text:
\section{introduction}

Thanks to the high-controllability of ultracold atoms in interatomic
interaction, geometry, purity, atomic species and lattice constant
(of optical lattices), ultracold atomic gases have already become
one of the most important footstones in modern physics, and can be
used in simulating various strongly correlated many-body models in
solid state physics \cite{Bloch2008M,Giorgini2008T,Chin2010F}. Very
recently, another controllability of ultracold atoms is realized experimentally,
in which the spin degree of freedom of atoms (i.e., the atomic internal
hyperfine states) is coupled to the orbital degree of freedom (i.e.,
the momentum) by using a two-photon Raman process. This so-called
spin-orbit (SO) coupling was first created and detected in atomic
$^{87}$Rb Bose-Einstein condensates (BEC) \cite{Lin2011A,Lin2011S},
and then produced in atomic Fermi gases of $^{40}$K atoms \cite{Wang2012S}
and $^{6}$Li atoms \cite{Cheuk2012S}. The realization of SO coupled
atomic gases enables the simulation of charged particles by using
neutral atoms, which are cleaner and more controllable \cite{Dalibard2011A}.
These experiments open an entirely new way to study the celebrated
effects of SO coupling, such as topological insulators, topological
superconductors and new exotic superfluid phases \cite{Qi2010T,Hasan2010T,Zhai2012S}.

Radio-frequency (rf) spectroscopy has been widely applied in many
experiments to study fermionic pairing in a two-component Fermi gas
near Feshbach resonances when it crosses from a Bardeen-Cooper-Schrieffer
(BCS) superfluid of weakly interacting Cooper pairs over to a BEC
of tightly bound molecules \cite{Chin2004O,Schunck2008D}, and also
to investigate the properties of polarons in spin-imbalanced Fermi
gases \cite{Schirotzek2009O,Zhang2012P,Kohstall2012M,Koschorreck2012A}.
This technology allows a precise determination of the molecular binding
energy \cite{Chin2004O,Regal2003C} and pairing gap in a degenerate
Fermi gas \cite{Chin2004O}. In addition, the momentum-resolved rf
spectroscopy, i.e., the spectroscopy at a specific momentum, is also
available \cite{Stewart2008U} and provides important information
on low-energy excitations of novel exotic state of matter in ultracold
atoms. In this respect, it would be a powerful tool for characterizing
the recently realized spin-orbit coupled atomic Fermi gases. Indeed,
momentum-resolved rf spectroscopy of \emph{non-interacting} spin-orbit
coupled $^{40}$K and $^{6}$Li atoms has already been reported \cite{Wang2012S,Cheuk2012S}.

In this work, we aim to qualitatively predict the momentum-resolved
rf spectroscopy of a \emph{resonantly interacting} atomic Fermi gas
with equal Rashba and Dresselhaus spin-orbit coupling, a system already
realized experimentally at Shanxi University and also at MIT by using
broad Feshbach resonances \cite{ZhangPrivateComm,ZwierleinPrivateComm}.
As is well-known, a strongly interacting Fermi gas near Feshbach resonances
is notoriously difficult to handle theoretically \cite{Hu2008,Hu2010}.
Therefore, at the \emph{qualitative} level, we approximate the strongly
interacting Fermi gas as a mixture of non-interacting atoms and molecules.
All the properties of individual molecules are determined from the
two-particle solution of two-interacting atoms. Our approximation
may be justified at large temperatures well above the superfluid transition
temperature $T_{c}$, where molecules are formed below the Feshbach
resonance and have little correlations among themselves or with atoms.
This is exactly the situation in current experiments, for instance,
at Shanxi University \cite{Wang2012S}, the typical temperature of
spin-orbit coupled $^{40}$K atoms is now at about $0.6T_{F}$, where
$T_{F}$ is the Fermi degenerate temperature.

We consider separately the momentum-resolved rf-spectroscopy of atoms
and of molecules. Moreover, according to real experiments we take
into account the crucial trapping potential that prevents the atoms
and molecules from escaping. Within local density approximation (LDA),
the trapped Fermi cloud may be divided into many locally uniform subsystems.
For each subsystem, we calculate the rf responses of the atomic and
molecular components based on the one-particle and two-particle solutions
of uniform systems and Fermi's golden rule. We finally add up all
the local contributions to determine the total momentum-resolved rf
spectroscopy. We note that due to SO coupling, weakly bound molecules
with anisotropic mass and wavefunction may be formed \cite{Vyasanakere2011B,Hu2011P,Yu2011S}.
The bound to free rf transition of weakly bound molecules at rest
has been predicted \cite{Hu2012R}. In this work, we will consider
the rf transition of weakly bound molecules at motion with arbitrary
center-of-mass (COM) momentum.

The paper is arranged as follows. In the next section, we consider
the situation of current experiments, and introduce the model Hamiltonian
responsible for equal Rashba and Dresselhaus SO coupling. In addition,
we introduce the LDA formulism and present the calculation for the
chemical potential. Then, the single-particle and two-particle problems
of a local uniform subsystem are solved in Sec. \ref{sec:single--and-two-body}.
The general formulas for the binding energy and wavefunction of two-particle
bound state with non-zero COM momentum are also provided. In Sec.
\ref{sec:radio-frequency-spectroscopy}, we derive the momentum-resolved
rf transition signals for\emph{ }non-interacting\emph{ }atoms and
molecules, respectively, and then obtain the total rf spectroscopy
for the harmonically trapped\emph{ }ideal\emph{ }gas mixture of fermionic
atoms and bosonic molecules. Finally, our main results are concluded
in Sec.\ref{sec:conclusions}.

\section{models\label{sec:models}}

\subsection{Hamiltonian}

We consider a SO coupled Fermi gas with atomic mass $m$ in a harmonic
trap $V\left(\mathbf{r}\right)=m(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}+\omega_{z}^{2}z^{2})/2$.
The SO coupling is created by a two-photon Raman process \cite{Wang2012S,Cheuk2012S}.
The many-body Hamiltonian responsible for this process may be modeled
as $\mathcal{H}=\mathcal{H}_{0}+\mathcal{H}_{int}$, where
\begin{eqnarray}
\mathcal{H}_{0} & = & \sum_{\sigma}\int d\mathbf{r}\Psi_{\sigma}^{\dagger}\left(\mathbf{r}\right)\left[-\frac{\hbar^{2}\mathbf{\nabla}^{2}}{2m}+V\left(\mathbf{r}\right)-\mu\right]\Psi_{\sigma}\left(\mathbf{r}\right)+\nonumber \\
 &  & \int d\mathbf{r}\left[\Psi_{\uparrow}^{\dagger}\left(\mathbf{r}\right)\left(\frac{\Omega_{R}}{2}e^{+i2k_{R}x}\right)\Psi_{\downarrow}\left(\mathbf{r}\right)+H.c.\right]\label{eq:2.1.1}
\end{eqnarray}
 is the single-particle Hamiltonian and
\begin{equation}
\mathcal{H}_{int}=U_{0}\int d\mathbf{r}\Psi_{\uparrow}^{\dagger}\left(\mathbf{r}\right)\Psi_{\downarrow}^{\dagger}\left(\mathbf{r}\right)\Psi_{\downarrow}\left(\mathbf{r}\right)\Psi_{\uparrow}\left(\mathbf{r}\right)\label{eq:2.1.2}
\end{equation}
is the interaction Hamiltonian describing the contact interaction
between two spin states. Here, $\Psi_{\sigma}^{\dagger}\left(\mathbf{r}\right)$
is the creation field operator for atoms in the spin-state $\sigma$.
The second term in $\mathcal{H}_{0}$ describes the Raman coupling
between two spin states with strength $\Omega_{R}$ , $k_{R}$ is
the wave-number of two Raman laser beams and therefore $2\hbar k_{R}$
is the momentum transfer during the two-photon Raman process \cite{Wang2012S,Cheuk2012S}.
The interaction strength is denoted by the bare interaction parameter
$U_{0}$, which should be regularized in terms of the \textit{s}-wave
scattering length $a_{s}$, i.e., $1/U_{0}=m/\left(4\pi\hbar^{2}a_{s}\right)-(1/V)\sum_{\mathbf{k}}m/\left(\hbar^{2}k^{2}\right)$.
By using Feshbach resonances, the \textit{s}-wave scattering length
$a_{s}$ could be arbitrarily tuned, and the system may undergo a
crossover from a BCS superfluid of weakly interacting Cooper pairs
to a BEC of tightly bound molecules. Therefore, the atomic chemical
potential $\mu$ may decrease as a result of decreasing atomic population,
when weakly bound molecules are formed.

As argued in the Introduction section, we will treat the Fermi cloud
as a mixture of non-interacting atoms and molecules, although the
degree of freedom of molecules is not made explicitly in our model
Hamiltonian from the outset. We will take the molecule as a two-particle
bound state and determine all its properties by using two-particle
solution of two interacting atoms. In the harmonic trap, we assume
that these molecules experience the same harmonic trap as atoms, but
with double mass, i.e., $V_{M}\left(\mathbf{r}\right)=m(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}+\omega_{z}^{2}z^{2})$.
Note that, our treatment of ideal mixture of atoms and molecules becomes
exact for a Feshbach resonance with \emph{zero} resonance width. However,
for the broad Feshbach resonance used experimentally, this treatment
is valid at the qualitative level only.

\subsection{Local density approximation}

If the number of atoms is sufficiently large, it is reasonable to
assume that the trapped cloud may be divided into many locally uniform
subsystems with a local chemical potential \cite{Yi2006T,Liu2007M}.
Then, within LDA, the external trap $V\left(\mathbf{r}\right)$ in
the Hamiltonian (\ref{eq:2.1.1}) is absorbed into the chemical potential,
and we can define \emph{effective} local chemical potential
\begin{equation}
\mu\left(\mathbf{r}\right)=\mu-\frac{1}{2}m(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}+\omega_{z}^{2}z^{2}).\label{eq:2.2.1}
\end{equation}
Note that the global chemical potential $\mu$, which is determined
by the total number of atoms, can be regarded as the local chemical
potential in the trap center ($\mathbf{r}=0$).

In order to evaluate the chemical potential $\mu$, let us consider
the ideal gas mixture of fermionic atoms and bosonic molecules at
finite temperatures. The distributions of the atoms and molecules
are respectively determined by the Fermi-Dirac and Bose-Einstein distributions,
\begin{equation}
f_{A}\left(\epsilon\right)=\frac{1}{e^{\left(\epsilon-\mu\right)/\left(k_{B}T_{A}\right)}+1},\label{eq:2.2.2}
\end{equation}
\begin{equation}
f_{M}\left(\epsilon\right)=\frac{1}{e^{\left(\epsilon-\mu_{M}\right)/\left(k_{B}T_{M}\right)}-1},\label{eq:2.2.3}
\end{equation}
where $k_{B}$ is Boltzmann's constant, and $T_{A,M}$, $\mu$ and
$\mu_{M}$ are the temperatures, chemical potentials of atoms and
molecules, respectively. From the thermal and chemical equilibrium
conditions, we have the following relations, $T_{A}=T_{M}\equiv T$
and $\mu_{M}=2\mu+\varepsilon_{B}$, where $\varepsilon_{B}$ is the
binding energy of the molecules relative to the threshold of two free
atoms. To ease the numerical workload, we will approximate it as $\varepsilon_{B}\simeq E_{B}\equiv\hbar^{2}/\left(ma_{s}^{2}\right)$.
Here, $E_{B}$ is the binding energy in the absence of SO coupling.

The populations of the atoms $N_{A}$ (in a single spin component)
and \emph{noncondensed} molecules $N_{M}$ are given by
\begin{equation}
N_{A}=\int_{0}^{\infty}d\epsilon\rho_{A}\left(\epsilon\right)f_{A}\left(\epsilon\right),\label{eq:2.2.6}
\end{equation}
\begin{equation}
N_{M}=\int_{0}^{\infty}d\epsilon\rho_{M}\left(\epsilon\right)f_{M}\left(\epsilon\right),\label{eq:2.2.7}
\end{equation}
where $\rho_{A,M}\left(\epsilon\right)$ are the density of states
(DOS) of the atoms and molecules, respectively. Qualitatively, for
atoms we shall use the well-known expression for DOS in harmonically
trapped systems without SO coupling,
\begin{equation}
\rho_{A}\left(\epsilon\right)d\epsilon=\frac{\epsilon^{2}}{2\left(\hbar\omega\right)^{3}}d\epsilon,\label{eq:2.2.8}
\end{equation}
where $\omega\equiv(\omega_{x}\omega_{y}\omega_{z})^{1/3}$. For molecules,
we instead use,
\begin{equation}
\rho_{M}\left(\epsilon\right)d\epsilon=\frac{\sqrt{\gamma}\epsilon^{2}}{2\left(\hbar\omega\right)^{3}}d\epsilon.\label{eq:2.2.9}
\end{equation}
The factor $\sqrt{\gamma}$ appearing in the DOS of molecules is due
to the SO coupling, which induces an effective mass $M_{x}=\gamma\cdot(2m)$
as we will make clear later. Combining Eqs. (\ref{eq:2.2.2}) and
(\ref{eq:2.2.3}) and (\ref{eq:2.2.6})-(\ref{eq:2.2.9}), the populations
of the atoms and \emph{noncondensed} molecules are given by
\begin{equation}
N_{A}=-\left(\frac{k_{B}T}{\hbar\omega}\right)^{3}Li_{3}\left(-z_{A}\right),\label{eq:2.2.10}
\end{equation}
\begin{equation}
N_{M}=\sqrt{\gamma}\left(\frac{k_{B}T}{\hbar\omega}\right)^{3}Li_{3}\left(z_{M}\right),\label{eq:2.2.11}
\end{equation}
where $z_{A}=e^{\mu/\left(k_{B}T\right)}$ and $z_{M}=e^{\left(2\mu+E_{B}\right)/\left(k_{B}T\right)}$
are the fugacities of the atoms and molecules, respectively, and $Li_{n}\left(z\right)$
is the polylogarithm function. Therefore, the chemical potential $\mu$
should satisfy the following equation,
\begin{equation}
2N_{A}+2\left[N_{M}+N_{M}^{(0)}\right]=N,\label{eq:2.2.12}
\end{equation}
where $N_{M}^{(0)}$ is the population of the \emph{condensed} molecules,
and $N$ is the total number of atoms. Note that one molecule is counted
as two atoms. Above the critical temperature, i.e., $T>T_{c}$, there
is no condensed molecules ($N_{M}^{(0)}=0$), therefore Eqs. (\ref{eq:2.2.10})-(\ref{eq:2.2.12})
yield
\begin{equation}
-2Li_{3}\left(-z_{A}\right)+2\sqrt{\gamma}Li_{3}\left(z_{M}\right)=N\left(\frac{\hbar\omega}{k_{B}T}\right)^{3}.\label{eq:2.2.13}
\end{equation}
Then from Eq. (\ref{eq:2.2.13}), the chemical potential $\mu$ can
numerically be evaluated at a specific temperature and interaction
strength $E_{B}$. The molecule BEC occurs below the critical temperature,
i.e., $T<T_{c}$. In this case, the chemical potential of molecules
vanishes, i.e., $\mu_{M}=0$, which in turn gives $\mu=-E_{B}/2$.

With the chemical potential $\mu$ obtained, the properties of the
trapped systems may be calculated by integrating over the whole space,
based on the local solutions of uniform systems. In the following
sections, we will firstly solve single-particle and two-particle problems
in uniform systems, and then investigate the rf spectroscopy of trapped
gases under the LDA approximation.

\section{single- and two-particle problems for uniform systems\label{sec:single--and-two-body}}

\subsection{Single-particle solution}

The single-particle problem in a uniform system has been discussed
in detail in our previous work \cite{Hu2012R,Liu2012M}. Here, for
self-containment we summarize briefly the results. We focus on the
following non-interacting Hamiltonian for the subsystem,
\begin{eqnarray}
\mathcal{H}_{0} & = & \sum_{\sigma}\int d\mathbf{r}\Psi_{\sigma}^{\dagger}\left(\mathbf{r}\right)\frac{\hbar^{2}\mathbf{\hat{k}}^{2}}{2m}\Psi_{\sigma}\left(\mathbf{r}\right)+\nonumber \\
 &  & \frac{\Omega_{R}}{2}\int d\mathbf{r}\left[\Psi_{\uparrow}^{\dagger}\left(\mathbf{r}\right)e^{i2k_{R}x}\Psi_{\downarrow}\left(\mathbf{r}\right)+H.c.\right],\label{eq:3.1.1}
\end{eqnarray}
where $\mathbf{\hat{k}}\equiv-i\nabla$. Under the transformation
\cite{Hu2012R,Liu2012M}, $\Psi_{\uparrow}\left(\mathbf{r}\right)=e^{+ik_{R}x}\psi_{\uparrow}\left(\mathbf{r}\right)$
and $\Psi_{\downarrow}\left(\mathbf{r}\right)=e^{-ik_{R}x}\psi_{\downarrow}\left(\mathbf{r}\right)$,
the spatial dependence of the Raman coupling term in Eq.(\ref{eq:3.1.1})
is removed and yields
\begin{eqnarray}
\mathcal{H}_{0} & = & \int d\mathbf{r}\psi_{\uparrow}^{\dagger}\left(\mathbf{r}\right)\left[\frac{\hbar^{2}\left(k^{2}+k_{R}^{2}\right)}{2m}+\lambda k_{x}\right]\psi_{\uparrow}\left(\mathbf{r}\right)+\nonumber \\
 &  & \int d\mathbf{r}\psi_{\downarrow}^{\dagger}\left(\mathbf{r}\right)\left[\frac{\hbar^{2}\left(k^{2}+k_{R}^{2}\right)}{2m}-\lambda k_{x}\right]\psi_{\downarrow}\left(\mathbf{r}\right)+\nonumber \\
 &  & h\int d\mathbf{r}\left[\psi_{\uparrow}^{\dagger}\left(\mathbf{r}\right)\psi_{\downarrow}\left(\mathbf{r}\right)+H.c.\right],\label{eq:3.1.4}
\end{eqnarray}
 where $\mathbf{e}_{x}$ is the unit vector along $x$ direction.
For convenience, the effective Zeeman field $h\equiv\Omega_{R}/2$
and the SO coupling strength $\lambda\equiv\hbar^{2}k_{R}/m$ are
introduced. This single-particle Hamiltonian (\ref{eq:3.1.4}) can
be easily diagonalized to yield two eigenvalues \cite{Hu2012R,Liu2012M},
\begin{equation}
E_{\mathbf{k}\pm}=\frac{\hbar^{2}k_{R}^{2}}{2m}+\frac{\hbar^{2}k^{2}}{2m}\pm\sqrt{h^{2}+\lambda^{2}k_{x}^{2}}.\label{eq:3.1.5}
\end{equation}
 The symbols ``$\pm$'' stands for the two helicity branches, and
the corresponding single-atom eigenstates in helicity basis take the
form
\begin{equation}
\left[\begin{array}{c}
\left|\mathbf{k}+\right\rangle \\
\left|\mathbf{k}-\right\rangle
\end{array}\right]=\left[\begin{array}{cc}
\cos\theta_{\mathbf{k}} & \sin\theta_{\mathbf{k}}\\
-\sin\theta_{\mathbf{k}} & \cos\theta_{\mathbf{k}}
\end{array}\right]\left[\begin{array}{c}
\left|\mathbf{k}\uparrow\right\rangle \\
\left|\mathbf{k}\downarrow\right\rangle
\end{array}\right],\label{eq:3.1.6}
\end{equation}
 where $\theta_{\mathbf{k}}=\arctan\left[\left(\sqrt{h^{2}+\lambda^{2}k_{x}^{2}}-\lambda k_{x}\right)/h\right]$.
It is obvious that the lowest single-particle energy occurs at $k_{\perp}=\sqrt{k_{y}^{2}+k_{z}^{2}}=0$
and $k_{x}=\sqrt{m^{2}\lambda^{2}/\hbar^{4}-h^{2}/\lambda^{2}}$ if
$h<m\lambda^{2}/\hbar^{2}$, and is given by,
\begin{equation}
E_{min}=\frac{\hbar^{2}k_{R}^{2}}{2m}-\frac{m\lambda^{2}}{2\hbar^{2}}-\frac{\hbar^{2}h^{2}}{2m\lambda^{2}}=-\frac{\hbar^{2}h^{2}}{2m\lambda^{2}}.\label{eq:3.1.7}
\end{equation}

\subsection{Two-body bound state}

The two-body problem with \emph{zero} COM momentum in a uniform system
has already been solved in \cite{Hu2012R}. However, real experiments
are carried out well above the critical temperature \cite{Wang2012S,Cheuk2012S},
where two-particle states with a \emph{non-zero} COM momentum may
become important. As we shall see, at the quantitive level, molecules
with a nonzero COM momentum can have small contribution to the (momentum-resolved)
rf spectroscopy. Here, we outline briefly the general solution with
arbitrary COM momenta.

Under the gauge transformation for the field operators, the form of
the interatomic interaction (\ref{eq:2.1.2}) is invariant, $\mathcal{H}_{int}=U_{0}\int d\mathbf{r}\psi_{\uparrow}^{\dagger}\left(\mathbf{r}\right)\psi_{\downarrow}^{\dagger}\left(\mathbf{r}\right)\psi_{\downarrow}\left(\mathbf{r}\right)\psi_{\uparrow}\left(\mathbf{r}\right)$.
For a two-body problem in the presence of SO coupling, the COM momentum
is a good quantum number, and therefore the two-body wavefunction
at a given COM momentum $\mathbf{q}$ can generally be written as
(we set the volume $V=1$) \cite{Yu2011S} \begin{widetext}
\begin{multline}
\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle =\frac{1}{\sqrt{2\mathcal{C}}}\sum_{\mathbf{k}}\left[\psi_{\uparrow\downarrow}\left(\mathbf{k}\right)c_{\mathbf{q}/2+\mathbf{k}\uparrow}^{\dagger}c_{\mathbf{q}/2-\mathbf{k}\downarrow}^{\dagger}+\psi_{\downarrow\uparrow}\left(\mathbf{r}\right)c_{\mathbf{q}/2+\mathbf{k}\downarrow}^{\dagger}c_{\mathbf{q}/2-\mathbf{k}\uparrow}^{\dagger}+\right.\\
\left.\psi_{\uparrow\uparrow}\left(\mathbf{k}\right)c_{\mathbf{q}/2+\mathbf{k}\uparrow}^{\dagger}c_{\mathbf{q}/2-\mathbf{k}\uparrow}^{\dagger}+\psi_{\downarrow\downarrow}\left(\mathbf{k}\right)c_{\mathbf{q}/2+\mathbf{k}\downarrow}^{\dagger}c_{\mathbf{q}/2-\mathbf{k}\downarrow}^{\dagger}\right]\left|vac\right\rangle ,\label{eq:3.2.2}
\end{multline}
where $c_{\mathbf{k}\uparrow}^{\dagger}$ and $c_{\mathbf{k}\downarrow}^{\dagger}$
are creation field operators of spin-up and spin-down atoms with momentum
$\mathbf{k}$, and $\mathcal{C}=\sum_{{\bf k}}[\left|\psi_{\uparrow\downarrow}\left({\bf k}\right)\right|^{2}+\left|\psi_{\downarrow\uparrow}\left({\bf k}\right)\right|^{2}+\left|\psi_{\uparrow\uparrow}\left({\bf k}\right)\right|^{2}+\left|\psi_{\downarrow\downarrow}\left({\bf k}\right)\right|^{2}]$
is the normalization factor of the two-body wavefunction. The Schr\"{o}dinger
equation $\mathcal{H}\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle =\varepsilon_{B}\left(\mathbf{q}\right)\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle $
is equivalent to the following set of coupled equations,

\begin{eqnarray}
\left[\varepsilon_{B}\left(\mathbf{q}\right)-\left(\frac{\hbar^{2}k_{R}^{2}}{m}+\frac{\hbar^{2}k^{2}}{m}+\frac{\hbar^{2}q^{2}}{4m}+2\lambda k_{x}\right)\right]\psi_{\uparrow\downarrow}\left({\bf k}\right) & = & +(U_{0}/2)\sum\nolimits _{{\bf k}^{\prime}}\left[\psi_{\uparrow\downarrow}\left({\bf k}^{\prime}\right)-\psi_{\downarrow\uparrow}\left({\bf k}^{\prime}\right)\right]+h\psi_{\uparrow\uparrow}\left({\bf k}\right)+h\psi_{\downarrow\downarrow}\left({\bf k}\right),\label{eq:3.2.4}\\
\left[\varepsilon_{B}\left(\mathbf{q}\right)-\left(\frac{\hbar^{2}k_{R}^{2}}{m}+\frac{\hbar^{2}k^{2}}{m}+\frac{\hbar^{2}q^{2}}{4m}-2\lambda k_{x}\right)\right]\psi_{\downarrow\uparrow}\left({\bf k}\right) & = & -(U_{0}/2)\sum\nolimits _{{\bf k}^{\prime}}\left[\psi_{\uparrow\downarrow}\left({\bf k}^{\prime}\right)-\psi_{\downarrow\uparrow}\left({\bf k}^{\prime}\right)\right]+h\psi_{\uparrow\uparrow}\left({\bf k}\right)+h\psi_{\downarrow\downarrow}\left({\bf k}\right),\label{eq:3.2.5}\\
\left[\varepsilon_{B}\left(\mathbf{q}\right)-\left(\frac{\hbar^{2}k_{R}^{2}}{m}+\frac{\hbar^{2}k^{2}}{m}+\frac{\hbar^{2}q^{2}}{4m}-\lambda q_{x}\right)\right]\psi_{\uparrow\uparrow}\left({\bf k}\right) & = & h\psi_{\uparrow\downarrow}\left({\bf k}\right)+h\psi_{\downarrow\uparrow}\left({\bf k}\right),\label{eq:3.2.6}\\
\left[\varepsilon_{B}\left(\mathbf{q}\right)-\left(\frac{\hbar^{2}k_{R}^{2}}{m}+\frac{\hbar^{2}k^{2}}{m}+\frac{\hbar^{2}q^{2}}{4m}+\lambda q_{x}\right)\right]\psi_{\downarrow\downarrow}\left({\bf k}\right) & = & h\psi_{\uparrow\downarrow}\left({\bf k}\right)+h\psi_{\downarrow\uparrow}\left({\bf k}\right).\label{eq:3.2.7}
\end{eqnarray}
\end{widetext}Here, $\varepsilon_{B}\left(\mathbf{q}\right)$ is
the energy of the two-body wavefunction which is dependent on the
COM momentum $\mathbf{q}$. By introducing
\begin{equation}
A_{{\bf kq}}\equiv\varepsilon_{B}\left(\mathbf{q}\right)-\left(\frac{\hbar^{2}k_{R}^{2}}{m}+\frac{\hbar^{2}k^{2}}{m}+\frac{\hbar^{2}q^{2}}{4m}\right),\label{eq:3.2.8}
\end{equation}
$\psi_{s}\left(\mathbf{k}\right)=[\psi_{\uparrow\downarrow}\left({\bf k}\right)-\psi_{\downarrow\uparrow}\left({\bf k}\right)]/\sqrt{2}$
and $\psi_{a}\left({\bf k}\right)=[\psi_{\uparrow\downarrow}\left({\bf k}\right)+\psi_{\downarrow\uparrow}\left({\bf k}\right)]/\sqrt{2}$,
it is easy to show that,

\begin{eqnarray}
\psi_{\uparrow\uparrow}\left({\bf k}\right) & = & \frac{\sqrt{2}h}{A_{{\bf kq}}-\lambda q_{x}}\psi_{a}\left({\bf k}\right),\label{eq:3.2.11}\\
\psi_{\downarrow\downarrow}\left({\bf k}\right) & = & \frac{\sqrt{2}h}{A_{{\bf kq}}+\lambda q_{x}}\psi_{a}\left({\bf k}\right),\label{eq:3.2.12}\\
\psi_{a}\left({\bf k}\right) & = & \frac{2\lambda k_{x}}{4h^{2}+\lambda^{2}q_{x}^{2}}\times\nonumber \\
 &  & \left(\frac{\lambda^{2}q_{x}^{2}}{A_{{\bf kq}}}+\frac{4h^{2}A_{{\bf kq}}}{A_{{\bf kq}}^{2}-4h^{2}-\lambda^{2}q_{x}^{2}}\right)\psi_{s}\left({\bf k}\right),\label{eq:3.2.13}
\end{eqnarray}
and
\begin{equation}
\left[A_{{\bf kq}}-\frac{4\lambda^{2}k_{x}^{2}/A_{{\bf kq}}}{1-4h^{2}/\left(A_{{\bf kq}}^{2}-\lambda^{2}q_{x}^{2}\right)}\right]\psi_{s}\left({\bf k}\right)=U_{0}\sum_{{\bf k}^{\prime}}\psi_{s}\left({\bf k}^{\prime}\right).\label{eq:3.2.14}
\end{equation}
From Eq. (\ref{eq:3.2.14}), the un-normalized spin-singlet wavefunction
$\psi_{s}\left({\bf k}\right)$ is given by,
\begin{equation}
\psi_{s}\left({\bf k}\right)=\left[A_{{\bf kq}}-\frac{4\lambda^{2}k_{x}^{2}/A_{{\bf kq}}}{1-4h^{2}/\left(A_{{\bf kq}}^{2}-\lambda^{2}q_{x}^{2}\right)}\right]^{-1}.\label{eq:3.2.15}
\end{equation}
 Thus, the two-body binding energy is determined by $U_{0}\sum_{\mathbf{k}}\psi_{s}\left(\mathbf{k}\right)=1$
, or
\begin{equation}
\frac{m}{4\pi\hbar^{2}a_{s}}-\sum_{{\bf k}}\left[\psi_{s}\left({\bf k}\right)+\frac{m}{\hbar^{2}k^{2}}\right]=0.\label{eq:3.2.16}
\end{equation}
Here, the bare interaction $U_{0}$ has been regularized by the \textit{s}-wave
scattering length $a_{s}$ . A bound state exists if its energy satisfies
$\varepsilon_{B}\left(\mathbf{q}\right)<2E_{min}$, where $E_{min}$
is the lowest single-atom energy given by Eq. (\ref{eq:3.1.7}). The
normalization factor for the wavefunction is given by

\begin{eqnarray}
\mathcal{C} & = & \sum_{\mathbf{k}}\left|\psi_{s}\left({\bf k}\right)\right|^{2}\left\{ 1+\left[\frac{4h^{2}\left(A_{\mathbf{kq}}^{2}+\lambda^{2}q_{x}^{2}\right)}{\left(A_{\mathbf{kq}}^{2}-\lambda^{2}q_{x}^{2}\right)^{2}}+1\right]\times\right.\nonumber \\
 &  & \left.\frac{4\lambda^{2}k_{x}^{2}/A_{{\bf kq}}^{2}}{\left[1-4h^{2}/\left(A_{{\bf kq}}^{2}-\lambda^{2}q_{x}^{2}\right)\right]^{2}}\right\} .\label{eq:3.2.17}
\end{eqnarray}
\begin{figure}
\includegraphics[width=1\columnwidth]{fig1}

\caption{(Color online) Binding energies $-\varepsilon_{B}/E_{\lambda}$ of
two atoms with zero COM momentum $\mathbf{q}=0$ as functions of the
\textit{s}-wave scattering length $1/\left(k_{R}a_{s}\right)$ for
three typical values of the Zeeman field. The horizontal dotted lines
are the threshold energies $2E_{min}$ where the bound states disappear.
Here, we have introduced $k_{R}=m\lambda/\hbar^{2}$ and $E_{\lambda}=m\lambda^{2}/\hbar^{2}$
as the units of the wavevector and energy, respectively.}

\label{fig1}
\end{figure}
The bound state energy can be easily solved from Eq. (\ref{eq:3.2.16})
and the results are plotted as a function of the \textit{s}-wave scattering
length in Fig.\ref{fig1} for zero COM momentum at three typical values
of the Zeeman field ($h/E_{\lambda}=0.2,0.5,0.8$). Here, we have
introduced $k_{R}=m\lambda/\hbar^{2}$ and $E_{\lambda}=m\lambda^{2}/\hbar^{2}$
as the units of the wavevector and energy, respectively. We find that
the bound state only exists for $a_{s}>0$.

Due to the SO coupling, the effective mass of the two-body bound state
is affected. In the limit of small COM momentum, i.e., $\mathbf{q}\approx0$,
the energy of two-body bound state would have a well-defined dispersion,
\begin{equation}
\varepsilon_{B}\left(\mathbf{q}\right)=\varepsilon_{B}\left(0\right)+\frac{\hbar^{2}q_{x}^{2}}{2M_{x}}+\frac{\hbar^{2}q_{\perp}^{2}}{2M_{\perp}},\label{eq:3.2.18}
\end{equation}
where $\varepsilon_{B}\left(0\right)$ is the two-body binding energy
with zero COM momentum, and $M_{x,\perp}$ are the effective masses
in $x$ and transverse ($y$ and $z$) directions, respectively. The
transverse effective mass $M_{\perp}$ is not affected by the SO coupling,
i.e., $M_{\perp}=2m$, since the SO coupling is only applied along
$x$ direction, therefore
\begin{equation}
A_{{\bf kq}}=A_{\mathbf{k}}+\frac{\hbar^{2}q_{x}^{2}}{4m}\left(\frac{2m}{M_{x}}-1\right),\label{eq:3.2.19}
\end{equation}
where $A_{\mathbf{k}}=\varepsilon_{B}\left(0\right)-\hbar^{2}k_{R}^{2}/m-\hbar^{2}k^{2}/m$.
Substituting Eq. (\ref{eq:3.2.19}) into Eq. (\ref{eq:3.2.16}), expanding
up to the second order of $q_{x}$, and comparing the corresponding
coefficients, we can easily obtain
\begin{equation}
\frac{1}{\gamma}\equiv\frac{2m}{M_{x}}=1-\frac{m\lambda^{2}}{\hbar^{2}}\cdot\frac{I_{2}}{I_{1}},\label{eq:3.2.21}
\end{equation}
 where
\begin{eqnarray}
I_{1} & = & -\sum_{\mathbf{k}}\frac{\left(A_{\mathbf{k}}^{2}-4h^{2}\right)^{2}+4\lambda^{2}k_{x}^{2}\left(A_{\mathbf{k}}^{2}+4h^{2}\right)}{4A_{\mathbf{k}}^{2}\left(A_{\mathbf{k}}^{2}-4h^{2}-4\lambda^{2}k_{x}^{2}\right)^{2}},\label{eq:3.2.23}\\
I_{2} & = & \sum_{\mathbf{k}}\frac{16\lambda^{2}h^{2}k_{x}^{2}}{A_{\mathbf{k}}^{3}\left(A_{\mathbf{k}}^{2}-4h^{2}-4\lambda^{2}k_{x}^{2}\right)^{2}}.\label{eq:3.2.24}
\end{eqnarray}
The effective mass ratio $\gamma\equiv M_{x}/\left(2m\right)$ as
a function of the interaction strength is shown in Fig.\ref{fig2}
for three typical values of the Zeeman field ($h/E_{\lambda}=0.2,0.5,0.8$).
We find in the BEC limit, the effective mass is independent of the
values of the Zeeman field, and $\gamma$ approaches unity, since
two atoms are deeply bound and form a tightly bound molecule. When
the interaction is tuned towards the resonance, the two-body bound
state becomes looser, and the SO coupling will induce a relatively
large effective mass. However, the large SO coupling could also destroy
the bound state at a critical \textit{s}-wave scattering length as
we have seen in Fig.\ref{fig1}.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig2}

\caption{(Color online) The effective mass of weakly bound molecules, $\gamma\equiv M_{x}/\left(2m\right)$,
as functions of the interaction strength $1/\left(k_{R}a_{s}\right)$
for three typical values of the Zeeman field. }

\label{fig2}
\end{figure}

\section{radio-frequency spectroscopy\label{sec:radio-frequency-spectroscopy}}

In this section, we will firstly present the basic idea of the rf
transition based on the Fermi's golden rule, and then derive the Frank-Condon
factor for the rf transition of atoms as well as weakly bound molecules.
Finally, under LDA, we theoretically investigate the rf-spectroscopy
of a harmonically trapped ideal gas mixture of fermionic atoms and
bosonic molecules, which is observable in current experiments.

\subsection{Radio-frequency transition and the Fermi's golden rule}

The basic idear of the rf transition is simple. An rf field is applied
to an atomic Fermi gas with two hyperfine states (denoted as $\left|1\right\rangle =\left|\uparrow\right\rangle $
and $\left|2\right\rangle =\left|\downarrow\right\rangle $), and
drives one of the hyperfine states (i.e., $\left|\downarrow\right\rangle $)
to an upper state $\left|3\right\rangle $ with a bare atomic hyperfine
energy difference $\hbar\omega_{3\downarrow}$ due to the magnetic
field splitting. The Hamiltonian of this rf coupling may be written
as,
\begin{equation}
\mathcal{V}_{rf}=V_{0}\int d\mathbf{r}\left[\psi_{3}^{\dagger}\left(\mathbf{r}\right)\Psi_{\downarrow}\left(\mathbf{r}\right)+H.c.\right],\label{eq:4.1.1}
\end{equation}
 where $\psi_{3}^{\dagger}\left(\mathbf{r}\right)$ creates an atom
in the third state at position $\mathbf{r}$, and $V_{0}$ is the
strength of the rf drive and is related to the Rabi frequency $\omega_{R}$
by $V_{0}=\hbar\omega_{R}/2$. Using the field operator after the
gauge transformation, the rf coupling is
\begin{equation}
\mathcal{V}_{rf}=V_{0}\int d\mathbf{r}\left[e^{-ik_{R}x}\psi_{3}^{\dagger}\left(\mathbf{r}\right)\psi_{\downarrow}\left(\mathbf{r}\right)+H.c.\right],\label{eq:4.1.2}
\end{equation}
 or in the momentum space,
\begin{equation}
\mathcal{V}_{rf}=V_{0}\sum_{\mathbf{k}}\left(c_{\mathbf{k}-k_{R}\mathbf{e}_{x},3}^{\dagger}c_{\mathbf{k}\downarrow}+H.c.\right)\label{eq:4.1.3}
\end{equation}
 for later convenience. Note that, because of the transformation,
effectively there is a momentum loss of $k_{R}\mathbf{e}_{x}$ for
the transferred atoms.

According to the Fermi's golden rule, the general strength of the
rf transition is proportional to the Franck-Condon factor \cite{Chin2005R},
\begin{equation}
\Gamma\left(\Omega\right)=\left|\left\langle \psi_{f}\right|\mathcal{V}_{rf}\left|\psi_{i}\right\rangle \right|^{2}\delta\left(\Omega-\frac{E_{f}-E_{i}}{\hbar}\right),\label{eq:4.1.4}
\end{equation}
where $\left|\psi_{i,f}\right\rangle $ denote the initial and final
states with corresponding energy $E_{i,f}$ , respectively, and $\Omega$
is the frequency of the rf field. The delta function keeps the energy
conserved during the transition.

\subsection{The single-particle rf transition}

In the single-particle picture, the eigenstates of a SO coupled atom
are the helicity states $\left|\mathbf{k}+\right\rangle $ ($\left|\mathbf{k}-\right\rangle $),
other than the original spin states $\left|\mathbf{k}\uparrow\right\rangle $
($\left|\mathbf{k}\downarrow\right\rangle $). If an atom is initially
prepared in the state $\left|\mathbf{k}-\right\rangle $ with energy
$E_{\mathbf{k}-}$, the rf photon will transfer this atom to a third
empty state $\left|3\right\rangle $. In order to obtain the final
state of the rf transition, it's useful to calculate $\mathcal{V}_{rf}\left|\mathbf{k}-\right\rangle $.
Using Eqs. (\ref{eq:3.1.6}) and (\ref{eq:4.1.3}), we can easily
obtain
\begin{equation}
\mathcal{V}_{rf}\left|\mathbf{k}-\right\rangle =V_{0}\cdot\cos\theta_{\mathbf{k}}c_{\mathbf{k}-k_{R}\mathbf{e}_{x},3}^{\dagger}\left|vac\right\rangle .\label{eq:4.2.1}
\end{equation}
This means after the rf transition, there is a probability of $\left|\cos\theta_{\mathbf{k}}\right|^{2}$
that transfers the atom to the final state $\left|3\right\rangle $
with a momentum $\mathbf{k}-k_{R}\mathbf{e}_{x}$. Similarly, the
final state should be
\begin{equation}
\mathcal{V}_{rf}\left|\mathbf{k}+\right\rangle =V_{0}\cdot\sin\theta_{\mathbf{k}}c_{\mathbf{k}-k_{R}\mathbf{e}_{x},3}^{\dagger}\left|vac\right\rangle ,\label{eq:4.2.2}
\end{equation}
if the atom occupies the state $\left|\mathbf{k}+\right\rangle $
before the rf transition. It also means the probability that transfers
an atom to the final state $\left|3\right\rangle $ with a momentum
$\mathbf{k}-k_{R}\mathbf{e}_{x}$ is $\left|\sin\theta_{\mathbf{k}}\right|^{2}$.
Generally, the atoms are initially prepared in the mixture states
of $\left|\mathbf{k}-\right\rangle $ and $\left|\mathbf{k}+\right\rangle $
with the probability $f\left(E_{\mathbf{k}\pm}-\mu\right)$, respectively,
where $f\left(E_{\mathbf{k}\pm}-\mu\right)$ is the Fermi-Dirac distribution,
and $\mu$ is the chemical potential. According to the Fermi's golden
rule, the total rf transition strength therefore is given by ($V_{0}=1$)
\begin{eqnarray}
\Gamma_{A}\left(\Omega\right) & = & \sum_{\mathbf{k}}\sin^{2}\theta_{\mathbf{k}}f\left(E_{\mathbf{k}+}-\mu\right)\delta\left(\Omega-\frac{\Delta\epsilon_{+}}{\hbar}\right)+\nonumber \\
 &  & \sum_{\mathbf{k}}\cos^{2}\theta_{\mathbf{k}}f\left(E_{\mathbf{k}-}-\mu\right)\delta\left(\Omega-\frac{\Delta\epsilon_{-}}{\hbar}\right),\label{eq:4.2.3}
\end{eqnarray}
 where
\begin{eqnarray}
\Delta\epsilon_{\pm} & = & \hbar\omega_{3\downarrow}+\frac{\hbar^{2}\left(\mathbf{k}-k_{R}\mathbf{e}_{x}\right)^{2}}{2m}-E_{\mathbf{k}\pm}\nonumber \\
 & = & -\lambda k_{x}\mp\sqrt{h^{2}+\lambda^{2}k_{x}^{2}}\label{eq:4.2.4}
\end{eqnarray}
 is the probable energy difference between the final and initial states.
Without any confusion, we have ignored the bare hyperfine splitting,
i.e., $\omega_{3\downarrow}=0$.

\subsection{Two-body bound to free radio-frequency transition}

The picture of the two-body bound to free rf transition is that the
rf photon breaks a weakly bound molecule and transfers one of the
two atoms to the third state $\left|3\right\rangle $. The final state
is determined by \cite{Hu2012R}
\begin{equation}
\mathcal{V}_{rf}\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle =V_{0}\sum_{\mathbf{k}}c_{\mathbf{k}-k_{R}\mathbf{e}_{x},3}^{\dagger}c_{\mathbf{k}^{\prime}\downarrow}\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle .\label{eq:4.3.1}
\end{equation}
 Substituting the wavefunction of the two-body bound state (\ref{eq:3.2.2})
into Eq. (\ref{eq:4.3.1}), and after some straightforward algebra,
we shall easily obtain
\begin{multline}
\mathcal{V}_{rf}\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle =-\sqrt{\frac{1}{\mathcal{C}}}V_{0}\sum_{\mathbf{k}}c_{\mathbf{q}/2-\mathbf{k}-k_{R}\mathbf{e}_{x},3}^{\dagger}\times\\
\left\{ \left[\psi_{s}\left(\mathbf{k}\right)+\psi_{a}\left(\mathbf{k}\right)\right]c_{\mathbf{q}/2+\mathbf{k}\uparrow}^{\dagger}+\sqrt{2}\psi_{\downarrow\downarrow}\left(\mathbf{k}\right)c_{\mathbf{q}/2+\mathbf{k}\downarrow}^{\dagger}\right\} \left|vac\right\rangle .\label{eq:4.3.2}
\end{multline}
 Using Eq. (\ref{eq:3.1.6}), the final state can be written in terms
of the helicity basis as
\begin{multline}
\mathcal{V}_{rf}\left|\Psi_{2B}\left(\mathbf{q}\right)\right\rangle =-\sqrt{\frac{1}{\mathcal{C}}}V_{0}\sum_{\mathbf{k}}c_{\mathbf{q}/2-\mathbf{k}-k_{R}\mathbf{e}_{x},3}^{\dagger}\times\\
\left(s_{\mathbf{q}/2+\mathbf{k}+}c_{\mathbf{q}/2+\mathbf{k}+}^{\dagger}-s_{\mathbf{q}/2+\mathbf{k}-}c_{\mathbf{q}/2+\mathbf{k}-}^{\dagger}\right)\left|vac\right\rangle ,\label{eq:4.3.3}
\end{multline}
where
\begin{eqnarray}
s_{\mathbf{q}/2+\mathbf{k}+} & = & \left[\psi_{s}\left(\mathbf{k}\right)+\psi_{a}\left(\mathbf{k}\right)\right]\cos\theta_{\mathbf{q}/2+\mathbf{k}}+\nonumber \\
 &  & \sqrt{2}\psi_{\downarrow\downarrow}\left(\mathbf{k}\right)\sin\theta_{\mathbf{q}/2+\mathbf{k}},\label{eq:4.3.4}
\end{eqnarray}
\begin{eqnarray}
s_{\mathbf{q}/2+\mathbf{k}-} & = & \left[\psi_{s}\left(\mathbf{k}\right)+\psi_{a}\left(\mathbf{k}\right)\right]\sin\theta_{\mathbf{q}/2+\mathbf{k}}-\nonumber \\
 &  & \sqrt{2}\psi_{\downarrow\downarrow}\left(\mathbf{k}\right)\cos\theta_{\mathbf{q}/2+\mathbf{k}}.\label{eq:4.3.5}
\end{eqnarray}
Eq. (\ref{eq:4.3.3}) could be understood as follows. During the rf
transition, a molecule with COM momentum $\mathbf{q}$ is broken by
absorbing the rf photon, and then the spin-down atom is transferred
to the third state. Finally, there are two possibilities.  We may
have two atoms in the third state and upper helicity state, respectively,
with a probability $\left|s_{\mathbf{q}/2+\mathbf{k}+}\right|^{2}/\mathcal{C}$.
Also, we may have a probability of $\left|s_{\mathbf{q}/2+\mathbf{k}-}\right|^{2}/\mathcal{C}$
for having two atoms in the third state and lower helicity state,
respectively. Taking into account these two final states, we should
have the following transfer strength or the Franck-Condon factor,
according to the Fermi's golden rule ($V_{0}=1$),

\begin{eqnarray}
\Gamma_{M}\left(\Omega\right) & = & \frac{1}{\mathcal{C}}\sum_{\mathbf{k}}\left|s_{\mathbf{q}/2+\mathbf{k}+}\right|^{2}\delta\left(\Omega-\frac{\Delta\varepsilon_{+}}{\hbar}\right)+\nonumber \\
 &  & \frac{1}{\mathcal{C}}\sum_{\mathbf{k}}\left|s_{\mathbf{q}/2+\mathbf{k}-}\right|^{2}\delta\left(\Omega-\frac{\Delta\varepsilon_{-}}{\hbar}\right),\label{eq:4.3.6}
\end{eqnarray}
 where
\begin{equation}
\Delta\varepsilon_{\pm}=\frac{\hbar^{2}\left(\mathbf{q}/2-\mathbf{k}-k_{R}\mathbf{e}_{x}\right)^{2}}{2m}+E_{\mathbf{q}/2+\mathbf{k},\pm}-\varepsilon_{B}\left(\mathbf{q}\right)\label{eq:4.3.7}
\end{equation}
 is the probable energy difference between the final and initial states.
Here, we have already ignored the bare hyperfine splitting, i.e.,
$\omega_{3\downarrow}=0$ as well. Eqs.(\ref{eq:4.3.6}) and (\ref{eq:4.3.7})
recover the results of \cite{Hu2012R} when $\mathbf{q}=0$ .

\subsection{Rf spectroscopy for a harmonically trapped\emph{ }gas mixture}

In this subsection, we are ready to investigate theoretically the
rf response signals of the \emph{harmonically trapped} \emph{ideal}
gas mixture of the atoms and molecules, as well as the total rf spectroscopy.

\subsubsection{The signals of the atomic component}

The rf transition strength for atoms in a uniform system is determined
by Eq. (\ref{eq:4.2.3}). In the harmonic trap, the rf transition
strength of atoms at the position $\mathbf{r}$ should be $\Gamma_{A}\left(\Omega,\mathbf{r}\right)$
, in which we shall use the local chemical potential,
\begin{equation}
\mu\left(\mathbf{r}\right)=\mu-\frac{m}{2}(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}+\omega_{z}^{2}z^{2}),\label{eq:4.4.13}
\end{equation}
where $\mu$ is the chemical potential in the trap center determined
by Eq. (\ref{eq:2.2.13}). Thus, using LDA the total rf transition
strength of the atomic component in a harmonic trap is $\Gamma_{A}\left(\Omega\right)=\int d\mathbf{r}\Gamma_{A}\left(\Omega,\mathbf{r}\right)$.

In the actual calculation, the summation $\sum_{\mathbf{k}}$ in $\Gamma_{A}\left(\Omega,\mathbf{r}\right)$
may be replaced by the integral $\left(2\pi\right)^{-3}\int dk_{x}\int d^{2}\mathbf{k}_{\perp}$.
After some tedious but straightforward derivation, the total atomic
rf transition strength has the following form,
\begin{eqnarray}
\frac{E_{\lambda}\Gamma_{A}\left(\Omega\right)}{N} & = & -\frac{3}{2\sqrt{2\pi}}\cdot\left(\frac{E_{\lambda}}{E_{F}}\cdot\frac{T}{T_{F}}\right)^{3/2}\frac{h^{2}}{\Omega^{2}}\times\nonumber \\
 &  & \int_{0}^{\infty}dk_{\perp}^{2}Li_{3/2}\left[-z_{A}e^{-E_{k_{\perp}}/\left(k_{B}T\right)}\right],\label{eq:4.4.14}
\end{eqnarray}
 where
\begin{equation}
E_{k_{\perp}}=\frac{\hbar^{2}\left(k_{\perp}^{2}+k_{R}^{2}\right)}{2m}+\frac{\hbar^{2}}{2m\lambda^{2}}\left(\frac{h^{2}-\Omega^{2}}{2\Omega}\right)^{2}-\frac{h^{2}+\Omega^{2}}{2\Omega},\label{eq:4.4.15}
\end{equation}
and $E_{F}=\hbar\omega\left(3N\right)^{1/3}$ and $T_{F}=E_{F}/k_{B}$
are the Fermi energy and Fermi temperature, respectively. At a given
temperature $T$ and interaction strength $E_{B}$, the atomic chemical
potential $\mu$ is obtained from Eq. (\ref{eq:2.2.13}). Then, with
a fixed total atom number $N$ (which in turn gives the Fermi energy
$E_{F}$), the rf transition strength of the atomic component can
numerically be calculated from Eq. (\ref{eq:4.4.14}). The results
are presented in Fig. \ref{fig3} for different values of Zeeman field
at the given parameters, $E_{B}/E_{\lambda}=2$, $E_{F}/E_{\lambda}=1$
and $T/T_{F}=1$. We find that there is a bimodal structure in the
rf spectroscopy of the atomic component. The stronger rf signals at
the positive frequency $\Omega$ correspond to the transition out
of the lower helicity state, while those with weaker strength at the
negative frequency correspond to the transition out of the upper helicity
state. On the other hand, there is a red-shift in the peak positions
as the strength of the Zeeman field decreases. As we may anticipate,
the peak transition frequency $\Omega$ will reach the bare hyperfine
splitting $\omega_{3\downarrow}=0$ in the absence of SO coupling.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig3}

\caption{(Color online) The rf spectroscopy of the atomic component in a harmonic
trap at different values of the Zeeman field, i.e., $h/E_{\lambda}=0.6,0.8,1.0$.
Here, we have taken $E_{B}/E_{\lambda}=2$, $E_{F}/E_{\lambda}=1$,
and $T/T_{F}=1$. }

\label{fig3}
\end{figure}

In addition to the total strength of the atomic rf transition, we
can also calculate the transition strength of atoms at a specific
momentum $k_{x}$. This momentum-resolved rf spectroscopy has already
been investigated at Shanxi University \cite{Wang2012S} and at MIT
\cite{Cheuk2012S}. According to Eq. (\ref{eq:4.2.3}), we can carry
out the integral over $\mathbf{k}_{\perp}$ and retain that of $k_{x}$.
Therefore, the momentum-resolved rf transition strength is determined
by \begin{widetext}
\begin{eqnarray}
\frac{k_{R}E_{\lambda}\Gamma_{A}\left(\Omega,k_{x}\right)}{N} & = & -\frac{3\sqrt{\pi}}{2\sqrt{2}}\left(\frac{E_{\lambda}}{E_{F}}\cdot\frac{T}{T_{F}}\right)^{3/2}\left(1-\frac{\lambda k_{x}}{\sqrt{h^{2}+\lambda^{2}k_{x}^{2}}}\right)\delta\left(\Omega-\frac{\Delta\epsilon_{+}}{\hbar}\right)\int_{0}^{\infty}dk_{\perp}^{2}Li_{3/2}\left(-z_{A}e^{-E_{\mathbf{k}+}/k_{B}T}\right)+\nonumber \\
 &  & -\frac{3\sqrt{\pi}}{2\sqrt{2}}\left(\frac{E_{\lambda}}{E_{F}}\cdot\frac{T}{T_{F}}\right)^{3/2}\left(1+\frac{\lambda k_{x}}{\sqrt{h^{2}+\lambda^{2}k_{x}^{2}}}\right)\delta\left(\Omega-\frac{\Delta\epsilon_{-}}{\hbar}\right)\int_{0}^{\infty}dk_{\perp}^{2}Li_{3/2}\left(-z_{A}e^{-E_{\mathbf{k}-}/k_{B}T}\right).\label{eq:4.4.16}
\end{eqnarray}
\end{widetext} Corresponding to experiments, the delta function in
Eq. (\ref{eq:4.4.16}) can be replaced by the Lorentz-line shape as,
\begin{equation}
\delta\left(x\right)\rightarrow\frac{1}{\pi}\cdot\frac{\sigma}{x^{2}+\sigma^{2}},\label{eq:4.4.17}
\end{equation}
where $\sigma$ is the line width.

Fig. \ref{fig4} reports the momentum-resolved rf spectroscopy of
the atomic component for different values of the Zeeman field and
interaction strength. Here, we define $K_{x}=k_{x}-k_{R}$ to make
sure the spectra is a symmetric function of $K_{x}$ in the absence
of SO coulping \cite{Liu2012M}. As we anticipate, there are two branches
in which one is relatively weaker than the other one. Since the energy
of the upper helicity state $\left|\mathbf{k}+\right\rangle $ is
larger than that of the lower helicity state $\left|\mathbf{k}-\right\rangle $
, the atoms will occupy the lower helicity state first and then the
upper one before the rf transition. Thus, the initial atomic population
of the lower helicity state should be larger. Therefore, similar to
the integrated rf spectroscopy, the brighter branch of the momentum-resolved
rf spectroscopy corresponds to the rf transition out of the state
$\left|\mathbf{k}-\right\rangle $, while the weaker one corresponds
to the rf transition out of the state $\left|\mathbf{k}+\right\rangle $.
Therefore, the contribution from two initial states could be well
identified experimentally. We can also see that the gap between two
branches becomes larger when the strength of the Zeeman field increases.
This is because the transition frequency deviates from the bare hyperfine
splitting $\omega_{3\downarrow}$ more obviously with a stronger SO
coupling.

All these features, discussed in the above, are in good agreement
with the experimental observation for a non-interacting spin-orbit
coupled atomic Fermi gas; see, for example, Fig. 4 in Ref. \cite{Wang2012S}
and Fig. 2 in Ref. \cite{Cheuk2012S}.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig4}

\caption{(Color online) The momentum-resolved rf spectroscopy of the harmonically
trapped atomic component for different values of the Zeeman field
and interaction strength. Here, we have taken $E_{F}/E_{\lambda}=2$,
$T/T_{F}=1$ and $\sigma=0.1$. In addition, we have also defined
$K_{x}=k_{x}-k_{R}$ (see text), which moves the whole spectra to
the right by an amount $k_{R}$.}

\label{fig4}
\end{figure}

\subsubsection{The rf spectroscopy of the molecular component}

With the two-body binding energy solved from Eq. (\ref{eq:3.2.16})
at a given COM momentum $\mathbf{q}$, the rf transition strength
of a single molecule can be calculated directly from Eq. (\ref{eq:4.3.6}).
We plot the rf response of a single molecule with zero COM momentum,
as well as non-zero COM momentum (i.e., $q_{x}/k_{R}=0.5$) in Fig.
\ref{fig5}. Here, the interatomic interaction strength, denoted by
$E_{B}/E_{\lambda}=\left[\hbar^{2}/\left(m\lambda a_{s}\right)\right]^{2}$,
is set to be unity. Since the SO coupling is only along the $x$ direction,
we can set the transverse COM momentum $\mathbf{q}_{\perp}=0$, which
is physically irrelevant and arbitrary. From Fig. \ref{fig5}, we
find that there is a slight red-shift in the peak position with a
non-zero COM momentum. This is because the energy gap between the
final and initial states becomes smaller at non-zero COM momentum,
and consequently a little higher transition peak appears.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig5}

\caption{(Color online) The rf spectroscopy of a single molecule for different
values of the Zeeman field at given COM momenta $\mathbf{q}/k_{R}=0$
and $\mathbf{q}/k_{R}=0.5$. Here, we have chosen an interaction strength
$E_{B}/E_{\lambda}=1$.}

\label{fig5}
\end{figure}

The momentum-resolved rf spectroscopy of a single molecule can be
obtained from Eq. (\ref{eq:4.3.6}) by integrating over $\mathbf{k}_{\perp}$,
and is given by
\begin{multline}
\Gamma_{M}\left(\Omega,k_{x}\right)=\frac{m}{8\pi^{2}\hbar\mathcal{C}}\left[\left|s_{\mathbf{q}/2+\mathbf{k}+}\right|^{2}\Theta\left(k_{\perp,+}^{2}\right)+\right.\\
\left.\left|s_{\mathbf{q}/2+\mathbf{k}-}\right|^{2}\Theta\left(k_{\perp,-}^{2}\right)\right],\label{eq:4.4.18}
\end{multline}
 where $\Theta\left(x\right)$ is the step function and
\begin{multline}
k_{\perp,\pm}^{2}=\frac{m}{\hbar}\left[\Omega+\varepsilon_{B}\left(\mathbf{q}\right)\right]-\left[\frac{k_{x}^{2}+\left(k_{x}+k_{R}\right)^{2}}{2}+\right.\\
\left.\frac{k_{R}^{2}+\left(k_{R}+q_{x}\right)^{2}}{4}\pm\frac{m}{\hbar^{2}}\sqrt{h^{2}+\lambda^{2}\left(\frac{q_{x}}{2}+k_{x}\right)^{2}}\right].\label{eq:4.4.19}
\end{multline}
 The momentum-resolved rf spectroscopy of a single molecule with zero
and non-zero COM momentum ($q_{x}/k_{R}=0.5$) are presented in Fig.
\ref{fig6}. In this plot, we have defined $K_{x}=-k_{x}-k_{R}$,
as we require that in the absence of SO coupling the spectra is an
even function of $K_{x}$ \cite{Hu2012R}. We find that the contribution
from two final states are well separated in different frequency domain
and therefore should be easily observed experimentally.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig6}

\caption{(Color online) The momentum-resolved rf spectroscopy of a single molecule
for different values of the Zeeman field at given COM momenta $\mathbf{q}_{x}/k_{R}=0$
and $\mathbf{q}/k_{R}=0.5$. Here, we have chosen the interaction
strength $E_{B}/E_{\lambda}=1$.}

\label{fig6}
\end{figure}

\subsubsection{The total rf spectroscopy of the gas mixture of atoms and molecules}

By increasing the attractive interaction strength of a two-component
Fermi gas at finite temperatures, the system will evolve into a mixture
of fermionic atoms and bosonic molecules. We have so far discussed
the integrated and momentum-resolved rf responses for both atoms and
molecules. The total rf spectroscopy of a harmonically trapped\emph{
}ideal\emph{ }gas mixture of atoms and molecules can now be easily
obtained.

For the atomic component, the rf transition strength is already given
by Eq. (\ref{eq:4.4.14}). As to the molecules, the population $N_{M}$
can be calculated from Eq. (\ref{eq:2.2.11}) at given temperature
$T/T_{F}$ and interaction strength $E_{B}/E_{\lambda}$. In real
experiments, the temperature is low so that the distribution of the
molecular COM momentum $\mathbf{q}$ is very narrow around $\mathbf{q}=0$.
As already seen from Figs. 5 and 6, the rf spectroscopy of molecules
depends weakly on $\mathbf{q}$. Thus, as a good qualitative approximation,
we may focus on the rf transition of molecules with zero COM momentum
only. Then, the total rf transition strength of molecular component
is given by $N_{M}\Gamma_{M}\left(\Omega\right)$. In the end, we
obtain the total rf spectroscopy,
\begin{equation}
\Gamma_{t}\left(\Omega\right)=\Gamma_{A}\left(\Omega\right)+N_{M}\Gamma_{M}\left(\Omega\right).\label{eq:4.4.20}
\end{equation}

Fig. \ref{fig7} shows the total rf spectroscopy of the atom-molecule
mixture at different Zeeman field and at fixed temperature $T/T_{F}=0.2$
and interaction strength $E_{B}/E_{\lambda}=1$. It is obvious that
there is a double-peak structure in the spectroscopy. The peak at
higher frequency is responsible for weakly bound molecules, since
more energy is required for pair breaking. With increasing the Zeeman
field $h$, the energy difference between the final and initial states
of atoms during the rf transition becomes larger, thus a decrease
in the atomic peak is observed. On the other hand, since the energy
difference between the molecular final and initial states becomes
smaller as the Zeeman field increases, the rf signal of molecules
grows, as we may anticipate.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig7}

\caption{(Color online) The total rf spectroscopy of a harmonically trapped
ideal gas mixture of fermionic atoms and bosonic molecules at different
Zeeman field. Here, we take $E_{B}/E_{\lambda}=1$, $E_{F}/E_{\lambda}=1$
and $T/T_{F}=0.2$ .}

\label{fig7}
\end{figure}

In order to show the temperature dependence, we also calculate the
total rf response of the ideal gas mixture at different temperatures
and at a fixed Zeeman field $h/E_{\lambda}=0.5$, as reported in Fig.
\ref{fig8}. As the temperature decreases, the population of molecules
becomes larger. Thus, a reduction of the atomic signal is observed.
The molecular response should dominate when temperature decreases
to zero.

\begin{figure}
\includegraphics[width=1\columnwidth]{fig8}

\caption{(Color online) The total rf spectroscopy of a harmonically trapped
ideal gas mixture of fermionic atoms and bosonic molecules as a function
of temperature at a fixed Zeeman field strength $h/E_{\lambda}=0.5$.
Here, we use $E_{B}/E_{\lambda}=1$ and $E_{F}/E_{\lambda}=1$. }

\label{fig8}
\end{figure}

\begin{figure}[b]
\includegraphics[width=1\columnwidth]{fig9}

\caption{(Color online) The total momentum-resolved rf spectroscopy of a harmonically
trapped ideal gas mixture of fermionic atoms and bosonic molecules.
The left panel shows the spectra at different Zeeman field strength
at a fixed temperature $T/T_{F}=0.2$, while the right panel shows
the temperature dependence of spectroscopy at a fixed Zeeman field
$h/E_{\lambda}=0.5$. Here, we take $E_{B}/E_{\lambda}=1$ and $E_{F}/E_{\lambda}=1$.
Note the different color scale in the left and right panels.}

\label{fig9}
\end{figure}
Finally, we present the total momentum-resolved rf spectroscopy in
Fig. \ref{fig9}, with parameters correpsonding to Fig. \ref{fig7}
(see the left panel) and Fig. \ref{fig8} (right panel).

\section{conclusions\label{sec:conclusions}}

In conclusion, we have theoretically investigated the momentum-resolved
radio-frequency spectroscopy of a harmonically trapped\emph{ }ideal\emph{
}gas mixture of fermionic atoms and bosonic molecules with spin-orbit
coupling. This is a \emph{qualitative} theoretical model of spin-orbit
coupled atomic Fermi gases near broad Feshbach resonances.

In particular, for local uniform systems, general formulas for the
wavefunction and binding energy of weakly bound bosonic molecules
with \emph{non-zero} center-of-mass momentum have been derived. The
influence of this non-zero center-of-mass momentum on the radio-frequency
spectroscopy of molecules has been discussed in detail. Based on few-body
solutions in uniform systems, the radio-frequency responses of atomic
and molecular components in harmonic traps have been calculated within
local density approximation, respectively, by using Fermi's golden
rule. Our results provide a qualitative picture of momentum-resolved
radio-frequency spectroscopy of a strongly interacting spin-orbit
coupled Fermi gas in harmonic traps, which could be directly observed
in future experiments at Shanxi University \cite{Wang2012S,ZhangPrivateComm}
and MIT\cite{Cheuk2012S,ZwierleinPrivateComm}.

In real experiments, where strongly interacting Fermi gases are created
in the vicinity of Feshbach resonances, the atoms and weakly bound
molecules must strongly interact with each other. An in-depth analysis
of radio-frequency spectroscopy would rely on much more complicated
many-body theories \cite{Hu2006}, beyond the simple few-body picture
presented in this work.
\begin{acknowledgments}
Shi-Guo Peng and Kaijun Jiang are supported by China Postdoctoral
Science Foundation (Grant No. 2012M510187), the NSFC projects (Grant
No. 11004224 and No.11204355) and the NFRP-China (Grant No. 2011CB921601).
Xia-Ji Liu and Hui Hu are supported by the ARC Discovery Projects
(Grant No. DP0984637 and No. DP0984522) and the NFRP-China (Grant
No. 2011CB921502). \end{acknowledgments}

redpajama_set_name: RedPajamaArXiv


Machu Picchu is the most famous destination in Peru. Its mystery, allure, and mind blowing setting bring millions of people to visit every year. For a once ancient city that was never meant to be found, Machu Picchu is quite popular. Visiting this massive complex can be a bit tricky. We broke down the best ways to reach the UNESCO World Heritage site.
Start in Cusco. Then, take a scenic 3.5 hour train along the Urumbamba River in the Sacred Valley to Aguas Calientes.
Start in Cusco. If tickets are sold out from Cusco, then depart to Aguas Calientes from the town of Ollantaytambo in the Sacred Valley, or vice versa. Cusco to Ollantaytambo is just over an hour each way so you can also take a taxi or a minivan. Then, from Ollantaytambo take a train to Aguas Calientes.
Either way, there are three main train companies that offer trips to Aquas Calientes: Perurail, Inca Rail, and the Machu Picchu Train. The most popular of these three is Perurail because it offers 3 levels of class to choose from: the Backpacker, the Vistadome, and the ultra-luxurious Hiram Bingham train. The Vistadome is quite popular due to its expansive glass windows that offer endless views of the wild landscape. Regardless of which train option you choose, book in advance as tickets sell out really fast. Tickets sell out months in advance during high season.
You made it to Aguas Calientes. Now what? If you are fit enough to take on 4 day /3 night hike, then challenge yourself to reaching the ruins via the Inca Trail. Hiking the Inca Trail is certainly not an easy trek, but it is the most memorable way to reach Machu Picchu.
It is important to note that it is prohibited to do any form of an independent hike. The only options available to tourists on the Inca Trail is to travel with a sanctioned group or tour agency. This also means that you must book your travel plans well in advance as these tours sell out quickly.
There are several hiking trails to get to Machu Picchu so do your research and choose the hike that best fits you.
Not wanting to rough it in the mountains of Peru hiking the Inca trail? No problem. Buses run every 15-20 mins to the ruins. The bus station is a simple 5 minute walk from the train station. We recommend buying tickets in advance if possible. Also, Aguas Calientes is an ancient town that is the closest form of civilization to Machu Picchu. It is well worth staying a night or two to explore the area.
If you have the option, choose the earliest possible bus ticket time. Arrive as early as possible to the site to enjoy sunrise light and beat the tour buses. Especially in high season, the ruins can fill up quickly with hundreds of tourists. Going in the off-season or arriving early in the day can really make a difference in your experience.


Annual performance reviews have been a staple in performance management for decades. It wasn't until recent years that companies started to question the validity of such an archaic process and started looking at more efficient practices. Working alongside HR professionals and industrial psychologists, business owners have been able to get rid of antiquated structures and implement systems that truly encourage high performance. Evidence shows that when leaders incorporate real-time, ongoing feedback throughout the year, they are able to increase engagement, promote growth and development, and address concerns before they become problems.
With the help of a brilliant Engagement Specialist, Performance Software has developed a customized feedback system, coined People First, to replace the status quo. People First consists of frequent, casual, 1:1s that occur monthly with the employee and the manager. Not only will this improve relationships between the various levels in the organization, but it's a cultural infusion for some of the values that we uphold at the organization.
Ownership: The employee is the one to schedule the check-in; they own their performance management. We believe that each individual should be able to solicit as much or as little feedback as they're looking for and receive coaching on the areas they're striving to improve.
Drive: Looking to move up or get that promotion? Take a role in leadership? Mentor? Whatever it may be, the employees are the ones to verbalize and set their goals alongside their manager to work toward them as a team.
Appreciation: What better time to thank the employee for their hard work? Although we infuse as much real-time appreciation into our culture, it's always nice to receive some extra validation for a job well done.
What is top leadership saying about the monthly check-ins?
How does your company encourage professional development, feedback, and growth plans? Share your thoughts with us.


We finally made it to Hawaii and it is B-E-A-utiful! and hot... We've been very busy trying to get things in order today Kapono is job hunting and figuring out what to do about a vehicle. Asa and Alice are loving it, well Alice doesn't have much of an opinion she sleeps a lot. But Asa is having a blast running around with his cousins and being outside without a jacket. Today out of habit he asked for his jacket before going out side, he put it on went out and after 30 sec came back in and threw it on the living room floor. Great News! the Bartons are set up on a network here and no other computers can be added which means we can not use our computer to update our blog and since our usb drive is broken we can not add pictures, so once again only a text blog nothing else. Not to brag but it really is great being back here, my skin instantly felt better. I had chapped lips when we boarded the plane in Utah and not even a week later its completely better without using moisturizer. the best part is that there are 5 cooks in the house, its not like I did a lot of cooking in Utah but now I dont even have to make dessert, my sister in law, the oldest barton boy Kaumana's wife Viana's mom owned a bakery so she picked up on her skills.


Well, we can't say we didn't see this one coming. Having just opened MP Taverna on Driggs and North 10th Streets (pretty much ground zero for Williamsburg's shiny new condos) chef and restauranteur Michael Psilakis is now getting into the business of venues. What better way to connect to the neighborhood and a now nearly mythologized North Brooklyn cultural movement and indie rock scene? The Hall, a new music venue that boasts a mission to "put the focus back on supporting local arts, musicians, creatives, students and neighborhood residents," won't officially open till later this month. But we attended a preview event last night to see what we're in for.


text:
Biz website security provider Sitelock acquires Dutch startup Patchman
Host Papa
Scottsdale (Arizona) headquartered business website security solutions provider SiteLock has acquired Dutch startup Patchman for an undisclosed amount. The latter provides web security solutions to hosting providers and detects vulnerabilities found commonly in content management systems such as WordPress, Drupal and Joomla. According to an official communication, SiteLock, with this acquisition will be able to have a global footprint and expand its threat database. It will also be able to extend its services to the hosting provider community. Read More: Cyber threats have been evolving, why hasn't security education? "The vision of Patchman aligns with SiteLock very well. We're excited to join forces and execute on our shared goal of securing every website on the internet," said Neill Feather, President, SiteLock.
"This acquisition will double the number of websites we protect, allowing us to increase our knowledge base, better identify malware patterns and more effectively mitigate threats. In addition, we can now provide more options for hosting providers around the world to work with us, and ensure high quality solutions are available to their end customers."
Founded in 2008, SiteLock claims to protect over 12,000,000 websites worldwide. With the addition of Patchman, the websites' security company will now have more than 500 global partners, including Host Papa, UK2 Group and more. Patchman was founded in 2014. "The Patchman team is thrilled to join forces with SiteLock, as combining our unique technologies presents many new and exciting opportunities," said Wouter de Vries, Founder of Patchman. "We look forward to accelerating our already impressive growth and continuing to deliver top notch web security solutions worldwide."
Big Data: The 10 Big challenges that are hindering its adoption
Bitcoin has laid a strong foundation for other cryptocurrencies: Sathvik Vishwanath, Unocoin
24 Haymarket
Privitar raises $16M to secure data analytics processes

redpajama_set_name: RedPajamaCommonCrawl


text:
Advertising On Connected TV
November 2, 2017 saucemaster
If you ask any Millennial or Gen Z'er what it is that they do for fun, a healthy majority will list off curling up on a couch and binge-watching their favorite tv show.
More and more people are turning away from traditional TV and are tuning in via Connected TV (Netflix, Hulu, Sling, YouTube, HBO GO, etc.).
What is Connected TV?
According to Centro, Connected TV is defined as a television set that is connected to the internet via built-in capabilities, a media streaming device, gaming console, or Blu-ray player, and facilitates the delivery of streaming video content.
Right now, 69% of households have a Connected TV, with that number projected to increase year-over-year. Similar to traditional TV's audience, Connected TV users are using the platforms to watch full episodes of their favorite shows and to tune-in to live events like The World Series and The Emmys.
While the audience that consumes traditional TV is skewed older, Connected TV's audience is younger and more affluent with a median age of 31. So, if your advertising tactics are strictly focused on traditional TV, you're missing out on reaching an important audience.
Right now, there are three different ways to reach your audience using Connected TV advertising: in-stream video, interactive pre-roll and home screen placements.
In-stream video ads are :15s or :30s non-skippable ads that play either before the user's program or during breaks. This kind of advertisement is a really great way to repurpose spots created for TV to earn impressions and reach other demographics.
Interactive Pre-Roll
Just like in-stream video, interactive pre-roll ads play before and during the user's show, but it gives the user the option to interact with the ad. In the example below, Snickers made a game out of their interactive pre-roll ad for YouTube which ends up not feeling like an ad at all, but rather a fun and memorable interaction with a brand.
Home Screen Placements
This option is the least intrusive ad type on Connected TV. These ads are placed on the home screen of a Connected TV app and have the capabilities to feature interactive calls-to-action like, "Click to Learn More."
So where do you start? First, define your audiences. Are your clients or consumers viewing more traditional or Connected TV?
Then, figure out the goal of your messaging. Do you want to send your audience members to a landing page and collect leads or are you more interested in earning impressions?
Next, take a look at your options for Connected TV platforms. For example, Netflix is still an advertising-free platform whereas Hulu and YouTube are not.
Finally, it's time to put together a creative and memorable ad to serve to your audience. That's where TAG comes in. Let's talk about your digital strategy and figure out a campaign to deliver measurable results.
Uncategorized	ad, ads, advertising, Agency, Audience, binge, binge watching, binge-watch, built-in, Campaign, collect, connect, connected, connected tv, define, device, digital, gen z, genz, hbo, hbo go, home screen placements, Hulu, impression, impressions, in-stream, in-stream video, interactive pre-roll, internet, lead, leads, marketing, media, millennial, Netflix, popular, pre-roll, results, show, shows, Sling, strategy, stream, streaming, streaming video, Television, traditional, tv, videow, watch, youtube	Leave a comment
Build Brand Loyalty With Relationship Marketing
April 14, 2017 saucemaster
Let's say a customer spends an average of $60 in your store every 3 months. They live in the same place for 20 years. Their lifetime value is $4,800.
Now let's say, the first time they went into your store they didn't have a great experience. You didn't take interest in them, they didn't feel valued, and you never followed up after they left. You didn't lose $60, you lost $4,800. (Calculate the lifetime value of your customers here)
That's why relationship marketing is so important. Relationship marketing is the act of developing customer loyalty and building long-term relationships through interaction and engagement.
IPSY and Caseology excel at forming relationships with their customers. You don't have to be a monthly subscription service or a phone case company to engage with your customers. Here are a few easy things you can do to create relationships with your customers.
1. Develop an emotional connection.
Brand loyalty is built when your customers feel like you und erstand and value them. By personalizing your customer's experience, they're more likely to return AND they're more likely to leave you a positive review – which is really important because 81% of shoppers conduct online research before buying!
2. Share exclusive offers and content.
Make your customers feel like they're part of a special club! Offering them coupons and promotions through an email or on your website encourages them to continue to shop with you. Specials on their birthdays, reward points when they review a product or your brand, and discounts on premium products are great ways to do this!
3. Use technology to your advantage.
Social media is your friend! You can gauge your customers' perceptions of your brand by what they say on social media. Use these platforms to communicate with your customers and really listen to their feedback. Regular communication with your customers, digitally or otherwise, is the best way to cultivate relationships. By doing so, you're adding value to your brand.
At the end of the day it's most important to remember that your customers are humans who want to feel valued. After all, they are why you do what you do!
If you're looking for guidance to begin building relationships with your customers, contact TAG!
Tell us what you think about Donuts & Digital and TAG Hot Sauce by leaving a Facebook review or commenting below!
Brand, Digital Marketing, Email Marketing, Social Media, Uncategorized, Video	awareness, Brand, build, caseology, company, company's, connect, connection, consumer, content, customer, digital media, emotional, Engagement, exclusive, ipsy, loyalty, marketing, media, offers, relationship, relationship marketing, relationships, review, social, social media, survey, value	Leave a comment


text:
Adrien Hardy, född den 30 juli 1978 i Nîmes i Frankrike, är en fransk roddare.

Han tog OS-guld i dubbelsculler tillsammans med Sébastien Vieilledent i samband med de olympiska roddtävlingarna 2004 i Aten.

Referenser

Noter

Externa länkar
olympedia.org

Franska roddare
Franska olympiska guldmedaljörer
Olympiska guldmedaljörer 2004
Tävlande vid olympiska sommarspelen 2000 från Frankrike
Tävlande i rodd vid olympiska sommarspelen 2000
Tävlande vid olympiska sommarspelen 2004 från Frankrike
Tävlande i rodd vid olympiska sommarspelen 2004
Tävlande vid olympiska sommarspelen 2008 från Frankrike
Tävlande i rodd vid olympiska sommarspelen 2008
Tävlande vid olympiska sommarspelen 2012 från Frankrike
Tävlande i rodd vid olympiska sommarspelen 2012
Personer från Nîmes
Födda 1978
Levande personer
Män

redpajama_set_name: RedPajamaWikipedia


text:
Q: How to run mediapipe with c++ on M1 mac environment
Macbook air M1 2020.
Big Sur 11.3.1.
What I wanna do
I want to succeed to run "mediapipe c++ hello_world" below.
https://google.github.io/mediapipe/getting_started/install.html#installing-on-macos
I tried the following tries and failed.
try1
do
Just follow the guideline
https://google.github.io/mediapipe/getting_started/install.html#installing-on-macos

*

*4.Install OpenCV and FFmpeg. => I used option1.

bug
In lesson 6, I faced a bug below.
// run lesson 6
mediapipe % bazel run --define MEDIAPIPE_DISABLE_GPU=1 \
    mediapipe/examples/desktop/hello_world:hello_world
2021/05/23 21:06:14 Downloading https://releases.bazel.build/3.7.2/release/bazel-3.7.2-darwin-arm64...
2021/05/23 21:06:16 could not download Bazel: HTTP GET https://releases.bazel.build/3.7.2/release/bazel-3.7.2-darwin-arm64 failed with error 404
mediapipe %

try2
"try1 bug" said that bazel tried to Download bazel-3.7.2-darwin-arm64 from github release.
But, release version 3.7.2 don't have the type of darwin-arm64.
https://github.com/bazelbuild/bazel/releases/tag/3.7.2
Fortunately, release version 4.1.0 have the type of darwin-arm64.
https://github.com/bazelbuild/bazel/releases/tag/4.1.0
Then I trid to Download bazel-4.1.0-darwin-arm64.
do
In root repository "mediapipe", fix ".bazelversion" file as following.
4.1.0

Then, retried lesson 6 and the download started.
bug
But the download didn't complete successfully: I got the other bug as below.
// run lesson 6 again
mediapipe % bazel run --define MEDIAPIPE_DISABLE_GPU=1 \
    mediapipe/examples/desktop/hello_world:hello_world

~~download process~~

ERROR: /Users/<some_mid_path>/mediapipe/mediapipe/framework/BUILD:971:11: Compiling mediapipe/framework/scheduler_queue.cc failed: (Aborted): wrapped_clang failed: error executing command external/local_config_cc/wrapped_clang '-D_FORTIFY_SOURCE=1' -fstack-protector -fcolor-diagnostics -Wall -Wthread-safety -Wself-assign -fno-omit-frame-pointer -O0 -DDEBUG '-std=c++11' ... (remaining 60 argument(s) skipped)

Use --sandbox_debug to see verbose messages from the sandbox
mediapipe/framework/scheduler_queue.cc:212:3: error: expected expression
  AUTORELEASEPOOL {
  ^
mediapipe/framework/scheduler_queue.cc:29:25: note: expanded from macro 'AUTORELEASEPOOL'
#define AUTORELEASEPOOL @autoreleasepool
                        ^
mediapipe/framework/scheduler_queue.cc:226:5: error: use of undeclared identifier 'is_idle'
    is_idle = IsIdle();
    ^
mediapipe/framework/scheduler_queue.cc:228:7: error: use of undeclared identifier 'is_idle'
  if (is_idle && idle_callback_) {
      ^
3 errors generated.
Error in child process '/usr/bin/xcrun'. 1
Target //mediapipe/examples/desktop/hello_world:hello_world failed to build
Use --verbose_failures to see the command lines of failed build steps.
INFO: Elapsed time: 1.430s, Critical Path: 1.26s
INFO: 4 processes: 3 internal, 1 darwin-sandbox.
FAILED: Build did NOT complete successfully
FAILED: Build did NOT complete successfully

mediapipe %

Now situation
Then I gave up temporarily and am questioning to you!
Could someone solve this problem??
Thanks in advance.

A: I successfully built the HelloWorld.cc fine.
I also managed to compile Android on M1 MAC. Had some issues with NDK version and permissions but in the end it compiled and ran on my S8 Plus.
My setup

*

*MacBook Pro (13-inch, M1, 2020)

*macOS Big Sur 11.5.1

*ndk version used : r20b (download manually and edit WORKSPACE path to ndk)

During first compile you will get a lot of permission error issues, go to and keep open during build > System Preferences > Security and Privacy > general.
When an error pops up allow to continue and restart build. Will take about 8-10 permissions before it builds.
Below is the last bit of the build process.
Warning: skipping import of repository
'build_bazel_apple_support' because it already exists.

DEBUG: /private/var/tmp/_bazel_jeranio/794c1b4078a0aeb97ae7a379447a99a6/external/org_tensorflow/third_party/repo.bzl:108:14:
Warning: skipping import of repository 'bazel_skylib' because it already exists.

DEBUG: /private/var/tmp/_bazel_jeranio/794c1b4078a0aeb97ae7a379447a99a6/external/org_tensorflow/third_party/repo.bzl:108:14:
Warning: skipping import of repository 'pybind11' because it already exists.

DEBUG: /private/var/tmp/_bazel_jeranio/794c1b4078a0aeb97ae7a379447a99a6/external/tf_runtime/third_party/cuda/dependencies.bzl:51:10: The following command will download NVIDIA proprietary software. By using the software you agree to comply with the terms of the license agreement that accompanies the software. If you do not agree to the terms of the license agreement, do not use the software.

INFO: Analyzed target //mediapipe/examples/android/src/java/com/google/mediapipe/apps/handtrackinggpu:handtrackinggpu (0 packages loaded, 0 targets configured).
INFO: Found 1 target...

Target //mediapipe/examples/android/src/java/com/google/mediapipe/apps/handtrackinggpu:handtrackinggpu up-to-date:
  bazel-bin/mediapipe/examples/android/src/java/com/google/mediapipe/apps/handtrackinggpu/handtrackinggpu_deploy.jar
  bazel-bin/mediapipe/examples/android/src/java/com/google/mediapipe/apps/handtrackinggpu/handtrackinggpu_unsigned.apk
  bazel-bin/mediapipe/examples/android/src/java/com/google/mediapipe/apps/handtrackinggpu/handtrackinggpu.apk
INFO: Elapsed time: 2.984s, Critical Path: 0.97s
INFO: 1 process: 1 internal.
INFO: Build completed successfully, 1 total action

jeranio at Jerrys-MacBook-Pro in ~/mediapipe on master [!?]

redpajama_set_name: RedPajamaStackExchange


text:
Cadillac, Toyota capture top consumer scores
By Georgia Wells
Published: Aug 17, 2011 12:01 a.m. ET
GeorgiaWells
The 2011 Cadillac STS. Cadillac garnered a high score in a survey of customer satisfaction.
SAN FRANCISCO (MarketWatch) — U.S. drivers' satisfaction with cars made by Japanese and Korean companies rose over the past year, while they were less pleased with domestic car markers — though Cadillac tied Toyota and Toyota Lexus for top marks, according to an annual study.
Some Detroit-based U.S. car companies saw their customers' satisfaction decrease in the same period, according to the University of Michigan's American Customer Satisfaction Index, or ACSI.
Overall, the auto industry's customer-satisfaction score improved by 1.2%, bringing it to 83 on a 100-point sale. But that increase is unlikely to signal an increase in car purchases.
"While demand generally shifts to companies that do a good job of satisfying their customers, aggregate demand in times of economic distress is hampered by other factors, such as doubt about the future, job uncertainty, and lack of discretionary income," said Claes Fornell, a professor of business at the University of Michigan, and founder of the ACSI, in the report.
Of the top three car companies in the ACSI, all with a score of 87, two are Japanese brands: Toyota Motor Corp. TM, -0.04% and Toyota's Lexus. The third is General Motors Co.'s Cadillac GM, +0.14% .
The eight-passenger Lexus LX570. Lexus was one of three car lines that notched the top score in a recent customer-satisfaction survey.
Toyota's customer-satisfaction score rose four percentage points and Toyota's Lexus brand added two percentage points from a year earlier. The company surprised analysts Aug. 2 when executives reported that they managed to squeeze out a first-quarter profit despite the March 11 earthquake that hampered production. Read more about Toyota's earnings report.
Six of seven Japanese and Korean car companies improved their customer-satisfaction scores. Mazda Motor Corp. MZDAF, -0.46% was the sole decliner in that group, dropping to an ACSI of 79, from 80 a year ago.
Of the European brands, Daimler AG's DDAIF, -1.14% Mercedes-Benz scored 86, the same score it earned a year ago, while Volkswagen's VLKAF, -0.99% score rose to 84, from 81, and BMW BAMXF, -0.35% (BMW) lost three percentage points to a score of 83 this year.
Last year, Lincoln, owned by Ford Motor Co. F, -0.11%, and General Motors's Buick earned the top scores of 89 and 88, respectively. But the scores of both companies lost three percentage points this year, putting Lincoln at 86 and Buick at 85. Lincoln and Buick joined Chrysler and General Motors's GMC as the four American brands with declining customer satisfaction. All told, six car markers saw their scores fall.
When the score for each of the U.S. car company's various divisions is averaged, Ford's overall score was highest, at 85, with General Motors close behind at 84. Chrysler came in at 78. Ford reported July sales rose 8.9% from a year ago, while General Motors said sales climbed 7.6%. Chrysler's sales added 20% in its 16th straight month of year-over-year improvements. Read more about U.S. car company July sales.
Luxury brands dominated the industry's top scores. Five of the seven brands with a score above 85 were luxury brands: General Motors's Cadillac and Buick, Toyota's Lexus, Daimler's Mercedes-Benz and Ford's Lincoln.
The companies with the worst customer satisfaction were Mazda, Chrysler, Chrysler's Dodge and Chrysler's Jeep, although Dodge and Jeep made small improvements. The Chrysler brand fell for a second straight year — down four percentage points. Mazda was the only Japanese brand not to beat the industry average.
Here's the complete list of scores, with the percentage change from a year earlier:
GM Cadillac, 87, up 1.2% from 2010
Toyota Lexus, 87, up 2.4%
Toyota, 87, up 3.6%
Ford Lincoln, 86, down 3.4%
Daimler Mercedes-Benz, 86, unchanged
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text:
Pakistan court grants stay of execution to paraplegic prisoner
A court in Pakistan today stayed the execution of a paraplegic man who was set to hang tomorrow (Wednesday).
Abdul Basit, 43, was convicted and sentenced to death for murder in 2009. In 2010, he contracted tubercular meningitis in prison, which left him paralysed from the waist down. Despite being unable to stand, and reliant on a wheelchair, a 'Black Warrant' issued last week scheduled his execution for July 29th.
The Lahore High Court today upheld an appeal by lawyers for Basit who argued that his execution would constitute cruel and unusual punishment, violating the fundamental right to human dignity enshrined in Pakistan's Constitution.
The Pakistan Prison Rules of 1978 – the statute regulating executions – state that the rope for hanging must be the correct length, in order to avoid prisoners facing protracted strangulation (if it is too long) or decapitation (if it is too short). The rules state that the rope's length is determined by measuring it from "the lower jaw of the condemned prisoner as he stands on the scaffold." This and other procedures set out in the Prison Rules cannot be followed in Basit's case, leaving open the possibility of a botched hanging.
The Court has now given the government two weeks to respond to the appeal with a hearing scheduled for August 17th.
Pakistan's law makes provisions for mercy to be granted in cases where prisoners are suffering from severe "ill-health". The Government's failure to acknowledge this and commute Basit's sentence appears to form part of a worrying trend involving the blanket dismissal of all mercy petitions considered since executions resumed in 2014. Over 180 prisoners have been hanged in Pakistan's recent rush to the gallows and recent reports suggest that many more who have now had their mercy petitions dismissed without proper consideration may be next in line.
Among them is Shafqat Hussain, convicted and sentenced to death when under 18, who was yesterday issued with a 'black warrant'. His execution has been set for August 4th despite widespread concerns over torture and the government covering up evidence – notably a school record – that could prove his age.
Kate Higham, caseworker at human rights NGO Reprieve, said: "We are enormously relieved by the court's decision today. To allow Pakistan's government to continue with the hanging of a paraplegic man would have been in clear violation of Pakistani law, not to mention an affront to basic common decency. The government must now commute Basit's sentence."
1. For further information, please contact Reprieve's press office: +44 (0) 207 553 8166 / clemency[DOT]wells[AT]reprieve.org.uk
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text:
Art, History & CultureLatest in Lebanon
A New Chapter: The Lebanese National Library
After more than a decade of renovations, the Lebanese National Library reopened its doors to the public earlier this month at its new location in Sanayeh.
The library was first founded in 1922 when the 19th-century Lebanese philanthropist Philippe de Tarrazi donated around 200,000 documents to the Prussian Deaconess school in Downtown. Although the collection grew over the years, efforts to establish the library were abandoned with the onset of the Lebanese Civil War in 1975.
Many of the documents were destroyed during the war and efforts were more focused on their preservation, with many being moved to a storehouse in the Beirut Port area for restoration. The current collection is said to hold 7,000,000 books, 10,000 drawings, 20,000 maps, 25,000 16th-century editions and 8,000 manuscripts.
The new library is now located at the old Ottoman Sanayeh complex, which has been renovated with a $25 million grant from Qatar. Formerly the law school of the Lebanese University, the atrium has been transformed into a large reading room.
Construction of the new library finished in November 2017 but it wasn't until June this year that the various collections were transferred into the library. In the time taken to renovate the new building, the Lebanese Foundation for the National Library was busy conserving and restoring over 3,000 older documents in the collection.
"We started helping in 2000 and […] just like we have a national museum, we should have a national library, like all countries," LFNL president Randa Daouk said.
It's our heritage gathered in one place. The old books and manuscripts I saw demand a small space in the National Library.
"We had a restoration set up in the port over 15 years and we put all our effort and finances into this project," she added. "Now the government is responsible for it and we wish them the very best in this endeavor for the public."
bnl.gov.lb
Text: Maghie Ghali
A New Chapter: The Lebanese National Library was last modified: December 14th, 2018 by LBTAdmin
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"I wish to thank our witnesses for appearing today. The threat of sequestration continues to cast a shadow of uncertainty over the federal budget, and this committee is right to assess the harms that this uncertainty may have on: our national security, the defense industrial base, American job security, and the country's overall economic vitality. The insight offered by private industry is helpful to our understanding of the potential impact of sequestration.
"For nearly a year, Members of Congress have raised the alarm about the impact that automatic cuts would have on the Department of Defense and its supporting industrial base. They are justified in their concern. If implemented, sequestration could jeopardize our ability to meet national security needs. Large, indiscriminate cuts could seriously challenge our ability to operate militarily and threaten to disrupt the industrial base that supports our service men and women in their brave efforts to deter or defeat current and future threats. Congress must act soon to prevent sequestration from occurring. Until it does, our national security will remain at risk.
"We should also be mindful of the reality that the Department of Defense and the defense industrial base would not be the only targets of sequestration. Other essential federal programs vital to our economic wellbeing and our security are also at risk. National security involves much more than defense; it includes prudent and meaningful investments in the country's workforce, infrastructure, education, health care, innovation, business development, public safety and law enforcement means, energy resources and many other facets of enduring national strength. We need a comprehensive solution to this pervasive threat to our future.
"I voted against the Budget Control Act, which put us on the excruciating path to sequestration. We need a long-term plan for curbing our debt and for getting our deficits under control, but I disagreed with the BCA's approach. Deficit-reduction goals cannot be achieved through spending cuts alone, especially if those cuts are exclusive to non-defense programs. Everything needs to be carefully considered in devising a balanced approach. Revenues need to be increased, mandatory programs need to be brought in line with what we can afford over the long term, and domestic spending should be carefully examined to find real and substantial savings over time. Unfortunately, instead of working seriously to find such a balanced solution, the majority in Congress has refused to consider revenue increases and focused instead on measures to kick the threat of sequester slightly down the road. That is just not good enough.
"With national security, the economy, and our future at stake, it is my hope that reason will prevail. The solution to this problem is simple. Let's put realistic revenue options on the table and find the $1.2 trillion in savings mandated by the Budget Control Act. I stand ready to work with each of my colleagues in reaching a timely and sensible compromise.


A garage area for your off-road toys, a separate kitchen and living room, plus a queen bed, and two entries will have you taking this unit out of storage each chance you get. Once the cargo is unloaded, you can take off to have fun all day long knowing you will have a comfortable trailer to come back to. From the dual fold down sofas with an indoor/outdoor table for dining, and an optional queen electric lift bed, the garage will allow the kids to have their own space. Slide the interior patio door back to enter the main living and kitchen area or use the exterior side door. There is more to enjoy, and options to choose from such as a party deck or a larger air conditioner, so grab your helmets, off-road gear, and toiletries!
The Freedom Express Blast by Coachmen RV offers a toy hauler travel trailer with a raised roof line at 7'6" of cargo height for your four wheelers, motorcross bike, or even a Fat Boy. You will enjoy a smooth ride from the EZ-Lube axles, four season of traveling with a heated/enclosed underbelly, plus fresh air inside with the magnetic screen wall at the ramp door. Come see what else makes this the perfect model for you!


text:
Q: Image out of the container

I do not have a definite css yet, I still think how I can do this
How can I make the image full width inside a container, just like in the example using bootstrap, The black border is the bootstrap container
What I try to do is the content in a container and the image outside of that container,

A: Just set these properties--
.imagen img{
  width: 100%;
  max-width: 100%;
}

A: Here you go with jsfiddle https://jsfiddle.net/9drawa3h/1/.

I know you are looking for text-content in the second column, but I specified something else.
As I not sure whether you are going to use img tag ro background-image in the 1st column, so I have provided both.
1st Column with background-image and 2nd column with img tag.


text:
Q: why doesn't dot notation work in function parameter? This code is simple it is used to get the scanned Items and return their value if the scannedItems aren't there  then it will return undefined, but for some reason the dot notation doesn't work for my function but only the bracket notation! i want to know out of curiosity why isn't it working for dot notation with detailed explanation
The code with dot notation

let foods = {
  apples: 25,
  oranges: 32,
  plums: 28,
  bananas: 13,
  grapes: 35,
  strawberries: 27
};

function checkInventory(scannedItem) {

var inventory = foods.scannedItem;

return inventory;

}

console.log(checkInventory("apples"));

The code with Bracket Notation

let foods = {
  apples: 25,
  oranges: 32,
  plums: 28,
  bananas: 13,
  grapes: 35,
  strawberries: 27
};

function checkInventory(scannedItem) {

var inventory = foods[scannedItem];

return inventory;

}

console.log(checkInventory("apples"));

A: What comes after the dot is the name of the key which means you must know the key in advance and that is a limitation.  Likely we have bracket notation that take a string letral for key name or a variable that holds your requested key for the cases when key is provided by user  for example or key is passed in function args in your case.


text:
The Sport Digest
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International Paralympic Committee (IPC) President Andrew Parsons will pay tribute to the late British scientist Professor Stephen Hawking during his speech at the Closing...
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Pyeongchang 2018 admitted that more than 25 percent of people with tickets for the Winter Paralympic Games have failed to turn up for events. The...
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text:
Q: How do I extract the values from the string i.e a syntax tree? I am working on the movies dataset and a column for Genres has a very complicated format. It is of type Object with an embedded dictionary. The format looks like JSON format. SO the genres column is a series of json values .A typical value in the first 2 cells are as follows:
Cell1:    '[{"id": 12, "name": "Adventure"}, {"id": 14, "name": "Fantasy"}, {"id": 28, "name": "Action"}]'
Cell2:    '[{"id": 28, "name": "Action"}, {"id": 12, "name": "Adventure"}, {"id": 80, "name": "Crime"}]'
Now I want to extract these genres i.e values for the key name into a list. For eg: the first cell should contain [Adventre,Fantasy,Action]
second cell should contain [Action,Adventure,Crime]
I tried with some string methods, but it didn't work. I also read about the library ast, but I have no idea about the same. How do I achieve the above result??
Thanks