Sensing microwave photons with a Bose–Einstein condensate

Abstract

We consider the interaction of a magnetically trapped Bose–Einstein condensate of Rubidium atoms with the stationary microwave radiation field sustained by a coplanar waveguide resonator. This coupling allows for the measurement of the magnetic field of the resonator by means of counting the atoms that fall out of the condensate due to hyperfine transitions to non-trapped states. We determine the quantum efficiency of this detection scheme and show that weak microwave fields at the single-photon level can be sensed.

Appendices

Appendix 1: Transitions between the $F=1$ and $F=2$ hyperfine manifolds

Transitions between the $F=2$ and $F=1$ manifolds of the $5 ^{2}\mathrm{S}_{1/2}$ ground state of ⁸⁷Rb are magnetic dipole transitions. As the quantum numbers L and $m_{L}$ are both zero and the nuclear Landé factor $g_{\mathrm{I}}$ is much smaller than the spin Landé factor $g_{\mathrm{S}}$, it is the spin magnetic moment $\boldsymbol{\mu}_{S}=-g_{\mathrm{S}}\mu_{\mathrm{B}}\hat{\mathbf {S}}$ which determines the coupling of the atomic magnetic moment with a resonant, oscillating magnetic field that is orthogonal to the offset field. The x and y components of $\boldsymbol{\mu}_{S}$ can be expressed by the spin raising and lowering operators $\hat{S}_{+}=\hat {S}_{x}+i\hat{S}_{y}$ and $\hat{S}_{-}=\hat{S}_{x}-i\hat{S}_{y}$, therefore, it is the nonzero matrix elements of $\hat{S}_{+}$ and $\hat{S}_{-}$ which determine the possible transitions. (This can be seen by expressing the basis states $\lbrace \vert F,m_{F} \rangle, F=1,2, m_{F}=-F,\ldots,F \rbrace$ common to the operators $\hat{\mathbf{S}}^{2},\hat{\mathbf{I}}^{2},\hat {\mathbf{F}}^{2},\hat{F}_{z}$ with the basis states $\lbrace \vert m_{S},m_{I} \rangle, m_{S}=-1/2,1/2, m_{I}=-3/2,\ldots,3/2 \rbrace$ common to the operators $\hat {\mathbf{S}}^{2},\hat{\mathbf{I}}^{2},\hat{S}_{z},\hat{I}_{z}$.) We focus on the $\vert 2,1 \rangle$ → $\vert 1,0 \rangle$ transition, because the transition matrix element $\langle1,0 \vert \hat{S}_{-} \vert 2,1 \rangle\approx-\sqrt{3}/2\sqrt{2}$ is the largest possible involving a field-insensitive final state. (We note that if one chooses the state $\vert 1,-1 \rangle$ as the trapped state and the $\vert 2,0 \rangle$ state as the untrapped one, then the corresponding transition matrix element of the $\hat {S}_{+}$ operator is a factor of $1/\sqrt{3}$ smaller.)

Appendix 2: Outline of the solution of the outcoupling problem

Here we outline the solution to the partial differential equation (1), following the method presented in Ref. [18]. We use the approximation that the mean-field potential $N_{0}g_{s} \varPhi^{2}_{\mathrm{BEC}}(\mathbf {r})$ is negligible compared to the gravitational potential, and consider the resulting inhomogeneous differential equation as a scattering problem, i.e., use the Green’s function of the corresponding free problem to determine the outcoupled wave function. The stationary solution of Eq. (1), including the undepleted BEC assumption, is then given by

$$\begin{aligned} \hat{\varPsi}_{ \vert 1,0 \rangle}(\mathbf {r},t)= - \hbar\eta e^{i\Delta t} \int d ^{3}\mathbf {r}^{\prime\prime} G^{\mathrm{3D}}_{+} \bigl(\mathbf {r},\mathbf {r}^{\prime \prime}; -\hbar\Delta\bigr)\varPhi_{\mathrm{BEC}}\bigl( \mathbf {r}^{\prime\prime}\bigr) =i \hbar\eta e^{i\Delta t} F(\mathbf {r}), \end{aligned}$$

(7)

where $G^{\mathrm{3D}}_{+}(\mathbf {r},\mathbf {r}^{\prime\prime}; -\hbar\Delta )$ is the advanced (outgoing) energy-dependent Green’s function (corresponding to the energy $E=-\hbar\Delta$), which can be written as

$$\begin{aligned} G^{\mathrm{3D}}_{+}\bigl(\mathbf {r},\mathbf {r}^{\prime\prime}; -\hbar\Delta\bigr)={}& \frac{1}{(2 \pi)^{2}} \int_{-\infty}^{\infty}\,dk_{x} \int_{-\infty }^{\infty}\,dk_{z} e^{-ik_{x}(x-x^{\prime\prime})} e^{-ik_{z}(z-z ^{\prime\prime})} \\ &{}\times G^{\mathrm{1D}}_{+} \biggl(y,y^{\prime\prime};-\hbar\Delta - \frac{\hbar^{2}}{2M} \bigl(k_{x}^{2}+k_{z}^{2} \bigr) \biggr), \end{aligned}$$

(8)

$G^{\mathrm{1D}}_{+}$ being the Green’s function of the 1D free-fall problem

$$\begin{aligned} G^{\mathrm{1D}}_{+}\bigl(y,y^{\prime\prime};E\bigr)={}&{-} \frac{\pi}{Mgl_{0}^{2}} \mathrm{Ai} \biggl(\frac{y+y^{\prime\prime}+ \vert y-y^{\prime \prime} \vert }{2l_{0}}-\frac{E}{Mgl_{0}} \biggr) \\ &{}\times\mathrm{Ci} \biggl(\frac{y+y^{\prime\prime}- \vert y-y^{ \prime\prime} \vert }{2l_{0}}-\frac{E}{Mgl_{0}} \biggr), \end{aligned}$$

(9)

and Ci being the complex Airy function $\mathrm{Ci}(x)=\mathrm{Bi}(x)+i \mathrm{Ai}(x)$. In the above expressions $\mathbf {r}^{\prime\prime}$ refers to the coordinates of the source (i.e., the atomic cloud), $k_{x}$ and $k_{z}$ are the wave numbers in the x and z direction, and $l_{0}= (\hbar^{2}/2M^{2}g )^{1/3}$ is the natural length of the Airy function.

The outcoupled atom density at a position r in this monochromatic outcoupling case is

$$\begin{aligned} N(\mathbf {r})= \bigl\langle \hat{\varPsi}^{\dagger}_{\vert1,0 \rangle}(\mathbf {r},t) \hat{\varPsi} _{\vert1,0 \rangle}(\mathbf {r},t) \bigr\rangle =(\hbar\eta )^{2} \bigl\vert F(\mathbf {r}) \bigr\vert ^{2}= \biggl(\frac{\hbar\eta}{Mgl_{0}} \biggr)^{2} D(\mathbf {r}), \end{aligned}$$

(10)

where

$$\begin{aligned} D(\mathbf {r})=(Mgl_{0})^{2} \bigl\vert F(\mathbf {r}) \bigr\vert ^{2} \end{aligned}$$

(11)

has the dimension of 1/volume and is dependent on the geometry of the BEC cloud. We note that when the outcoupling magnetic field has a spectral distribution, then $D(\mathbf {r})$ is a function of the frequency, and may be considered the so-called spectral resolution function of the BEC as a measuring device [18].

For coordinates below a spherical BEC with radius a, in the Tomas–Fermi approximation

$$\begin{aligned} F(\mathbf {r}) ={}& {-}\frac{\pi}{Mgl_{0}} \sqrt{ \frac{\mu}{N g_{s}}} \int_{0}^{\infty} \,d\bar{k}_{\perp} \bar{k}_{\perp} J_{0}(\bar{k}_{\perp} \bar{r}_{ \perp}) \mathrm{Ci} \biggl(\bar{y}-\frac{E_{y}(\bar{k}_{\perp})}{Mgl _{0}} \biggr) \\ &{}\times \int_{0}^{1} \,d\bar{r}^{\prime\prime}_{\perp} \bar{r}^{\prime\prime}_{\perp} J _{0}\bigl(\bar{k}_{\perp} \bar{r}^{\prime\prime}_{\perp}\bigr) \int _{-\bar{a}\sqrt{1- (\bar{r}^{\prime\prime}_{\perp} ) ^{2}}}^{\bar{a}\sqrt{1- (\bar{r}^{\prime\prime}_{\perp } ) ^{2}}} \,d\bar{y}^{\prime\prime} \sqrt {1- \bigl(\bar{r}_{\perp}^{\prime \prime} \bigr)^{2}- \frac{ (\bar{y}^{\prime\prime} ) ^{2}}{\bar{a}^{2}}} \\ &{}\times\mathrm{Ai} \biggl(\bar{y}^{\prime\prime}-\frac{E _{y}(\bar{k}_{\perp})}{Mgl_{0}} \biggr) . \end{aligned}$$

(12)

Here $\bar{k}_{\perp}$ is the length of the dimensionless wave vector perpendicular to the direction of gravity, with $\bar{k}_{\perp}^{2}=a ^{2}(k_{x}^{2}+k_{z}^{2})=2Ma^{2}(E_{x}+E_{z})/\hbar^{2}$, $E_{y}( \bar{k}_{\perp})=-\hbar\Delta-\hbar^{2}\bar{k}_{\perp}^{2}/2Ma ^{2}$, $\bar{r}^{\prime\prime}_{\perp}=\sqrt{ (x^{\prime \prime} )^{2}+ (z^{\prime\prime} )^{2}}/a$ is the length of the dimensionless position vector perpendicular to gravity, while $\bar{y}^{\prime\prime}=y/a$ is the dimensionless coordinate in the direction of gravity inside the BEC. $\bar{r}_{\perp}$ refers to the dimensionless length of the position vector r perpendicular to gravity [18].

Appendix 3: Analytical formulas for the magnetic field of a CPW

If the transverse size of the conductors and their distance is small relative to the mode wavelength of the cavity, then the quasi-static approximation may be applied [26]. Let us introduce the coordinates $x\rq=x$, $y\rq=y-d$, $z\rq=z+L/2$ (see Fig. 1) to describe the magnetic field components of the CPW, where d is the distance between the center of the BEC and that of the CPW. For reasons of symmetry, it suffices to restrict to one half of the structure (i.e., $0\leq x\rq\leq b$, $0\leq z\rq\leq L$), and consider the CPW as coupled slots [27]. We assume magnetic walls at $x\rq=0$ and $x\rq=b=S/2+W+w$ and open-circuit boundary conditions at $z\rq=0$ and $z\rq=L=\lambda_{g}/2$, with $\lambda_{g}$ being the mode wavelength [27, 28]. As a result of the longitudinal boundary condition, the maxima of the electric and magnetic fields are shifted in space. The magnetic field is maximal at $z\rq=L/2$, where the electric field is zero, and zero at the two ends, where the electric field is maximal. Here we only present the components of the magnetic field for the case of odd (transverse) excitations, according to Ref. [22]. On the air side of the structure ($y\rq\leq0$) they can be approximated as

$$\begin{aligned} B_{x}(\mathbf {r}\rq) & = p \Biggl[\sum_{n>0}^{\infty} \frac{s_{n}}{F _{n}}\cos \biggl(\frac{n\pi x\rq}{b} \biggr) e^{-\gamma_{n} \vert y \rq \vert } \Biggr] \sin \biggl(\frac{\pi z\rq}{L} \biggr) , \\ B_{y}(\mathbf {r}\rq) & = p \Biggl[\sum_{n>0}^{\infty}s_{n} \sin \biggl(\frac{n\pi x\rq}{b} \biggr) e^{-\gamma_{n} \vert y \rq \vert } \Biggr] \sin \biggl( \frac{\pi z\rq}{L} \biggr) , \\ B_{z}(\mathbf {r}\rq) & = p \frac{2b}{\lambda_{g}} \Biggl[\sum _{n>0} ^{\infty}q\frac{s_{n}}{n F_{n}}\sin \biggl( \frac{n\pi x\rq }{b} \biggr) e^{-\gamma_{n} \vert y\rq \vert } \Biggr] \cos \biggl(\frac{\pi z\rq}{L} \biggr) , \end{aligned}$$

(13)

where

$$\begin{aligned} &p = -i \mu_{0}\mu_{r}\frac{4 V_{0}}{\eta b} \frac{\lambda}{\lambda_{g}}, \end{aligned}$$

(14)

$$\begin{aligned} &s_{n} = \frac{\sin (\frac{n\pi\delta}{2} )}{\frac{n\pi\delta }{2}}\sin \biggl(\frac{n\pi\bar{\delta}}{2} \biggr), \end{aligned}$$

(15)

$$\begin{aligned} &q = 1- \biggl(\frac{\lambda}{\lambda_{g}} \biggr)^{2}, \end{aligned}$$

(16)

$$\begin{aligned} &F_{n} = \frac{b\gamma_{n}}{n\pi}=\sqrt{1+ \biggl(\frac{2bv}{n\lambda } \biggr) ^{2}}, \end{aligned}$$

(17)

with $\delta=W/b$, $\bar{\delta}=(S+W)/b$, $v=\sqrt{(\lambda/ \lambda_{g})^{2}-1}$, $\mu_{0}$ is the permeability of free space, $\mu_{r}\approx1$ is the relative permittivity of the conductor, $\eta=\sqrt{\mu_{0}/\varepsilon_{0}}$ is the impedance of free space, and $V_{0}$ is the voltage directly across the slot between the central and ground electrodes. Let us note that the complex factor i in the expression of p is due to the fact that there is a $\pi/2$ phase difference in the time dependence of the magnetic field and the electric field, i.e., when the magnetic field is maximal in time, then the electric field is zero and vice versa.

For the case when $b\ll\lambda, h$, where h is the thickness of the substrate, it can be shown that the magnetic field on the substrate side of the structure is approximately given by

$$\begin{aligned} \begin{aligned} &B^{0\leq y\rq\leq h}_{x}(x\rq,y\rq,z\rq)\approx -B_{x}(x \rq,-y \rq,z\rq), \\ &B^{0\leq y\rq\leq h}_{y}(x\rq,y\rq,z\rq)\approx B_{y}(x \rq,-y\rq,z\rq), \end{aligned} \end{aligned}$$

(18)

while the z components of the magnetic field can be neglected on both sides of the CPW, i.e., $\vert B_{z}(\mathbf {r}\rq) \vert , |B^{0\leq y \rq\leq h}_{z}(\mathbf {r}\rq)| \ll \vert B_{x}(\mathbf {r}\rq) \vert , \vert B _{y}(\mathbf {r}\rq) \vert$.

Appendix 4: Effective dielectric constant and the capacitance of a CPW

For a conventional CPW on a substrate of relative permittivity $\varepsilon_{r}$, the effective dielectric constant $\varepsilon_{\text{eff}}$ and the capacitance per unit length $c_{\mathrm{CPW}}$ can be analitically approximated using conformal mapping techniques as [22]

$$\begin{aligned} &\varepsilon_{\mathrm{eff}} =1+\frac{\varepsilon_{r}-1}{2}\frac{\kappa _{1}}{\kappa_{0}}, \end{aligned}$$

(19)

$$\begin{aligned} &c_{\mathrm{CPW}} =4 \varepsilon_{0} \varepsilon_{\mathrm{eff}} \kappa_{0}, \end{aligned}$$

(20)

where $\kappa_{i}=K (k_{i} )/K (k\rq_{i} )$, K being the complete elliptic integral, $k_{0}=S/(S+2W)$, $k_{1}=\sinh (\frac{\pi S}{4 h} )/\sinh (\frac{ \pi (S+2W )}{4 h} )$, $k\rq_{i}= \sqrt{1-k_{i}^{2}}$ for ($i=0,1$).