Two-level Gibbs and noncommuting susceptibility

Write $r(h)=\sqrt{(ϵ\mathrm{/}2{)}^2+h^2}$. Direct matrix multiplication gives $H(h{)}^2=r(h{)}^{2}I$ and $\mathrm{Tr}H(h)=0$. Splitting the exponential power series into even and odd powers therefore gives $e^{-\beta H(h)}=\cosh (\beta r(h))I-(\sinh (\beta r(h))\mathrm{/}r(h))H(h)$, hence $Z(h)=2\cosh (\beta r(h))$. Also $\mathrm{Tr}(H(h){\sigma }_x)=-2h$, so $⟨{\sigma }_x{⟩}_h=h\tanh (\beta r(h))\mathrm{/}r(h)$. Dividing by $h$ and taking $h\to 0$ yields $\chi =\tanh (\beta ϵ\mathrm{/}2)\mathrm{/}(ϵ\mathrm{/}2)=\beta \tanh (x)\mathrm{/}x$.

At $h=0$, the Gibbs density is diagonal in the displayed basis, so $\mathrm{Tr}({\rho }_0{\sigma }_x)=0$. By the ordinary variance definition in [F1], ${\mathrm{Var}}_{{\rho }_0}({\sigma }_x)=\mathrm{Tr}({\rho }_0{\sigma }_x^2)-\mathrm{Tr}({\rho }_0{\sigma }_x{)}^2=1$. The commutator of $\mathrm{diag}(-ϵ\mathrm{/}2,ϵ\mathrm{/}2)$ with ${\sigma }_x$ has nonzero off-diagonal entries for $ϵ>0$, so the commuting equality case in [F1] does not apply. Finally, for $x>0$, $x-\tanh x={\int }_0^x(1-{\mathrm{sech}}^{2}t)dt>0,$ because the integrand is strictly positive for $t>0$. Thus $0<\chi <\beta$. At the zero-gap boundary, ${\lim }_{x\downarrow 0}\tanh (x)\mathrm{/}x=1$, so the response tends to the commuting value $\beta$.