Quantum optics to quantum matter

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Quantum optics to quantum matter
Richard Phillips
Peter Littlewood
Quanta of light Second quantisation for photons.
Classical correlations. Quantum states of the
field coherent states, number states, squeezed
states and the vacuum. Quantum correlations.
Problems of quantisation infinities the
Casimir effect. Beamsplitters.
Light and atoms Resonant interaction of a
classical electromagnetic field and a quantised
atom - optical Bloch equations. Coherence and
decoherence of the atomic system. Line broadening
in real systems Lorentz, Gaussian and Voigt
lineshapes. Dressed atoms Mollow triplet.
Coherent manipulation of quantum states.
Lasers Paraxial beams and the ABCD matrix
treatment of cavities. Mode structure.
Spontaneous and stimulated emission. Rate
equations for a four-level laser. Gain clamping.
Properties of laser light coherence, the
Schawlow-Townes limit. Examples of lasers.
Cold atoms Optical trapping, molasses and atom
cooling. Experiments on atomic Bose-Einstein
condensation, phase coherence, atom lasers,
rotating condensates and vortices. Optical
lattices. Cold Fermi gases resonant molecular
levels and the BCS/BEC crossover. Excitons in
semiconductors as a solid-state analogue of cold
atoms.
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1 Photons
1.1 Summary of results from previous courses
2 orthogonal transverse polarisations
spin 1 angular momentum h
rewrite in terms of photon creation and
annihilation operators
and
The state with n 0, the vacuum state, is the
state of lowest energy, and from this the other
number states can be arrived at by use of the
creation operator.
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Lets ask some simple questions
Spin 1 - shouldnt there be 2S1 3
states/polarisations?
A is a four-vector shouldnt there be 4 states?
We have spatial wavefunctions for electrons
in atoms, in harmonic oscillators - why
havent we ever done this for a photon?
If we take the quantisation seriously, can we do
anything with that vacuum state?
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1.2 How many states are there in the Coulomb
gauge?
The standard approach is to work with the
potentials
Recall
so we write
E and B remain unaltered under the gauge
transformation
The choice of gauge makes substantial changes to
the appearance of the result, but the physical
result should be the same in any gauge.
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Maxwells inhomogeneous equations can be written
in terms of potentials as
and
i.e.
this is the term we want to control via the gauge
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The derivation in Part II AQP at this point makes
the gauge choice
the Coulomb, or transverse, gauge
Poissons equation
This gives
Note that Poissons equation is not a wave
equation - so its solution is not retarded
In this gauge, a change in r at r
instantaneously alters f at r.
For A we now have
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We now have a wave equation for the components of
A, which can be separated into two parts by
virtue of Helmholtzs theorem for any vector
field (J, say)
is the position of the source
operates with respect to that position
for derivation see pages 93-97 of Arfken
Weber Maths Methods
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consider
apply the continuity equation
with
this becomes
so since
the longitudinal part of the current exactly
cancels the contribution from the potential f
Transverse currents can only lead to transverse
components of A there are just 2, transverse,
states in the Coulomb gauge.
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In the Coulomb gauge
has plane-wave solutions which look like
These lead to the fields
..and therefore to the energy in a particular
k-mode, Uk
Taking the cycle-averaged result gives
and with
are annihilation and creation operators for
transversely-polarised photons in mode k
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1.3 How many states are there in a Lorentz gauge?
This is a brief look at a different approach
drawing on Part II REL course.
Recall that the DAlembertian is a Lorentz
invariant
so we can impose a condition which is also
Lorentz-invariant by restricting the potential by
the Lorentz condition
or
Note that this has not yet restricted us to a
particular gauge
there are many
possible Lorentz gauges.
Maxwells equations rewritten for the potentials
now yield decoupled wave equations for f and A
In Lorentz gauges the potentials satisfy wave
equations with retarded solutions which are
explicitly causal .
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Tidying up with
And constructing the usual 4-vectors for the
potential and current,
This gives an equation entirely in terms of
4-vectors and Lorentz-invariant operators, the
manifestly covariant form
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Plane-wave solutions to this 4-dimensional wave
equation will exist - with 4 polarisation states
unit polarisation vector labelled by l

• in a Lorentz gauge there are 4 parts of A we
  need to specify
• there are 4 photon
  polarisations.

The price of covariance is the need to deal with
a longitudinal photon and a timelike or scalar
photon, as well as the transverse polarisations.
These have to be kept in the theory in Lorentz
gauges.
The longitudinal and scalar photons do not appear
in expressions for measurable quantities,
because their effects cancel in a gauge-dependent
way (Bleuler-Gupta theory).
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1.4 Spin, symmetry, transformations and the
number of states.
So we really do have only 2, transverse,
polarisations. How does that square with spin 1?
angle
angular momentum operator
With two polarisations transverse to k, we can
represent the fields as
This has to be rotated about k by 2p to restore
this figure to itself, consistent with spin 1.
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There are several ways to argue why there is no
state corresponding to a projection of spin on
the k-vector 0
Eg.
A state with s 0 is spherically symmetric
this is impossible for a
transverse vector function.
A particle with mass exhibits its intrinsic
symmetry properties in its rest frame. A
photon moves with the speed of light in any
frame therefore in every frame there is a
distinctive direction in space, the direction of
k. There can therefore be no symmetry with
respect to the full rotation group in 3D, only
axial rotation symmetry about the preferred axis.
In this case of axial symmetry only the
helicity about k is conserved this
corresponds to the two possible circular
polarisation states i.e. angular momentum
eigenstates - spin 1 and spin -1.
Are these the only angular momentum states for
the photon?
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1.5 Spin, angular momentum and beams of light
Lets go back to the fields, and use them to
calculate quantities which can also be described
in terms of photons
linear momentum density
angular momentum density
Transfer of angular momentum by light was shown
by Beth in 1936 by measuring a torque generated
by circular light.
Flips again at a reflector then passes back
through the plate.
Upward going circular light, flips in sense
passing the plate.
Each circular polarisation flip at the
l/2 involves angular momentum change
generating a torque
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What gave rise to the torque?
For a plane wave, EB is the same anywhere across
a phase front, then because r is odd, the
integral of the angular momentum over all space
must vanish.
The wave components, in Cartesians, can take the
form
The z-component is required in order to satisfy
Ñ.E0, and is assumed to vary only very slowly
with z, leading to
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This just implies that a wave with a transverse
intensity profile has an E-vector which tilts
toward or away from the axis of propagation
i.e. it diffracts.
To get to the angular momentum we use real
fields, from above
this gives the energy flux
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We find that contributions to the angular
momentum come from
this gives the angular momentum flux along k
which comes entirely from the lateral variation
the beam profile gives the energy transport
the gradient controls the angular momentum
transport
When intercepting a plane wave by a finite
object, the angular momentum coupling is
controlled by the edge.
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The angular momentum transported through a plane
z constant will be
odd
Finally you find that the rate of angular
momentum transport through the surface looks like
And the rate of energy transport is
The ratio, 1/w , is consistent with transport of
h momentum and hw energy per photon
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The angular momentum transported through a plane
z constant will be
Finally you find that the rate of angular
momentum transport through the surface looks like
And the rate of energy transport is
The ratio, 1/w , is consistent with transport of
h momentum and hw energy per photon
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So the distribution of the field pattern in the
beam of light has some bearing on the total
transfer of angular momentum
The photon is described by a vector (a spinor of
rank 2) in this sense the photon has spin 1.
If it were possible to make a distinction between
intrinsic spin and orbital angular momentum
consistently the spin and orbital functions must
be independent of each other.
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So the distribution of the field pattern in the
beam of light has some bearing on the total
transfer of angular momentum
The photon is described by a vector (a spinor of
rank 2) in this sense the photon has spin 1.
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Where is the angular momentum?
Weve seen that it is in the lateral structure
e.g. the spherical harmonics.
Laguerre-Gaussian
Hermite-Gaussian
plane wave l 0
s1
s-1
phase fronts
linear
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1.6 The wave function of a photon?
Does A constitute a wavefunction for the photon?
Light is quantised as field modes in k-space in
this picture k describes the entire spatial
extent of the mode. The modes can be any
orthonormal functions satisfying the spatial
boundary conditions.
The field is characterised by the structure of
the mode and the photon occupation number
Absorbing a photon from the field - reduces n
by 1 - annihilates a quantum in the whole
spatial mode
It is difficult (impossible?) to construct a
real-space photon wavefunction which specifies
where a photon is, so verbal physics like a
photon goes through this aperture or a photon
arrived here usually leads to internal
contradictions.
Even when n 0 the spatial modal volume contains
fluctuating fields corresponding to the
zero-point energy the vacuum state is an
important part of the physical system.
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So, in summary
Photon quantisation is gauge-dependent most
treatments neglect this and present only the
result in the transverse gauge.
Only two, transverse polarisations contribute
directly to observations.
The origin of the reduction in the number of
states lies in the facts that the photon has no
rest frame, and the theory has to be
gauge-invariant.
A photon is an excitation of a k-space mode the
operators act on the whole mode. Care is required
in any discussion of real-space effects.
Only the total angular momentum is meaningful
relativistically, but with a loss of generality
the split is made into spin and orbital parts
the latter is controlled by the field
distribution.
The vacuum state with no real photons present
is an important part of the physical description
of the electromagnetic field.
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