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The Hamiltonian for transverse Ising model is $$\hat{H}=-\sum_{j=0}^{N-1}(\lambda \hat{\sigma}_j^x\hat{\sigma}_{j+1}^x+\hat{\sigma}_j^z)$$ where $\hat{\sigma}$s are Pauli matrices. This model shows quantum phase transition at $\lambda=1$. The order parameter is transverse magnetization ($\langle \hat{\sigma}^x\rangle$). The value of $\langle \hat{\sigma}^x\rangle$ for $\lambda \geq 1$ is $$(1-\frac{1}{\lambda^2})^{\frac{1}{8}}$$ This expression comes from a lengthy calculation as done by P. Pfeuty [Annals of Physics: 57, 79-90 (1970)]. Is there any simpler way to understand the $\lambda$ dependence and the overall power $\frac{1}{8}$ based on some physical argument?

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$\lambda$ describes the strength of the spin-spin interaction ($J$) relative to (or against) the thermal fluctuation ($kT$); the former prefers an ordered phase whereas the latter a disorder phase. So, $\lambda$ is a function of $J$ and $kT$.

The power $\frac{1}{8}$ is called the critical exponent of the order parameter. It's a kind of signature to identify which universality class the phenomenon in question belongs to. Since the power is $\frac{1}{8}$, I believe the model describes a phase transition in $d=2$, i.e. 2D Ising model. In short we could say, a critical exponent is a manifestation of several factors (dimension, range of interaction, and spin dimension) that determines the universality class of a phase transition phenomenon.

More details about critical exponents: https://en.wikipedia.org/wiki/Critical_exponent

And universality class: https://en.wikipedia.org/wiki/Universality_class

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    $\begingroup$ I don't think that this answers the question by the OP. As far as I understand, he would like to have a heuristic explanation for the exponent $1/8$. The fact that one finds the same exponent as for the 2d Ising model is intuitively clear (given the well-known representation of the transverse chain as a $1+1$-dimensional semi-discrete Ising model). So the question can be restated, I believe, as "what is an intuitive reason for the value of the magnetization exponent in the 2d Ising model". I don't know of any intuitive argument and would also be interested in seeing one. $\endgroup$ Dec 1 '17 at 10:49

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