View source for Physics of the dipole force: fields and energy

=== Four physical pictures of the dipole force ===

Let us recall four different physical pictures for deriving the dipole force:

<blockquote>

* Optical Bloch equations.
* Dressed atom picture.
* Classical dipole model of atom, in electric field.
* Macroscopic dielectric object.

</blockquote>

==== Classical dipole ====

The classical dipole model is based on the fact that the force seen by
the dipole depends on whether the electric dipole is in-phase or
out-of-phase with the driving field <math>E</math>.  This gives either an
attractive or repulsive force.

==== Macroscopic dielectric object ====

The picture of a macroscopic dielectric object assumes, for example,
that you have a microscopic sphere of glass, with a laser beam which
is stronger in part of space than another:
::[[Image:dipforce2-dfig1.png|thumb|500px|none|]]
So if the microsphere were an atom, with index of refraction greater
than one, then the sphere moves toward the field maximum.  This is
because if you imagine the microsphere is a prism (with index greater
than one), then the upper beam is deflected downwards and the lower
beam is deflected upwards, such that the sphere acts like a lens.  The
sphere then is forced to move to compensate for the imbalance in the
change of photon momentum of the two refracted beams.

Optical tweezers have made an impact on biology almost as large as the
impact of laser cooling on atomic physics.  With optical tweezers, one
can now manipulate objects within a cell.  The people who demonstrated
optical tweezers first were the same people who demonstrated laser
cooling first: Art Ashkin, who teamed up with Steven Chu, at Bell
Labs, in the late 1980's.

=== Electric and magnetic components of the dipole force ===

Now a question.  We have discussed the stimulated light force, also
called the dipole force, or the reactive force.  Is this force an {\em
electric} force (the force on an electric dipole), a magnetic
force (Lorentz force), or both?  In other words, imagine an atom is in
the potential of the standing wave in an optical lattice.  Is the
potential due to electric forces or magnetic forces?  Intuition says
that the force is electric, because the electric force is generally
larger than the magnetic one.  However, that turns out to be
incorrect, and indeed it is the Lorentz force which gives rise to the
electric dipole potential.  At this point, you may completely
appreciate the electric dipole potential, since by writing
<math>\vec{d}\cdot \vec{E}</math> you actually include magnetic forces.

==== Static fields ====

In the limit of DC fields, everything does have to be electric, so let
us begin with electrostatics.  Consider two charges separated by
distance <math>d</math>:
::[[Image:dipforce2-dfig2.png|thumb|400px|none|]]
The energy of this configuration is <math>\alpha E^2/2</math>, due to the
polarizability of the atom.  Now imagine the charge is placed in an
inhomogeneous field:
::[[Image:dipforce2-dfig3.png|thumb|500px|none|]]
There is a force on the two halves of the dipole, and the two would
cancel unless there is a gradient, so the force on the dipole is
:<math> 
q \frac{\partial E_z}{\partial z} d
</math>
so the work done is
:<math> 
\frac{qd E_z}{2}
\,.
</math>
When we move the dipole not along the electric field, but rather,
perpendicular to the electric field, consider what happens:
::[[Image:dipforce2-dfig3a.png|thumb|500px|none|]]
Now, what is
the force on the dipole?  What matters is the <math>\hat{x}</math> component
of the force, which is caused by the <math>\hat{x}</math> component of the
electric field, or more specifically, it matters that the <math>\hat{x}</math>
component changes by the distance separating the two charges in the
dipole.

Now, we want to use the fact that in electrostatics, the curl of the
electric field is zero, and therefore by vector calculus,
:<math> 
\left|q\frac{\partial E_z}{\partial x} d \right| =
\left| q \frac{\partial E_z}{\partial z} d \right|
</math>
and the work done by moving the dipole in from infinity is <math>q d
E_z/2</math>.  This derivation relied on there being no magnetic fields,
because otherwise the curl of the electric field would not be zero.

==== Oscillating fields ====

Now consider an oscillating dipole in a standing wave, created by
counter-propagating laser fields, so that the electric field is
perpendicular:
::[[Image:dipforce2-dfig4.png|thumb|400px|none|]]
Recall that the potential from the AC Stark shift is <math>\alpha E^2/2</math>.
Let us understand what the force is, which provides this periodic
potential.  Well, if you have a plane standing wave, that means you
can draw the above picture extended up and down.  What is the electric
force on the dipole in that case?  The dipole moment and the gradient
of the electric field are orthogonal, in this configuration.  There is
thus no gradient in the field along the direction of the dipole.
Thus, in a plane standing wave, the electric force is identically
zero, <math>\vec{p} \perp \vec{\nabla}E</math>.  So where does the force come
from?

It must originate from the Lorentz force.  In this scenario, recall
that the Lorentz force is <math>\vec{v}\times\vec{B}</math>.  There is an
oscillating dipole in the scenario.  If you take the derivative of the
optical dipole potential, the Lorentz force is
:<math> 
F_{Lorentz} = \frac{d}{dz}\left(\frac{1}{2} \alpha E^2 \right)
\,.
</math>
The standing wave has both electric and magnetic components,
:<math>\begin{align} 
\vec{E} &= A_0 \hat{e}_x \cos kz \sin \omega t
\\ \vec{B} &= -A_0 \hat{e}_y \sin kz \cos \omega t
\end{align}</math>
The dipole potential, averaged over a cycle, is
:<math> 
\frac{1}{2} \alpha \bar{E}^2 = \frac{1}{4} \alpha A_0^2 \cos^2 kz
\,.
</math>
Recall that the dipole moment is <math>p=\alpha E</math>.  Thus, the Lorentz
force is
:<math> 
q \frac{\vec{v}}{c} \times \vec{B} = \frac{\dot{\vec{p}}}{c}\times \vec{B}
\,.
</math>
Inserting the expression above for the electric field, we thus obtain
this expression for the Lorentz force:
:<math> 
F_{lorentz} = \frac{\alpha}{c} A_0 \omega \cos kz \cos\omega_t (-A_0 \sin kz \cos
\omega t)
\,.
</math>
Now averaging, we find
:<math>\begin{align} 
\bar{F}_{lorentz} &= -\frac{1}{2} \alpha A_0^2 k \cos kz \sin kz
\\
&= \frac{d}{dz} \left(\frac{1}{4} A_0^2 \cos^2 kz\right)
\\
&= \frac{d}{dt} \left(\frac{1}{2} \alpha \bar{E}^2\right)
\,,
\end{align}</math>
where we now recognize the term on the right in the parentheses as the
dipole potential.

Thus, we see that the stimulated force can be either electrical,
that is of the form <math>p_z\nabla E_z</math>, or magnetic, that is, of the form
<math>\dot{p}\times B</math>, in nature.  Each of the two forces may not be curl
free, but the sum of both can be.

=== Energy conservation and dipole forces ===

When we considered the dipole force earlier, we looked at how it
arises from a redistribution of photons.  If you have a monochromatic
array of beams, the dipole does not arise due to spontaneous emission,
but rather, because photons <math>\hbar\omega</math> are taken from one beam and
put into another beam.  So how can this result in a change of energy
for the atom?

==== Energy conservation in blue molasses picture ====

Well, energy exchange does occur, because as we saw atoms flying along
a blue standing wave, in the picture of "blue molasses," sees a
damping force as it walks up the potential hills:
::[[Image:dipforce2-dfig5.png|thumb|400px|none|]]
When atoms slow down due to this viscous force, where has the kinetic
energy of the atoms gone?

If you have an ordinary molasses, the energy is radiated away.  The
basic ingredients available, therefore, are photon absorption,
stimulated emission, and spontaneous emission.  Recall that in
Sisyphus cooling, an atom does emit a photon after climbing up a hill.
From the upper dressed level, it emits a blue detuned photon, when the
sideband splitting is larger.  From the lower dressed level, it emits
at the top of the potential, at which the sideband splitting is
smaller.  The upper sideband is emitted at a larger Rabi frequency
than the lower sideband.

The principle behind this observation is that spontaneous emission is
once again responsible for the removal of entropy involved in the
laser cooling process, and energy is removed along with the entropy
change.

One way for providing dissipation is thus radiation into the
sidebands.

==== Transient energy conservation picture ====

Another scenario for understanding energy conservation is the
following transient experiment.  Imagine that you have an atom in the
harmonic potential of an optical dipole trap.
::[[Image:dipforce2-dfig6.png|thumb|500px|none|]]
In extremely far detuned optical lattices, there is essentially no
spontaneous emission.  The atom oscillating in the dipole trap, say at
1 Hz, going from high to low energy.  It is clear this energy must be
transformed into the light field, and the question is what actually
happens.  But it turns out the energy does not emit through
spontaneous emission.

What happens is that the moment the atoms reach the bottom of the
trap, they have a maximum in energy.  The photons which pass through
the atom at that point are red shifted.  When the atom climbs up the
potential hill, the photons are blue shifted.  While the atom
oscillates, the photons which arrive are blue and red shifted.  There
is no spontaneous emission in this transient picture, but there is
phase modulation of the laser which shifts the laser frequencies.

In principle, this frequency shift should be observable in an
experiment.  In practice, this is complicated by the fact that real
experiments involve spatially inhomogeneous beams, so there is some
momentum change as well as phase change of photons involved.

[[Category:Light forces]]