Taking advantage of spin squeezing

Question

I have $N$ 2-level systems (I call them atoms) on a Bloch sphere with the south pole when all of them are in the ground state (they are not interacting with each other) and the north pole when they are in the excited state (I ignore any decays) and I call these directions -z and z. A Dicke state is a state in which all atoms point on the same direction on the (N/2 dimensional) Bloch sphere (see for example the discussion in arxiv.org/abs/2106.13234). If I prepare them in an equal super of up and down, they will be on the equator of the Bloch sphere (say along the x axis). If I start measuring, each one of them has a 50/50 probability of ending up in z or -z and if I do this for all atoms and repeat this many times, I will get a distribution with a variance of N/2 i.e. the standard quantum limit.

However, I can create (doesn't matter how) a squeezed state, such that the variance along the z axis is smaller compared to the one on the y-axis, but the Dicke state is still pointing along the x axis. For this question I ignore any possible reduction of the Bloch vector during the squeezing or any information loss (i.e. the amount of squeezing along the z-axis is equal to the amount of anti-squeezing along the y-axis). Now, in principle, if I measure the population of the atoms, I would still get zero on average as before (as I am on the equator of the Bloch sphere), but the variance would be reduced relative to the initial N/2 value. The question I have is, how do I measure the population in practice, such that I can take advantage of this reduced variance? Say I have my squeezed state and somehow I can collapse each atom individually (e.g. I can resonantly ionize each atom, with the ionization happening only if the atom collapses in the higher energy level state). In the non-squeezed case, each atom will ionize or not independent of the others. How does this work in the squeezed case? Is the collapse of one atom somehow changing the wavefunction of the collective state of all the others, such that if I get, let's say, a spin-up measurement, the probability of getting a spin-up for the next measurement next time is lower (such that overall my final result is very close to equal amounts of spin up and spin down)? I am having some hard time understanding what is happening with the squeezed system when I start collapsing individual atoms.

Answer 1

You're question is a bit unclear and hopefully I'm not misunderstanding anything.

If all the atom points in the same direction $\hat{\mathbf{n}}$ then your state $$ \vert{\psi}\rangle=\vert\hat{\mathbf{n}}\rangle_1 \otimes \vert\hat{\mathbf{n}}\rangle_2\ldots \vert\hat{\mathbf{n}}\rangle_N $$ is actually a coherent state. There is a unique rotation $R(\hat{\mathbf{n}})$ that will take the spin-up state $\vert +\rangle$ (or the spin-down state $\vert -\rangle$ if you prefer) to $\vert\hat{\mathbf{n}}\rangle$ and then you have $$ \vert{\psi}\rangle=[R(\hat{\mathbf{n}})\vert+ \rangle_1] \otimes [R(\hat{\mathbf{n}})\vert+ \rangle_2]\ldots [R(\hat{\mathbf{n}})\vert+ \rangle_N]:= R(\hat{\mathbf{n}})\vert SS\rangle $$ where $S=N/2$ and there is no definition of squeezing where the coherent state is squeezed. In fact, the coherent state is used to define the standard quantum limit.

In addition, there are several definitions of squeezing but the only one that makes sense on the sphere is the so-called covariant squeezing. Given any state $\vert \phi\rangle$, you first compute the triple $(\langle S_x\rangle, \langle S_y\rangle,\langle S_z\rangle)$ and you use these expectation values to define the direction $\hat{\mathbf{m}}$ of your quantization axis. Thus, the relevant observables are $\hat S_k^\prime=R(\hat{\mathbf{m}})\hat S_k R^{-1}(\hat{\mathbf{m}})$ and you then measure squeezing by comparing the variances of $\hat S_x^\prime$ and $\hat S_y^\prime$ in the plane perpendicular to the direction $\hat{\mathbf{m}}$ on the sphere with the corresponding variance of $\hat S_y^\prime$ and $\hat S_x^\prime$ for the coherent state. This guarantees you get genuine squeezing rather than just a projection effect.

If you're into Wigner function of the sphere, the Wigner function for the relevant coherent state is a Gaussian-like blob pointing the $\hat{\mathbf{m}}$ direction and squeezing is manifested by a deformation of the blob, in particular by the loss of cylindrical symmetry about the $\hat{\mathbf{m}}$ axis.

Because coherent state are just rotated $\vert SS\rangle$ states (or $\vert S,-S\rangle$, you cannot get squeezing by simple rotation from the North or South pole. In particular, any Hamiltonian linear in the angular momenta $$ \hat H=\sum_k \alpha_k S_k $$ generates are rotation by exponentiation so such Hamiltonians cannot produce squeezing. This is why Kitagawa and Ueda generate their squeezing using Hamiltonians quadratic in the generators.

Now, using the definition of Klimov and Chumakov

Klimov, Andrei B., and Sergei M. Chumakov. A group-theoretical approach to quantum optics: models of atom-field interactions. John Wiley & Sons, 2009.

the Dicke state is NOT a state where all spins point in one direction. Rather it is a symmetrized sum $$ \vert k,A\rangle = \sqrt{\frac{k!(A-k)!}{A!}}\sum_p \vert j_1,1\rangle\ldots j_k,1\rangle \vert j_{k+1},0\rangle\ldots \vert j_A,0\rangle $$ where the sum is over all the possible permutations that interchange excited and non-excited atoms (basically, 0's and 1's).