1 Correction to: Journal of Statistical Physics (2020) 178:319–378 https://doi.org/10.1007/s10955-019-02434-w

The claimed bound \({{\,\mathrm{Ric}\,}}(\mathcal {A}, \nabla , \tau ) \ge \gamma \) in Theorem 10.6 in our paper [1] is unfortunately incorrect, as pointed out in [2]. A small modification of the proof shows that the weaker estimate \({{\,\mathrm{Ric}\,}}(\mathcal {A}, \nabla , \tau ) \ge \frac{\gamma }{2}\) holds.

This bound can be obtained by replacing (10.5) by the following computation, using the scalar inequalities \(\partial _1 \Lambda (a,b), \partial _2 \Lambda (a,b) \ge 0\) for \(a, b > 0\):

$$\begin{aligned} {{\,\mathrm{Hess}\,}}_\mathscr {K}{{\,\mathrm{Ent}\,}}(\rho )[A, A]&= - \tau [( \nabla \mathscr {L}A)^* \widehat{\rho }\# \nabla A] + \tau \big [(\nabla A)^* \mathcal {N}_{\rho ,\mathscr {L}^\dagger \rho }^{(\eta )} \# (\nabla A) \big ] \\&= \frac{\gamma }{2} \tau \big [(\nabla A)^* (\Lambda + \partial _1 \Lambda + \partial _2 \Lambda )(\rho , \rho ) \# (\nabla A) \big ] \\&\ge \frac{\gamma }{2} \tau \big [(\nabla A)^* \Lambda (\rho , \rho ) \# (\nabla A) \big ] \\&= \frac{\gamma }{2} \tau [(\nabla A)^* \widehat{\rho }\# \nabla A] = \frac{\gamma }{2} \langle {\mathscr {K}_\rho A, A}\rangle _{L^2(\tau )} \ . \end{aligned}$$

On the matrix algebra \({\mathbb M}_n({\mathbb C})\), it is possible to improve the curvature bound \(\frac{\gamma }{2}\) to \(\frac{\gamma }{2}(1 + \frac{1}{n})\), using the scalar inequality \(\partial _1 \Lambda (a,b) + \partial _2 \Lambda (a,b) \ge 1 \ge \frac{1}{n}\Lambda (a,b)\) for \(0 < a, b \le n\).

We thank Michael Brannan and Li Gao and Marius Junge for pointing out the error, and Melchior Wirth and Haonan Zhang for useful discussions.