Reduced state of transverse ising model

Question

Consider the model $$ H = - \sum_{i=1}^{N} (\lambda \sigma_i^x \sigma_{i+1}^x + \sigma_z) $$ with periodic boundary conditions $\sigma_1^x=\sigma_{N+1}$.

Equation 33 in this paper says the reduced ground state for a single spin (in the $N\to\infty$ and zero temperature limits) is $$ \rho_1 = \frac{I+\langle \sigma^x \rangle \sigma^x + \langle \sigma^z \rangle \sigma^z}{2}. $$ I'm trying to do verify this numerically using exact diagonalization for finite $N$, but I'm getting something completely different. In particular, my off-diagonal entries are nearly zero. I am really lost as to what's going on. Does anyone have experinece with this?

Answer 1

This problem is related to the absence of symmetry breaking in the finite-size quantum Ising model. In an infinite system, the ground state is twice degenerate if $\lambda > 1$. Therefore, states with broken $\hat{\sigma}_i^x\rightarrow -\hat{\sigma}_i^x$ symmetry are possible for which $\langle\hat{\sigma}_i^x\rangle\neq 0$. For any system of finite size, the ground state is non-degenerate, so it must be symmetric with respect to the transformation $\hat{\sigma}_i^x\rightarrow -\hat{\sigma} _i^x$, and it is impossible to have $\langle\hat{\sigma}_i^x\rangle\neq 0$. An indication of symmetry breaking at $N =\infty$ in finite-size systems is the exponential smallness of the energy gap, $E_1 - E_0\sim O(e^{-Nc(\lambda)})$ if $\lambda > 1$.

So, if $\lambda > 1$, I would advise you to perform calculations for the modified Hamiltonian $$ \hat{H}_1 = \hat{H} - h_1 \sum_{i = 1}^N\hat{\sigma}^x_i, $$ with the parameter $h_1$ satisfying the conditions $E_1 - E_0 \ll h_1 \ll E_2 - E_0$.

A short but fairly complete discussion of symmetry breaking in the quantum Ising model can be found in he first chapter of the Franchini's book