I've stumbled across an interesting extra question in an old exam. In my own words:

Consider the Magnetization of the 1D Ising model $$H=-J\sum_iS_{z,i}S_{z,i+1}-B\sum_iS_{x,i}$$ at $T=0$. We know that for $B=0$ the magnetization $M_z$ is finite. How does the magnetization behave for large $B$?

It's obvious that all the spins "flip" for large $B$ in $x$-direction and $M_z$ becomes zero. I estimate that this will happen at around $B\approx J$. Is there a way to solve this model more quantitatively? I'm struggling as $[S_z,S_x]\neq0$.


1 Answer 1


The model you described is known as the transverse field quantum Ising model. Yes, you are correct that for one phase ($B\gg J$), we have a paramagnetic phase, and for ($B\ll J$), we have a ferromagnetic phase. The quantum phase transition boundary is $B=J$, which can be shown as from Karmer's Duality relation.

What happens when $B\sim J$ or when we reach near criticality. Pretty much everything that happens for our ole classical ising model, like correlation length diverges. The traditional 2D ising model and Transverse field quantum Ising model belong to the same universality class.

What else happens? "Bulk gap" closes here, which has an interesting aspect related to the topological superconductor.

But let's come back to point "universality class." When I say two systems belong to the same universality class, it means that all critical exponents for two systems are identical. In classical Ising model, the magnetisation or the order parameter below criticality goes as $$M_z\sim (T-T_c)^{-\beta}$$ The same thing happens here.

In classical Ising model, the phase transtion is driven by thermal fluctuation. But, here, the transitions are driven by quantum fluctuations. So treatment is a different form of standard statistical mechanics treatment.

An important feature of this setup is that, in a quantum sense, the spin projection along the x axis and the spin projection along the z axis are not commuting observable quantities. That is, they cannot both be observed simultaneously. This means classical statistical mechanics cannot describe this model, and a quantum treatment is needed.

You can find treatment in any standard Quantum Phase Transitions textbooks

  • 1
    $\begingroup$ This is great, thanks Kartik! $\endgroup$
    – user224659
    Jun 24, 2020 at 13:54

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