When $T\ll\Delta$, the argument of the inverse hyperbolic sine is large. Using $\operatorname{arsinh}z=\log(2z)+1/(4z^2)+O(z^{-4})$ gives
$$
\boxed{
m(T)=\Delta+2T e^{-\Delta/T}
+O(Te^{-2\Delta/T}).}
$$
The mass remains the zero-temperature gap, with an exponentially small correction from thermally activated excitations.

When $T\gg\Delta$, expand around $z=1/2$. If
$$
\varphi=\frac{1+\sqrt5}{2}
$$
is the <golden ratio>, then $\operatorname{arsinh}(1/2)=\log\varphi$, and
$$
\boxed{
m(T)=2T\log\varphi+\frac{\Delta}{\sqrt5}
+O\left(\frac{\Delta^2}{T}\right).}
$$
The leading value $m/T=2\log\varphi$ is universal. This is the <quantum-critical regime>: temperature is the only leading energy scale and the <correlation length> is of order $v/T$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 / 1 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-337/1/f/solution