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Finished section on D-op scattering in chapter 3

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Wojciech Kozlowski 2016-07-17 22:34:59 +01:00
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Chapter3/chapter3.tex
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@ -160,6 +160,7 @@ Eq. \eqref{eq:D-3} into our expression for $R$ in Eq. \eqref{eq:R} and
we will make use of the shorthand notation we will make use of the shorthand notation
$A_i = u_1^*(\b{r}_i) u_0(\b{r}_i)$. The result is $A_i = u_1^*(\b{r}_i) u_0(\b{r}_i)$. The result is
\begin{equation} \begin{equation}
\label{eq:Rfluc}
R = |C|^2 \sum_{i,j}^K A^*_i A_j \langle \delta \hat{n}_i \delta R = |C|^2 \sum_{i,j}^K A^*_i A_j \langle \delta \hat{n}_i \delta
\hat{n}_j \rangle, \hat{n}_j \rangle,
\end{equation} \end{equation}
@ -176,35 +177,46 @@ measure the quadrature of the light fields which in section
case when both the scattered mode and probe are travelling waves the case when both the scattered mode and probe are travelling waves the
quadrature quadrature
\begin{equation} \begin{equation}
\label{eq:Xtrav}
\hat{X}^F_\beta = \frac{1}{2} \left( \hat{F} e^{-i \beta} + \hat{X}^F_\beta = \frac{1}{2} \left( \hat{F} e^{-i \beta} +
\hat{F}^\dagger e^{i \beta} \right) = \sum_i^K \hat{n}_i\cos[(\b{k}_1 - \b{k}_2) \cdot \hat{F}^\dagger e^{i \beta} \right) = \sum_i^K \hat{n}_i\cos[(\b{k}_0 - \b{k}_1) \cdot
\b{r}_i - \beta]. \b{r}_i + (\phi_0 - \phi_1) - \beta].
\end{equation} \end{equation}
Note that different light quadratures are differently coupled to the Note that different light quadratures are differently coupled to the
atom distribution, hence by varying the local oscillator phase, and atom distribution, hence by varying the local oscillator phase, and
thus effectively $\beta$, and/or the detection angle one can scan the thus effectively $\beta$, and/or the detection angle one can scan the
whole range of couplings. A similar expression exists for $\hat{D}$ whole range of couplings. A similar expression exists for $\hat{D}$
for a standing wave probe, where $\beta$ is replaced by $\varphi_0$, for a standing wave probe, where instead of varying $\beta$ scanning
and scanning is achieved by varying the position of the wave with is achieved by varying the position of the wave with respect to
respect to atoms. atoms. Additionally, the quadrature variance, $(\Delta X^F_\beta)^2$,
will have a similar form to $R$ given in Eq. \eqref{eq:Rfluc},
\begin{equation}
(\Delta X^F_\beta)^2 = |C|^2 \sum_{i.j}^K A_i^\beta A_j^\beta
\langle \dn_i \dn_j \rangle,
\end{equation}
where $A_i^\beta = (A_i e^{-i\beta} + A_i^* e^{i \beta})/2$. However,
this has the advantage that unlike in the case of $R$ there is no need
to subtract a spatially varying classical signal to obtain this
quantity.
The ``quantum addition'', $R$, and the quadrature variance, $(\Delta The ``quantum addition'', $R$, and the quadrature variance,
X^F_\beta)^2$, are both quadratic in $\a_1$ and both rely heavily on $(\Delta X^F_\beta)^2$, are both quadratic in $\a_1$ and both rely
the quantum state of the matter. Therefore, they will have a heavily on the quantum state of the matter. Therefore, they will have
nontrivial angular dependence, showing more peaks than classical a nontrivial angular dependence, showing more peaks than classical
diffraction. Furthermore, these peaks can be tuned very easily with diffraction. Furthermore, these peaks can be tuned very easily with
$\beta$ or $\varphi_l$. Fig. \ref{fig:scattering} shows the angular $\beta$ or $\varphi_l$. Fig. \ref{fig:scattering} shows the angular
dependence of $R$ for the case when the scattered mode is a standing dependence of $R$ for the case when the scattered mode is a standing
wave and the probe is a travelling wave scattering from bosons in a 3D wave and the probe is a travelling wave scattering from an ideal
optical lattice. The first noticeable feature is the isotropic superfluid in a 3D optical lattice. The first noticeable feature is
background which does not exist in classical diffraction. This the isotropic background which does not exist in classical
background yields information about density fluctuations which, diffraction. This background yields information about density
according to mean-field estimates (i.e.~inter-site correlations are fluctuations which, according to mean-field estimates (i.e.~inter-site
ignored), are related by $R = K( \langle \hat{n}^2 \rangle - \langle correlations are ignored), are related by
\hat{n} \rangle^2 )/2$. In Fig. \ref{fig:scattering} we can see a $R = K( \langle \hat{n}^2 \rangle - \langle \hat{n} \rangle^2 )/2$. In
significant signal of $R = |C|^2 N_K/2$, because it shows scattering Fig. \ref{fig:scattering} we can see a significant signal of
from an ideal superfluid which has significant density fluctuations $R = |C|^2 N_K/2$, because it shows scattering from an ideal
with correlations of infinte range. However, as the parameters of the superfluid which has significant density fluctuations with
correlations of infinte range. However, as the parameters of the
lattice are tuned across the phase transition into a Mott insulator lattice are tuned across the phase transition into a Mott insulator
the signal goes to zero. This is because the Mott insulating phase has the signal goes to zero. This is because the Mott insulating phase has
well localised atoms at each site which suppresses density well localised atoms at each site which suppresses density
@ -233,20 +245,125 @@ classical Bragg condition is actually not satisfied which means there
actually is no classical diffraction on top of the ``quantum actually is no classical diffraction on top of the ``quantum
addition'' shown here. Therefore, these features would be easy to see addition'' shown here. Therefore, these features would be easy to see
in an experiment as they wouldn't be masked by a stronger classical in an experiment as they wouldn't be masked by a stronger classical
signal. We can even derive the generalised Bragg conditions for the signal. This difference in behaviour is due to the fact that
peaks that we can see in Fig. \ref{fig:scattering}. classical diffraction is ignorant of any quantum correlations. This
signal is given by the square of the light field amplitude squared
\begin{equation}
|\langle \a_1 \rangle|^2 = |C|^2 \sum_{i,j} A_i^*
A_j \langle \n_i \rangle \langle \n_j \rangle,
\end{equation}
which is idependent of any two-point correlations unlike $R$. On the
other hand the full light intensity (classical signal plus ``quantum
addition'') of the quantum light does include higher-order
correlations
\begin{equation}
\langle \ad_1 \a_1 \rangle = |C|^2 \sum_{i,j} A_i^*
A_j \langle \n_i \n_j \rangle.
\end{equation}
Therefore, we see that in the fully quantum picture light scattering
not only depends on the diffraction structure due to the distribution
of atoms in the lattice, but also on the quantum correlations between
different lattice sites which will be dependent on the quantum state
of the matter. These correlations are imprinted in $R$ as shown in
Eq. \eqref{eq:Rfluc} and it highlights the key feature of our model,
i.e.~the light couples to the quantum state directly via operators.
\mynote{Derive and show these Bragg conditions below} We can even derive the generalised Bragg conditions for the peaks that
we can see in Fig. \ref{fig:scattering}. The exact conditions under
which diffraction peaks emerge for the ``quantum addition'' will
depend on the optical setup as well as on the quantum state of the
matter as it is the density fluctuation correlations,
$\langle \dn_i \dn_j \rangle$, that provide the structure for $R$ and
not the lattice itself as seen in Eq. \eqref{eq:Rfluc}. For classical
light it is straightforward to develop an intuitive physical picture
to find the Bragg condition by considering angles at which the
distance travelled by light scattered from different points in the
lattice is equal to an integer multiple of wavelength. The ``quantum
addition'' is more complicated and less intuitive as we now have to
consider quantum correlations which are not only nonlocal, but can
also be nagative.
As $(\Delta X^F_\beta)^2$ and $R$ are quadratic variables, We will consider scattering from a superfluid, because the Mott
the generalized Bragg conditions for the peaks are insulator has no ``quantum addition'' due to a lack of density
$2 \Delta \b{k} = \b{G}$ for quadratures of travelling waves, where fluctuations. The wavefunction of a superfluid on a lattice is given
$\Delta \b{k} = \b{k}_0 - \b{k}_1$ and $\b{G}$ is the reciprocal by
lattice vector, and $2 \b{k}_1 = \b{G}$ for standing wave $\a_1$ and \begin{equation}
travelling $\a_0$, which is clearly different from the classical Bragg \frac{1}{\sqrt{M^N N!}} \left( \sum_i^M \bd_i \right)^N |0 \rangle,
condition $\Delta \b{k} = \b{G}$. The peak height is tunable by the \end{equation}
local oscillator phase or standing wave shift as seen in Fig. where $| 0 \rangle$ denotes the vacuum state. This state has infinte
\ref{fig:scattering}b. range correlations and thus has the convenient property that all
two-point density fluctuation correlations are equal regardless of
their separation,
i.e.~$\langle \dn_i \dn_j \rangle \equiv \langle \dn_a \dn_b \rangle$
for all $(i \ne j)$, where the right hand side is a constant
value. This allows us to extract all correlations from the sum in
Eq. \eqref{eq:Rfluc} to obtain
\begin{equation}
\label{eq:RSF}
\frac{R}{|C|^2} = (\langle \dn^2 \rangle - \langle \dn_a \dn_b \rangle) \sum_i^K
|A_i|^2 + \langle \dn_a \dn_b \rangle |A|^2 = \frac{N}{M} \sum_i^K
|A_i|^2 - \frac{N}{M^2} |A|^2,
\end{equation}
where $A = \sum_i^K A_i$, and we have dropped the index from
$\langle \dn^2 \rangle$ as it is equal at every site. The second
equality follows from the fact that for a superfluid state
$\langle \dn^2 \rangle = N/M - N/M^2$ and
$\langle \dn_a \dn_b \rangle = -N/M^2$. Naturally, we get an identical
expression for the quadrature variance $(\Delta X^F_\beta)^2$ with the
coefficients $A_i$ replaced by the real $A_i^\beta$.
The ``quantum addition'' for the case when both the scattered and
probe modes are travelling waves is actually trivial. It has no peaks
and thus it has no generalised Bragg condition and it only consists of
a uniform background. This is a consequence of the fact that
travelling waves couple equally strongly with every site as only the
phase differs. Therefore, since superfluid correlations lack structure
as they're uniform we do no get a strong coherent peak. The
contribution from the lattice structure is included in the classical
Bragg peaks which we have subtracted in order to obtain the quantity
$R$. However, if we consider the case where the scattered mode is
collected as a standing wave using a pair of mirrors we get the
diffraction pattern that we saw in Fig. \ref{fig:scattering}. This
time we get strong visible peaks, because at certain angles the
standing wave couples to the atoms maximally at all lattice sites and
thus it uses the structure of the lattice to amplify the signal from
the quantum fluctuations. This becomes clear when we look at
Eq. \eqref{eq:RSF}. We can neglect the second term as it is always
negative and it has the same angular distribution as the classical
diffraction pattern and thus it is mostly zero except when the
classical Bragg condition is satisfied. Since in
Fig. \ref{fig:scattering} we have chosen an angle such that the Bragg
is not satisfied this term is essentially zero. Therefore, we are left
with the first term $\sum_i^K |A_i|^2$ which for a travelling wave
probe and a standing wave scattered mode is
\begin{equation}
\sum_i^K |A_i|^2 = \sum_i^K \cos^2(\b{k}_0 \cdot \b{r}_i + \phi_0) =
\frac{1}{2} \sum_i^K \left[1 + \cos(2 \b{k}_0 \cdot \b{r}_i + 2
\phi_0) \right].
\end{equation}
Therefore, it is straightforward to see that unless
$2 \b{k}_0 = \b{G}$, where $\b{G}$ is a reciprocal lattice vector,
there will be no coherent signal and we end up with the mean uniform
signal of strength $N_k/2$. When this condition is satisifed all the
cosine terms will be equal and they will add up constructively instead
of cancelling each other out. Note that this new Bragg condition is
different from the classical one $\b{k}_0 - \b{k}_1 = \b{G}$. This
result makes it clear that the uniform background signal is not due to
any coherent scattering, but rather due to the lack of structure in
the quantum correlations. Furthermore, we see that the peak height is
actually tunable via the phase, $\phi_0$, which is illustrated in
Fig. \ref{fig:scattering}b.
For light field quadratures the situation is different, because as we
have seen in Eq. \eqref{eq:Xtrav} even for travelling waves we can
tune the contributions to the signal from different sites by changing
the angle of measurement or tuning the local oscillator signal
$\beta$. The rest is similar to the case we discussed for $R$ with a
standing wave mode and we can show that the new Bragg condition in
this case is $2 (\b{k}_0 - \b{k}_1) = \b{G}$ which is different from
the condition we had for $R$ and is still different from the classical
condition $2 (\b{k}_0 - \b{k}_1) = \b{G}$. Furthermore, just like in
Fig. \ref{fig:scattering}b the peak height can be tuned using $\beta$.
A quantum signal that isn't masked by classical diffraction is very A quantum signal that isn't masked by classical diffraction is very
useful for future experimental realisability. However, it is still useful for future experimental realisability. However, it is still

1
Preamble/preamble.tex
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@ -205,6 +205,7 @@
\renewcommand{\H}{\hat{H}} \renewcommand{\H}{\hat{H}}
\newcommand{\n}{\hat{n}} \newcommand{\n}{\hat{n}}
\newcommand{\dn}{\delta \hat{n}}
\newcommand{\ad}{a^\dagger} \newcommand{\ad}{a^\dagger}
\newcommand{\bd}{b^\dagger} \newcommand{\bd}{b^\dagger}
\renewcommand{\a}{a} % in case we decide to put hats on \renewcommand{\a}{a} % in case we decide to put hats on