# Magnetization ($z$-basis) of a 1D Transverse Ising Model

I'm trying to find the magnetization $$\langle\sigma_{z} \rangle$$ of a 1D transverse Ising chain and plot it as a function of the transverse field $$\lambda$$. More specifically, I want to plot this for different values of $$N$$ (number of spins) for instance, $$N=4,8,16,32$$ etc. This is at zero temperature. The Hamiltonian of the system is this $$H = \sum_{i=1}^{n} \sigma_{i}^{x} \sigma_{i+1}^{x} + \lambda \sum_{i=1}^{n} \sigma_{i}^{z} .$$ Introducing periodic boundary conditions I have the Hamiltonian as $$H = \sum_{i=1}^{n} \sigma_{i}^{x} \sigma_{i+1}^{x} + \lambda \sum_{i=1}^{n} \sigma_{i}^{z} + \sigma_{1}^{y} \sigma_{2}^{z}... \sigma_{n-1}^{z} \sigma_{n}^{y} .$$ In particular I want to express the value of the magnetization solely as a function of $$\lambda$$ (if possible) and see how it changes with $$N$$. I'm expecting to see a kind of step function which becomes smoother as $$N$$ increases. Initially the spins starting in a paramagnetic phase and then moving to a ferromagnetic one.

• Sorry why are there $\sigma^y$'s? Jan 27 '21 at 18:48
• There will be no step function. Also, what is your question? Jan 27 '21 at 20:07
• @jacob1729 Presumably the boundary term couples to the global parity in the z basis. However, this is what I know from mapping to free fermions - then you get this effect in the fermion model. I would suspect that this is the kind of Ising model you get when transforming back a periodic free fermion chain. Jan 27 '21 at 20:14
• @NorbertSchuch What is the equation that relates $\lambda$ and $\langle \sigma_{z} \rangle$? Jan 27 '21 at 20:51
• Giving an equation would be a valid answer? (And it is $\langle\sigma_z\rangle$, not $\langle\sigma_x\rangle$?) In any case, this can be solved by mapping the system to free fermions via the Jordan-Wigner transformation. Jan 27 '21 at 21:45

In the thermodynamic limit, the magnetisation of the one-dimensional transverse field Ising model is $$m^2(\lambda) + c$$, where $$c\propto|r-r’|$$. However, this is only when $$N\to\infty$$. For finite $$N$$, there is no magnetisation ever.