,
(17)
(18)
3 b
−
4 2
and for mI + mJ = ±1
!
r
1
1
3 1
2
E=A − ±
+ (1 + b) ≈ A − ± (1 + b/4)
4
4 4
4
(19)
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.29/43
Level structure in a magnetic field
...and
1
E=A − ±
4
r
3 1
+ (1 − b)2
4 4
!
1
≈ A − ± (1 − b/4)
4
(20)
and for mI + mJ = 0
!
r
2
1
1
b
2
≈ A − ± 1 + b /8 . (21)
E =A − ± 1+
4
4
4
The approximate results are Taylor expansions with
small values of b i.e. |µB B| ∆Ehf .
This is relevant since, many experiments are carried out
in relatively small magnetic fields.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.30/43
Energy levels with I = 3/2 and J = 1/2
Hyperfine levels with I=3/2 in a magnetic field
4
3
2
E/A
1
0
−1
−2
−3
−4
0
1
2
3
4
5
b
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.31/43
Magnetic traps
In the regime which is linear in b some states have an
energy minima at small magnetic fields while for others
energy increases as the magnetic field becomes
smaller.
While spatially varying magnetic fields can be created
with ease, it is impossible to create a local maximum of |B| in
regions without electrical currents (i.e. inside the vacuum
chamber)
Of particular importance are the doubly polarized state
|mI = I, mJ = 1/2i and the maximally stretched state
with quantum numbers F = I − 1/2 and mF = −F in
zero magnetic field.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.32/43
Magnetic traps
Their importance stems from the fact that these states
have a negative magnetic moments µi and therefore
minimize their energy Ei = −µi B + constant at small
values of B
Atoms experience a force F = µz ∇B , which points
toward the field minimum whenever the magnetic
moment µz is negative. These states are called
weak-field seekers.
For magnetic trapping the sample must be prepared in
the weak-field seeking states.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.33/43
Magnetic traps
There are several types of magnetic traps, but for
concreteness I will just explain the principle behind
quadrupole traps and their extension into so called TOP
traps.
Maxwell’s third law ∇ · B = 0
One possible position dependent field configuration
satisfying this constraint is the quadrupole configuration
B = B 0 (x, y, −2z) .
(22)
Magnitude
of the magnetic field is then
p
B = B 0 x2 + y 2 + 4z 2 and it varies linearly with
distance.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.34/43
Magnetic traps
An important disadvantage for the quadrupole trap is
that the magnetic field vanishes at the origin.
This is a problem, because we want atoms to remain in
the same quantum state and in particular we do not
want the atoms to make transitions to states which are
high-field seeking and are therefore repelled away from
the trap.
An atom moving in the inhomogeneous magnetic field
sees a time-dependent magnetic field . If this variation
is “slow” atom can follow the changing magnetic field
and remain in the same quantum state relative to the
instantaneous direction of the magnetic field.
However, since atom does see time dependent
magnetic field this will induce transitions between
different states.
˜
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/ quantumgas/ – p.35/43
Magnetic traps
The question is: When are the transitions so frequent
that they pose a problem?
In turns out, that since the splitting between magnetic
sublevels are of order µi B , transition rates increases as
B approaches zero.
The quadrupole trap therefore has a hole in the origin where
atoms are lost from the trap.
This problem can be circumvented by plugging the hole
with a laser beam of appropriate frequency or by adding
some time-dependence to the magnetic field
configuration of the quadrupole trap.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.36/43
TOP trap
In the TOP (time-averaged orbiting potential) trap the
magnetic field is given by
0
0
0
B = B x + B0 cos ωt, B y + sin ωt, −2B z .
(23)
The magnetic field still has a zero somewhere, but this
zero is moving around.
It appears plausible that if the hole moves around
quickly enough, atom sees a time averaged trapping
potential without a hole.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.37/43
TOP trap
If the frequency ω is close to the frequencies of
transitions between magnetic substates, the time
varying field could induce transitions to non-trapped
states. To avoid this possibility ω is chosen much lower
than the transition frequencies (which are more than
1 M Hz ).
In this way atom stays in the same quantum state
relative to the instantaneous magnetic field
B(t) ≈ B0 + B 0 (x cos ωt + y sin ωt)
(24)
i
B 02 h 2
+
x + y 2 + 4z 2 − (x cos ωt + y sin ωt)2
2B0
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.38/43
TOP trap
Averaging this over one period from t = 0 to t = 2π/ω
we find
B 02 2
2
2
x + y + 8z
hBi ≈ B0 +
(25)
4B0
The hole has indeed disappeared and we got a
magnetic field enabling a harmonic trapping of atoms in
weak-field seeking states!
Note that trap aspect ratio ωz2 /ωx2 = 8... Use something
else than TOP to have more flexibility.
the potential energy felt by the atoms is not only due to
the magnetic field, but also gravitation can play role.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.39/43
Magnetic trapping and gravity etc etc...
Gravitational potential is linear in z and if the potential
due to the magnetic field can be approximated as
parabolic it will shift the position of the potential energy
minima away from the origin.
Learn more about magnetic bottles and Ioffe-Pritchard
traps from the literature or WWW
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.40/43
Exercise 1
1. Solve the hyperfine levels structure for a fermion with a
nuclear spin I = 1 and electronic spin J = 1/2 in a
magnetic field. This is valid for 6 Li, which is commonly
used in studies with degenerate fermion gases. Give
the energy levels in units of A from AJ · I term in the
Hamiltonian H = AJ · I + CJz + DIz , with C = gµB B and
D ≈ 0.
Given that the 6 Li hyperfine splitting is known to be
∆Ehf /h = 228 Mhz, what is the value of the constant A
in SI units? Can the states on corresponding to the
lower energy hyperfine manifold be trapped in a
magnetic trap? (Justify your answer...)
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.41/43
Exercise 2
1. Find out from the literature some typical values for the
depths of the optical potentials. Give your answers in
the units of µK . Give the typical values for at least
these people...
W. Ketterle in Phys. Rev. Lett. 1998
I. Bloch in Nature 2002
J. Dalibard in Phys. Rev. Lett. 2004
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.42/43
Next week
Two-body interaction between atoms
Microsocopic theory of a weakly interacting Bose gas.
Department of Physical Sciences, University of Helsinki
http://theory.physics.helsinki.fi/˜quantumgas/ – p.43/43