Magnetization in Quantum Transverse Ising Model: Mean Field Theory vs Reality One of the canonical examples of mean field theory concerns the ground state ($T=0$) of the transverse field Ising model, with Hamiltonian $$H = -J\sum_{<ij>} \sigma^z_i \sigma^z_j-h \sum_i\sigma^x_i.$$ The model has a phase transition as $h$ is increased from being ferromagnetic to being paramagnetic. When I refer to magnetization below, I mean the average magnetization per site, which varies from $1$ to $-1$. I'll also take $h$, $J$ non-negative.
In mean field theory, one finds that the longitudinal magnetization vanishes at the phase-transition, and the transverse magnetization is maxed out at $1$ everywhere in the paramagnetic phase. In the thermodynamic limit, does this really occur in dimensions where mean field theory does not apply? 
I expect the former to be true, but I'm not sure about the latter! I feel that maybe competitions with $J$ ruin the transverse magnetization equal to one in the paramagnetic phase - if I started with $J$ equal to zero and turned on $J<<h$, I expect that perturbation theory would find corrections to the ground state and thus the ground state transverse magnetization. I'm unsure whether those corrections would matter or not for the average magnetization in the thermodynamic limit.

Of course, mean field theory is an approximation, not the truth. For example, we know the phase transition for the 1D chain really occurs at $h=J$. Mean field theory predicts a phase transition to occur at $h=2dJ$ with $d$ the dimension which is off by a factor of 2 for $d=1$. My question is motivated by my curiosity about whether other predictions are correct or incorrect.
 A: I found the answer for the case of the $1$D transverse field Ising model - my expectation that the longitudinal magnetization $m_z = \langle \sigma_z \rangle$ vanishes at the phase transition but that $m_x$ doesn't saturate at the phase transition was correct. The answer can be found in Pierre Pfeuty's 1970 "The one-dimensional Ising model with a transverse field," and the citations therein, but I'll translate his notation here for a self-contained answer.
In the thermodynamic limit, we have $$m_z = \pm (1-\frac{h}{J})^{\frac{1}{8}}\text{ for }h<j$$ and just plain $0$ for $h>J$. The longitudinal magnetization thus indeed vanishes in the paramagnetic phase, just as mean field theory predicts. 
However, against the predictions of mean field theory, the transverse magnetization $m_x$ is not saturated in the paramagnetic phase. I've modified Pfeuty's graph of the transverse magnetization to jive with my notation in this problem.

I find the actual transverse magnetization interesting. It's never that far from the mean-field theory value, and the location of the non-analyticity isn't as obvious as in mean field theory. It's clear that the transverse magnetization is not saturated at $1$ in the paramagnetic phase, but instead is less than $2/3$ at the phase transition of $h=J$ and slowly asymptoting to $1$ in the limit $h/J \rightarrow \infty$.

For those interested, the transverse magnetization takes the formula $m_x = \frac{1}{\pi} \int_0^\pi dk \frac{1+\frac{J}{h}\cos(k)}{\sqrt{1+2\frac{J}{h}\cos(k)+\frac{J^2}{h^2}}}$, with a magnetization of $\frac{2}{\pi}$ at the phase transition.

