Finished section on D-op scattering in chapter 3
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@ -160,6 +160,7 @@ Eq. \eqref{eq:D-3} into our expression for $R$ in Eq. \eqref{eq:R} and
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we will make use of the shorthand notation
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$A_i = u_1^*(\b{r}_i) u_0(\b{r}_i)$. The result is
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\begin{equation}
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\label{eq:Rfluc}
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R = |C|^2 \sum_{i,j}^K A^*_i A_j \langle \delta \hat{n}_i \delta
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\hat{n}_j \rangle,
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\end{equation}
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@ -176,35 +177,46 @@ measure the quadrature of the light fields which in section
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case when both the scattered mode and probe are travelling waves the
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quadrature
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\begin{equation}
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\label{eq:Xtrav}
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\hat{X}^F_\beta = \frac{1}{2} \left( \hat{F} e^{-i \beta} +
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\hat{F}^\dagger e^{i \beta} \right) = \sum_i^K \hat{n}_i\cos[(\b{k}_1 - \b{k}_2) \cdot
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\b{r}_i - \beta].
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\hat{F}^\dagger e^{i \beta} \right) = \sum_i^K \hat{n}_i\cos[(\b{k}_0 - \b{k}_1) \cdot
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\b{r}_i + (\phi_0 - \phi_1) - \beta].
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\end{equation}
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Note that different light quadratures are differently coupled to the
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atom distribution, hence by varying the local oscillator phase, and
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thus effectively $\beta$, and/or the detection angle one can scan the
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whole range of couplings. A similar expression exists for $\hat{D}$
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for a standing wave probe, where $\beta$ is replaced by $\varphi_0$,
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and scanning is achieved by varying the position of the wave with
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respect to atoms.
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for a standing wave probe, where instead of varying $\beta$ scanning
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is achieved by varying the position of the wave with respect to
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atoms. Additionally, the quadrature variance, $(\Delta X^F_\beta)^2$,
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will have a similar form to $R$ given in Eq. \eqref{eq:Rfluc},
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\begin{equation}
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(\Delta X^F_\beta)^2 = |C|^2 \sum_{i.j}^K A_i^\beta A_j^\beta
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\langle \dn_i \dn_j \rangle,
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\end{equation}
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where $A_i^\beta = (A_i e^{-i\beta} + A_i^* e^{i \beta})/2$. However,
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this has the advantage that unlike in the case of $R$ there is no need
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to subtract a spatially varying classical signal to obtain this
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quantity.
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The ``quantum addition'', $R$, and the quadrature variance, $(\Delta
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X^F_\beta)^2$, are both quadratic in $\a_1$ and both rely heavily on
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the quantum state of the matter. Therefore, they will have a
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nontrivial angular dependence, showing more peaks than classical
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The ``quantum addition'', $R$, and the quadrature variance,
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$(\Delta X^F_\beta)^2$, are both quadratic in $\a_1$ and both rely
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heavily on the quantum state of the matter. Therefore, they will have
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a nontrivial angular dependence, showing more peaks than classical
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diffraction. Furthermore, these peaks can be tuned very easily with
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$\beta$ or $\varphi_l$. Fig. \ref{fig:scattering} shows the angular
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dependence of $R$ for the case when the scattered mode is a standing
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wave and the probe is a travelling wave scattering from bosons in a 3D
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optical lattice. The first noticeable feature is the isotropic
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background which does not exist in classical diffraction. This
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background yields information about density fluctuations which,
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according to mean-field estimates (i.e.~inter-site correlations are
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ignored), are related by $R = K( \langle \hat{n}^2 \rangle - \langle
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\hat{n} \rangle^2 )/2$. In Fig. \ref{fig:scattering} we can see a
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significant signal of $R = |C|^2 N_K/2$, because it shows scattering
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from an ideal superfluid which has significant density fluctuations
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with correlations of infinte range. However, as the parameters of the
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wave and the probe is a travelling wave scattering from an ideal
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superfluid in a 3D optical lattice. The first noticeable feature is
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the isotropic background which does not exist in classical
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diffraction. This background yields information about density
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fluctuations which, according to mean-field estimates (i.e.~inter-site
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correlations are ignored), are related by
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$R = K( \langle \hat{n}^2 \rangle - \langle \hat{n} \rangle^2 )/2$. In
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Fig. \ref{fig:scattering} we can see a significant signal of
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$R = |C|^2 N_K/2$, because it shows scattering from an ideal
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superfluid which has significant density fluctuations with
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correlations of infinte range. However, as the parameters of the
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lattice are tuned across the phase transition into a Mott insulator
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the signal goes to zero. This is because the Mott insulating phase has
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well localised atoms at each site which suppresses density
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@ -233,20 +245,125 @@ classical Bragg condition is actually not satisfied which means there
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actually is no classical diffraction on top of the ``quantum
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addition'' shown here. Therefore, these features would be easy to see
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in an experiment as they wouldn't be masked by a stronger classical
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signal. We can even derive the generalised Bragg conditions for the
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peaks that we can see in Fig. \ref{fig:scattering}.
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signal. This difference in behaviour is due to the fact that
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classical diffraction is ignorant of any quantum correlations. This
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signal is given by the square of the light field amplitude squared
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\begin{equation}
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|\langle \a_1 \rangle|^2 = |C|^2 \sum_{i,j} A_i^*
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A_j \langle \n_i \rangle \langle \n_j \rangle,
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\end{equation}
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which is idependent of any two-point correlations unlike $R$. On the
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other hand the full light intensity (classical signal plus ``quantum
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addition'') of the quantum light does include higher-order
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correlations
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\begin{equation}
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\langle \ad_1 \a_1 \rangle = |C|^2 \sum_{i,j} A_i^*
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A_j \langle \n_i \n_j \rangle.
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\end{equation}
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Therefore, we see that in the fully quantum picture light scattering
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not only depends on the diffraction structure due to the distribution
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of atoms in the lattice, but also on the quantum correlations between
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different lattice sites which will be dependent on the quantum state
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of the matter. These correlations are imprinted in $R$ as shown in
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Eq. \eqref{eq:Rfluc} and it highlights the key feature of our model,
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i.e.~the light couples to the quantum state directly via operators.
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\mynote{Derive and show these Bragg conditions below}
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We can even derive the generalised Bragg conditions for the peaks that
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we can see in Fig. \ref{fig:scattering}. The exact conditions under
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which diffraction peaks emerge for the ``quantum addition'' will
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depend on the optical setup as well as on the quantum state of the
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matter as it is the density fluctuation correlations,
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$\langle \dn_i \dn_j \rangle$, that provide the structure for $R$ and
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not the lattice itself as seen in Eq. \eqref{eq:Rfluc}. For classical
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light it is straightforward to develop an intuitive physical picture
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to find the Bragg condition by considering angles at which the
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distance travelled by light scattered from different points in the
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lattice is equal to an integer multiple of wavelength. The ``quantum
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addition'' is more complicated and less intuitive as we now have to
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consider quantum correlations which are not only nonlocal, but can
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also be nagative.
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As $(\Delta X^F_\beta)^2$ and $R$ are quadratic variables,
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the generalized Bragg conditions for the peaks are
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$2 \Delta \b{k} = \b{G}$ for quadratures of travelling waves, where
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$\Delta \b{k} = \b{k}_0 - \b{k}_1$ and $\b{G}$ is the reciprocal
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lattice vector, and $2 \b{k}_1 = \b{G}$ for standing wave $\a_1$ and
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travelling $\a_0$, which is clearly different from the classical Bragg
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condition $\Delta \b{k} = \b{G}$. The peak height is tunable by the
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local oscillator phase or standing wave shift as seen in Fig.
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\ref{fig:scattering}b.
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We will consider scattering from a superfluid, because the Mott
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insulator has no ``quantum addition'' due to a lack of density
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fluctuations. The wavefunction of a superfluid on a lattice is given
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by
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\begin{equation}
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\frac{1}{\sqrt{M^N N!}} \left( \sum_i^M \bd_i \right)^N |0 \rangle,
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\end{equation}
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where $| 0 \rangle$ denotes the vacuum state. This state has infinte
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range correlations and thus has the convenient property that all
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two-point density fluctuation correlations are equal regardless of
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their separation,
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i.e.~$\langle \dn_i \dn_j \rangle \equiv \langle \dn_a \dn_b \rangle$
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for all $(i \ne j)$, where the right hand side is a constant
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value. This allows us to extract all correlations from the sum in
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Eq. \eqref{eq:Rfluc} to obtain
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\begin{equation}
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\label{eq:RSF}
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\frac{R}{|C|^2} = (\langle \dn^2 \rangle - \langle \dn_a \dn_b \rangle) \sum_i^K
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|A_i|^2 + \langle \dn_a \dn_b \rangle |A|^2 = \frac{N}{M} \sum_i^K
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|A_i|^2 - \frac{N}{M^2} |A|^2,
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\end{equation}
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where $A = \sum_i^K A_i$, and we have dropped the index from
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$\langle \dn^2 \rangle$ as it is equal at every site. The second
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equality follows from the fact that for a superfluid state
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$\langle \dn^2 \rangle = N/M - N/M^2$ and
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$\langle \dn_a \dn_b \rangle = -N/M^2$. Naturally, we get an identical
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expression for the quadrature variance $(\Delta X^F_\beta)^2$ with the
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coefficients $A_i$ replaced by the real $A_i^\beta$.
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The ``quantum addition'' for the case when both the scattered and
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probe modes are travelling waves is actually trivial. It has no peaks
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and thus it has no generalised Bragg condition and it only consists of
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a uniform background. This is a consequence of the fact that
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travelling waves couple equally strongly with every site as only the
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phase differs. Therefore, since superfluid correlations lack structure
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as they're uniform we do no get a strong coherent peak. The
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contribution from the lattice structure is included in the classical
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Bragg peaks which we have subtracted in order to obtain the quantity
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$R$. However, if we consider the case where the scattered mode is
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collected as a standing wave using a pair of mirrors we get the
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diffraction pattern that we saw in Fig. \ref{fig:scattering}. This
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time we get strong visible peaks, because at certain angles the
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standing wave couples to the atoms maximally at all lattice sites and
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thus it uses the structure of the lattice to amplify the signal from
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the quantum fluctuations. This becomes clear when we look at
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Eq. \eqref{eq:RSF}. We can neglect the second term as it is always
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negative and it has the same angular distribution as the classical
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diffraction pattern and thus it is mostly zero except when the
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classical Bragg condition is satisfied. Since in
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Fig. \ref{fig:scattering} we have chosen an angle such that the Bragg
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is not satisfied this term is essentially zero. Therefore, we are left
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with the first term $\sum_i^K |A_i|^2$ which for a travelling wave
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probe and a standing wave scattered mode is
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\begin{equation}
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\sum_i^K |A_i|^2 = \sum_i^K \cos^2(\b{k}_0 \cdot \b{r}_i + \phi_0) =
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\frac{1}{2} \sum_i^K \left[1 + \cos(2 \b{k}_0 \cdot \b{r}_i + 2
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\phi_0) \right].
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\end{equation}
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Therefore, it is straightforward to see that unless
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$2 \b{k}_0 = \b{G}$, where $\b{G}$ is a reciprocal lattice vector,
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there will be no coherent signal and we end up with the mean uniform
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signal of strength $N_k/2$. When this condition is satisifed all the
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cosine terms will be equal and they will add up constructively instead
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of cancelling each other out. Note that this new Bragg condition is
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different from the classical one $\b{k}_0 - \b{k}_1 = \b{G}$. This
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result makes it clear that the uniform background signal is not due to
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any coherent scattering, but rather due to the lack of structure in
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the quantum correlations. Furthermore, we see that the peak height is
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actually tunable via the phase, $\phi_0$, which is illustrated in
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Fig. \ref{fig:scattering}b.
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For light field quadratures the situation is different, because as we
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have seen in Eq. \eqref{eq:Xtrav} even for travelling waves we can
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tune the contributions to the signal from different sites by changing
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the angle of measurement or tuning the local oscillator signal
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$\beta$. The rest is similar to the case we discussed for $R$ with a
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standing wave mode and we can show that the new Bragg condition in
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this case is $2 (\b{k}_0 - \b{k}_1) = \b{G}$ which is different from
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the condition we had for $R$ and is still different from the classical
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condition $2 (\b{k}_0 - \b{k}_1) = \b{G}$. Furthermore, just like in
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Fig. \ref{fig:scattering}b the peak height can be tuned using $\beta$.
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A quantum signal that isn't masked by classical diffraction is very
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useful for future experimental realisability. However, it is still
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@ -205,6 +205,7 @@
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\renewcommand{\H}{\hat{H}}
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\newcommand{\n}{\hat{n}}
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\newcommand{\dn}{\delta \hat{n}}
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\newcommand{\ad}{a^\dagger}
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\newcommand{\bd}{b^\dagger}
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\renewcommand{\a}{a} % in case we decide to put hats on
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