Quantum interferometry in multimode systems

We consider the situation when the signal propagating through each arm of an interferometer has a complicated multimode structure. We demonstrate that the shot-noise level for such a setup is the same as for the common two-mode case for as long as the interferometric transformation treats each arm as a whole and does not penetrate its inner structure. We find the relation between the particle entanglement and the possibility to surpass the shot-noise limit of the phase estimation. Our results are general—they apply to pure and mixed states of identical and distinguishable particles (or combinations of both) for a fixed and fluctuating number of particles. An important result is that the method for detecting the entanglement often used in a two-mode system—based on the measurement of the visibility of fringes and the population imbalance fluctuations—can give misleading results when applied to the multimode case. Our main result is that the additional modes do not boost the sensitivity for as long as they are not interacting. Therefore, the improvement above the shot-noise level established for the two-mode systems can safely be treated as an entanglement criterion also in the complex multimode case.

Figure 1

The two-arm multimode interferometer, illustrated here with the eigenmodes of the harmonic oscillator. Each arm (here a harmonic well) has the same ladder of states, just shifted by $d$. The interferometric transformation can either imprint the relative phase between the arms or move the particles from one arm to the other.

Figure 2

The single-particle density of atoms trapped in a double-well potential. (a) The standard two-mode setup. (b) The three-mode setup: the configuration after a coherent splitting of the $a$ mode into $a_1$ and $a_2$.

Figure 3

The single-particle far-field interference pattern, see Eq. (56), formed by a coherent spin-state (54) in the three-mode configuration shown in Fig. 2. The parameters are $z=0.91,\zeta =0.5,k=10$, and $\delta k=0.5$.