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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.

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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 $$m_z = \pm \left(1-\left(\frac{h}{J}\right)^2 \right)^{\frac{1}{8}}\text{ for }h<j$$ 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. Transverse magnetization

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.


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  • $\begingroup$ In any dimension, you should be able to compute $\langle \sigma^x \rangle$ perturbatively for small $J/h$, and one should find a relation like $\langle \sigma^x \rangle = 1 - k(J/h)^2 + \cdots$ where the constant $k$ depends on your lattice/dimension ($k = 1/4$ in the 1d case). $\endgroup$ Commented Apr 18, 2020 at 21:16
  • $\begingroup$ @SethWhitsitt Thanks for the suggestion. If I remember right, when I've tried to compute perturbatively in 1D, I usually get a length-of-the-chain $L$ dependence in $\langle \sigma^x \rangle$ that seems a little unphysical for very large $L$. I think I then tried simple matrix product states that matched to the same order in $J/h$ as I went to in perturbation theory, and then the large $L$ limit comes out nicely. I've only worked with MPS in 1D, so I'd need to get familiar with higher dimensions first, but I think I'll try that! Is that what you had in mind? $\endgroup$
    – user196574
    Commented Apr 18, 2020 at 21:25
  • $\begingroup$ I find that ordinary perturbation theory gives me a reasonable answer which correctly predicts $\langle \sigma^x \rangle = 1 - (1/4)(J/h)^2 + \cdots$ in one-dimension. Two possible issues which might trip you up: after obtaining the perturbed ground state, did you remember to normalize your state? Also, in your final expression, be sure to expand $(J/h)$ to leading order before taking any thermodynamic limit, since your expression should hold for arbitrary system sizes but only to leading order in the perturbation. $\endgroup$ Commented Apr 18, 2020 at 22:28
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    $\begingroup$ Mean-field theory correctly predicts the critical exponents above the upper critical dimension, but it doesn't correctly predict "non-universal" quantities, such as the transverse magnetization. $\endgroup$ Commented Apr 19, 2020 at 21:51
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    $\begingroup$ @Yumina弓那Nirvalen Regarding the symmetry (or duality) under exchanging h and J, recall that the Kramers-Wannier transformation maps the $\sigma^x$ operator to the product $\sigma^z_j \sigma^z_{j+1}$, so the magnetizations along the two directions don’t map directly to each other. Rather, you’d expect $m^z = \sqrt{m^x}|_{h \leftrightarrow J}$, which is exactly what you’re seeing. $\endgroup$ Commented Jul 3 at 5:12

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