Corrigendum: Quantum adiabatic Markovian master equations (2012 New J. Phys. 14 123016)

4.3. Master equation in the adiabatic limit with rotating wave approximation: Lindblad form

Following equation (52) when we discuss the rotating wave approximation, we implicitly assume a non-degeneracy condition. This should have been stated explicitly. We provide here the more general case of degenerate subspaces.

We note that ${\mu }_{{dc}}(t,0)+{\mu }_{{ba}}(t,0)={\displaystyle \int }_{0}^{t}{\rm{d}}\tau \left[{\omega }_{{dc}}(\tau )+{\omega }_{{ba}}(\tau )-({\phi }_{d}(\tau )-{\phi }_{c}(\tau ))+({\phi }_{b}(\tau )-{\phi }_{a}(\tau ))\right].$ One can now make the argument that when the $t\to \infty$ limit is taken, terms for which the integrand vanishes will dominate, thus enforcing the 'energy conservation' condition ${\omega }_{{ba}}=-{\omega }_{{dc}}$. This is a similar rotating wave approximation as made in the standard time-independent treatment, although here, the approximation of phase cancellation is made along the entire time evolution of the instantaneous energy eigenstates. Clearly, in light of the appearance of other terms involving t in equation (52b), this argument is far from rigorous. Nevertheless, we proceed from equation (52b) to write, in the $t\to \infty$ limit

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{t}{\rm{d}}\tau {A}_{\beta }(t-\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\approx \displaystyle \sum _{\omega }{A}_{\beta ,\omega }(t){A}_{\alpha ,\omega }(t){{\rm{\Pi }}}_{\omega }(0){\tilde{\rho }}_{S}(t){{\rm{\Pi }}}_{\omega }(0){{\rm{\Gamma }}}_{\alpha \beta }(\omega ),\end{eqnarray} \tag{ 53 }$$

where we have defined a new index ω such that:

$$\begin{eqnarray}&&{A}_{\alpha ,\omega }(t)=\displaystyle \sum _{{\varepsilon }_{b}(t)-{\varepsilon }_{a}(t)=\omega }\langle {\varepsilon }_{a}(t)| {A}_{\alpha }| {\varepsilon }_{b}(t)\rangle ,\quad {{\rm{\Pi }}}_{\omega }(0)=\displaystyle \sum _{{\varepsilon }_{b}(t)-{\varepsilon }_{a}(t)=\omega }| {\varepsilon }_{a}(0)\rangle \langle {\varepsilon }_{b}(0)| .\end{eqnarray} \tag{ 54 }$$

Note that the set of $\{\omega \}$'s involved in the sum ${\displaystyle \sum }_{\omega }$ is evolving in time since it corresponds to differences of the instantaneous energy eigenvalues, but we suppress the time dependence for notational brevity. We show in appendix G how, by performing a transformation back to the Schrödinger picture, along with a double-sided adiabatic approximation, we arrive from equation (53) at the Schrödinger picture adiabatic master equation in Lindblad form:

$$\begin{eqnarray}&&{\dot{\rho }}_{S}(t)=-{\rm{i}}\left[{H}_{S}(t)+{H}_{{\rm{LS}}}(t),{\rho }_{S}(t)\right]+\displaystyle \sum _{\alpha \beta }\displaystyle \sum _{\omega }{\gamma }_{\alpha \beta }(\omega )\left[{L}_{\omega ,\beta }(t){\rho }_{S}(t){L}_{\omega ,\alpha }^{\dagger }(t)-\displaystyle \frac{1}{2}\left\{{L}_{\omega ,\alpha }^{\dagger }(t){L}_{\omega ,\beta }(t),{\rho }_{S}(t)\right\}\right],\end{eqnarray} \tag{ 55 }$$

where the Hermitian Lamb shift term is

$$\begin{eqnarray}&&{H}_{{\rm{LS}}}(t)=\displaystyle \sum _{\alpha \beta }\displaystyle \sum _{\omega }{L}_{\omega ,\alpha }^{\dagger }(t){L}_{\omega ,\beta }(t){S}_{\alpha \beta }(\omega ),\end{eqnarray} \tag{ 56 }$$

and we have defined

$$\begin{eqnarray}&&{L}_{\omega ,\alpha }(t)\equiv \displaystyle \sum _{{\varepsilon }_{b}(t)-{\varepsilon }_{a}(t)=\omega }{L}_{{ab},\alpha }(t).\end{eqnarray} \tag{ 57 }$$

5.2. Correlation function analysis

Equation (70) should read:

$$\begin{eqnarray}&&{{\mathcal{B}}}_{{ab}}(\tau )=\displaystyle \frac{\eta }{{\beta }^{2}}{g}^{a}{g}^{b}(-{\pi }^{2}{\mathrm{csch}}^{2}\left(\displaystyle \frac{\pi \tau }{\beta }\right)+\displaystyle \sum _{n=1}^{\infty }\displaystyle \frac{{\psi }^{(n+1)}\left(1-\frac{{\rm{i}}\tau }{\beta }\right)+{\psi }^{(n+1)}\left(\frac{{\rm{i}}\tau }{\beta }\right)}{n!{\left(\beta {\omega }_{c}\right)}^{n}}),\end{eqnarray} \tag{ 70 }$$

Following the notation in this section, the expression for $| {{\mathcal{B}}}_{\alpha \beta }(\tau )|$ should not include a g², and equation (B8) should read:

$$\begin{eqnarray}&&\parallel {{\rm{\Delta }}}_{2}{\parallel }_{1}\lesssim \displaystyle \frac{\eta }{\beta {\omega }_{c}}{\displaystyle \int }_{{\tau }_{\mathrm{tr}}}^{\infty }{\rm{d}}\tau \ \displaystyle \frac{1}{{\tau }^{2}}\sim \displaystyle \frac{\eta }{{\beta }^{2}{\omega }_{c}\mathrm{ln}(\beta {\omega }_{c})},\end{eqnarray} \tag{ B8 }$$

Equation (E18) should read

$$\begin{eqnarray}Q(t,{t}^{\prime }) & \equiv & {U}_{S}^{{\rm{ad}}}(t,{t}^{\prime }){V}_{1}(t,{t}^{\prime }){U}_{S}^{{\rm{ad}}\dagger }(t,{t}^{\prime })\\ & = & \displaystyle \sum _{a\ne b}{{\rm{e}}}^{-{\rm{i}}{\mu }_{{ab}}(t,{t}^{\prime })}\left(-{\displaystyle \int }_{{t}^{\prime }}^{t}{\rm{d}}\tau {{\rm{e}}}^{{\rm{i}}{\mu }_{{ab}}(\tau ,{t}^{\prime })}\langle {\varepsilon }_{a}(\tau )| {\dot{\varepsilon }}_{b}(\tau )\rangle \right)| {\varepsilon }_{a}(t)\rangle \langle {\varepsilon }_{b}(t)| .\end{eqnarray} \tag{ E18 }$$

Equation (E19) should read:

$$\begin{eqnarray}&&{Q}_{{ab}}(t,{t}^{\prime })={{\rm{e}}}^{-{\rm{i}}{\mu }_{{ab}}(t,{t}^{\prime })}\left(-{\displaystyle \int }_{{t}^{\prime }}^{t}{\rm{d}}\tau {{\rm{e}}}^{{\rm{i}}{\mu }_{{ab}}(\tau ,{t}^{\prime })}\langle {\varepsilon }_{a}(\tau )| {\partial }_{\tau }| {\varepsilon }_{b}(\tau )\rangle \right).\end{eqnarray} \tag{ E19 }$$

Equation (E20) should read:

$$\begin{eqnarray}&&{\displaystyle \int }_{t^{\prime} }^{t}{\rm{d}}\tau {{\rm{e}}}^{{\rm{i}}{\mu }_{{ab}}(\tau ,{t}^{\prime })}\langle {\varepsilon }_{a}(\tau )| {\partial }_{\tau }| {\varepsilon }_{b}(\tau )\rangle ={\displaystyle \int }_{{t}^{\prime }/{t}_{f}}^{s}{\rm{d}}{s}^{\prime }\ {{\rm{e}}}^{{\rm{i}}{t}_{f}{\tilde{\mu }}_{{ab}}({s}^{\prime },{t}^{\prime }/{t}_{f})}\langle {\varepsilon }_{a}(s^{\prime} )| {\partial }_{{s}^{\prime }}| {\varepsilon }_{b}(s^{\prime} )\rangle ,\end{eqnarray} \tag{ E20 }$$

equation (E21) should read

$$\begin{eqnarray}{\displaystyle \int }_{{t}^{\prime }/{t}_{f}}^{s}{\rm{d}}{s}^{\prime }\ {{\rm{e}}}^{{\rm{i}}{t}_{f}{\tilde{\mu }}_{{ab}}({s}^{\prime },{t}^{\prime }/T)}\langle {\varepsilon }_{a}(s^{\prime} )| {\partial }_{s^{\prime} }| {\varepsilon }_{b}(s^{\prime} )\rangle & = & \displaystyle \frac{{\rm{i}}}{{t}_{f}{\omega }_{{ab}}(s^{\prime} )}{{\rm{e}}}^{{\rm{i}}{t}_{f}{\tilde{\mu }}_{{ab}}({s}^{\prime },{t}^{\prime }/{t}_{f})}\langle {\varepsilon }_{a}(s^{\prime} )| {\partial }_{{s}^{\prime }}| {\varepsilon }_{b}(s^{\prime} )\rangle {| }_{{t}^{\prime }/{t}_{f}}^{s}\\ & & -\displaystyle \frac{{\rm{i}}}{{t}_{f}}{\displaystyle \int }_{{t}^{\prime }/{t}_{f}}^{s}{\rm{d}}{s}^{\prime }\ {{\rm{e}}}^{{\rm{i}}{t}_{f}{\tilde{\mu }}_{{ab}}({s}^{\prime },{t}^{\prime }/{t}_{f})}\displaystyle \frac{{\rm{d}}}{{\rm{d}}{s}^{\prime }}\displaystyle \frac{\langle {\varepsilon }_{a}(s^{\prime} )| {\partial }_{{s}^{\prime }}| {\varepsilon }_{b}(s^{\prime} )\rangle }{{\omega }_{{ab}}(s^{\prime} )}.\end{eqnarray} \tag{ E21 }$$

Equation (E22) should read:

$$\begin{eqnarray}| {Q}_{{ab}}(t,{t}^{\prime })| & \leqslant & \displaystyle \frac{| \langle {\varepsilon }_{a}(s)| {\partial }_{s}{H}_{S}(s)| {\varepsilon }_{b}(s)\rangle | }{{\omega }_{{ab}}^{2}(s){t}_{f}}{| }_{{t}^{\prime }/{t}_{f}}^{t/{t}_{f}}\\ & & +\left|{\displaystyle \int }_{{t}^{\prime }/{t}_{f}}^{t/{t}_{f}}{\rm{d}}{s}^{\prime }\ \displaystyle \frac{{{\rm{e}}}^{{\rm{i}}{t}_{f}{\tilde{\mu }}_{{ab}}({s}^{\prime },{t}^{\prime }/{t}_{f})}}{{t}_{f}}\displaystyle \frac{{\rm{d}}}{{\rm{d}}{s}^{\prime }}\displaystyle \frac{\langle {\varepsilon }_{a}(s^{\prime} )| {\partial }_{{s}^{\prime }}{H}_{S}(s^{\prime} )| {\varepsilon }_{b}(s^{\prime} )\rangle }{{\omega }_{{ab}}^{2}(s^{\prime} )}\right|.\end{eqnarray} \tag{ E22 }$$

We provide the following rewrite of this appendix section that includes numerous fixes.

We wish to bound the error associated with neglecting Θ in equation (40), i.e., we wish to bound

$$\begin{eqnarray}&&\parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty }=\parallel {U}_{S}(t-\tau ,0)-{{\rm{e}}}^{{\rm{i}}\tau {H}_{S}(t)}{U}_{S}^{\mathrm{ad}}(t,0){\parallel }_{\infty }=\parallel {U}_{S}^{\mathrm{ad}\dagger }(t,0){{\rm{e}}}^{-{\rm{i}}\tau {H}_{S}(t)}{U}_{S}^{\dagger }(t,t-\tau ){U}_{S}(t,0)-{\mathbb{I}}{\parallel }_{\infty }.\end{eqnarray} \tag{ F1 }$$

Using that the operator $\hat{U}(\tau )={{\rm{e}}}^{-{\rm{i}}\tau {H}_{S}(t)}{U}_{S}^{\dagger }(t,t-\tau )$ satisfies:

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{d}}}{{\rm{d}}\tau }\hat{U}(\tau )=\left({H}_{S}(t)-{{\rm{e}}}^{-{\rm{i}}{H}_{S}(t)\tau }{H}_{S}(t-\tau ){{\rm{e}}}^{{\rm{i}}{H}_{S}(t)\tau }\right)\hat{U}(\tau ),\end{eqnarray} \tag{ F2 }$$

we can write a formal solution for $\hat{U}$ as:

$$\begin{eqnarray}&&\hat{U}(\tau )={\mathbb{I}}-{\rm{i}}{\displaystyle \int }_{0}^{\tau }{\rm{d}}{t}^{\prime }\left[{H}_{S}(t)-{{\rm{e}}}^{-{\rm{i}}{H}_{S}(t){t}^{\prime }}{H}_{S}(t-{t}^{\prime }){{\rm{e}}}^{{\rm{i}}{H}_{S}(t){t}^{\prime }}\right]\hat{U}({t}^{\prime }).\end{eqnarray} \tag{ F3 }$$

Therefore we can bound:

$$\begin{eqnarray}\parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty } & = & \| {U}_{S}^{\mathrm{ad}\dagger }(t,0){U}_{S}(t,0)\\ & & -{{\rm{i}}{\displaystyle \int }_{0}^{\tau }{\rm{d}}{t}^{\prime }{U}_{S}^{\mathrm{ad}}(t,0)\left[{H}_{S}(t)-{{\rm{e}}}^{-{\rm{i}}{H}_{S}(t){t}^{\prime }}{H}_{S}(t-{t}^{\prime }){{\rm{e}}}^{{\rm{i}}{H}_{S}(t){t}^{\prime }}\right]\hat{U}({t}^{\prime }){U}_{S}(t,0)-{\mathbb{I}}\| }_{\infty }\end{eqnarray} \tag{ F4a }$$

$$\begin{eqnarray}&&\leqslant \mathrm{min}\{2,\parallel Q(t,0){\parallel }_{\infty }+{\displaystyle \int }_{0}^{\tau }{\rm{d}}{t}^{\prime }\parallel \left[{H}_{S}(t)-{{\rm{e}}}^{-{\rm{i}}{H}_{S}(t){t}^{\prime }}{H}_{S}(t-{t}^{\prime }){{\rm{e}}}^{{\rm{i}}{H}_{S}(t){t}^{\prime }}\right]{\parallel }_{\infty }+O({t}_{f}^{-2})\},\end{eqnarray} \tag{ F4b }$$

where we used equation (E17) and the fact that supoperator norm between two unitaries is always upper bounded by 2 in the second line, and the standard adiabatic estimate to bound $\parallel Q(t,0){\parallel }_{\infty }$ (recall subsection E.2). While h of equation (26) is expressed in terms of a matrix element, a more careful analysis (e.g., [10]) would replace this with an operator norm. Thus we shall make the plausible assumption that $h\sim {t}_{f}{\mathrm{max}}_{{t}^{\prime }\in [t-\tau ,t]}\parallel {\partial }_{{t}^{\prime }}{H}_{S}({t}^{\prime }){\parallel }_{\infty },$ and, dropping the subdominant $O({t}_{f}^{-2}),$ we can write

$$\begin{eqnarray}&&\parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty }\leqslant \mathrm{min}\{2,\displaystyle \frac{h}{{{\rm{\Delta }}}^{2}{t}_{f}}+\displaystyle \frac{{\tau }^{2}\tilde{h}}{2{t}_{f}}\}.\end{eqnarray} \tag{ F5 }$$

where $\tilde{h}={\mathrm{max}}_{{t}^{\prime }\in [t-\tau ,t]}\frac{{t}_{f}}{{t}^{\prime }}\parallel \left[{H}_{S}(t)-{{\rm{e}}}^{-{\rm{i}}{H}_{S}(t){t}^{\prime }}{H}_{S}(t-{t}^{\prime }){{\rm{e}}}^{{\rm{i}}{H}_{S}(t){t}^{\prime }}\right]{\parallel }_{\infty }.$ We can now bound the error term in equation (42b). Let $X(t,\tau )\equiv {{\rm{e}}}^{{\rm{i}}\tau {H}_{S}(t)}{U}_{S}^{\mathrm{ad}}(t,0),$ so that ${U}_{S}(t-\tau ,0)=X(t,\tau )+{\rm{\Theta }}(t,\tau ).$ We can then write

$$\begin{eqnarray}&&{\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {A}_{\beta }(t-\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )={\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {X}^{\dagger }(t,\tau ){A}_{\beta }X(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\end{eqnarray} \tag{ F6a }$$

$$\begin{eqnarray}&&\quad +{\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {X}^{\dagger }(t,\tau ){A}_{\beta }{\rm{\Theta }}(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )+{\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau X{{\rm{\Theta }}}^{\dagger }(t,\tau ){A}_{\beta }X(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\end{eqnarray} \tag{ F6b }$$

$$\begin{eqnarray}&&\quad +{\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {{\rm{\Theta }}}^{\dagger }(t,\tau ){A}_{\beta }{\rm{\Theta }}(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau ).\end{eqnarray} \tag{ F6c }$$

The first term on the rhs of (F6a) is the approximation we have used in equation (42b). The terms in (F6b) and (F6c) can be bounded as follows, using equation (F5), the unitarity of X, the fact that $\parallel {\tilde{\rho }}_{S}{\parallel }_{\infty }\leqslant 1,$ and recalling that $\parallel {A}_{\alpha }{\parallel }_{\infty }=1.$ First we assume that equation (12) applies. Then:

$$\begin{eqnarray}&&{\| {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {X}^{\dagger }(t,\tau ){A}_{\beta }{\rm{\Theta }}(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\| }_{\infty },{\| {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau X{{\rm{\Theta }}}^{\dagger }(t,\tau ){A}_{\beta }X(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\| }_{\infty }\\ &&\quad \leqslant {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau \parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty }| {{\mathcal{B}}}_{\alpha \beta }(\tau )| \lesssim \mathrm{min}\{2{\tau }_{B},\displaystyle \frac{{\tau }_{B}h}{{{\rm{\Delta }}}^{2}{t}_{f}}+\displaystyle \frac{{\tau }_{B}^{3}\tilde{h}}{2{t}_{f}}\}\ ,\end{eqnarray} \tag{ F7a }$$

$$\begin{eqnarray}&&{\| {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {{\rm{\Theta }}}^{\dagger }(t,\tau ){A}_{\beta }{\rm{\Theta }}(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\| }_{\infty }\\ &&\quad \leqslant {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau \parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty }^{2}| {{\mathcal{B}}}_{\alpha \beta }(\tau )| \lesssim \mathrm{min}\{4{\tau }_{B},2\displaystyle \frac{{\tau }_{B}h}{{{\rm{\Delta }}}^{2}{t}_{f}}+2\displaystyle \frac{{\tau }_{B}^{3}\tilde{h}}{2{t}_{f}}\}\ ,\end{eqnarray} \tag{ F7b }$$

where in the last inequality we used the fact that if $x\leqslant 2$ then ${[\mathrm{min}(2,x)]}^{2}={x}^{2}\leqslant 2x,$ and if $x\geqslant 2$ then again ${[\mathrm{min}(2,x)]}^{2}=2\mathrm{min}(2,x)\leqslant 2x,$ with $x=\frac{h}{{{\rm{\Delta }}}^{2}{t}_{f}}+\frac{{\tau }^{2}\rangle h}{{t}_{f}},$ in order to avoid having to extend equation (12) to higher values of n. In all, then, the approximation error in equation (42b) is $O[\mathrm{min}\{{\tau }_{B},\frac{{\tau }_{B}h}{{{\rm{\Delta }}}^{2}{t}_{f}}+\frac{{\tau }_{B}^{3}\tilde{h}}{{t}_{f}}\}].$

Next we recall from the discussion in section 2.3 that equation (12) must, in the case of a Markovian bath with a finite cutoff, be replaced by the weaker condition (23), reflecting fast decay up to ${\tau }_{\mathrm{tr}},$ followed by power-law decay. In this case the terms in (F6b) can instead be bounded as follows:

$$\begin{eqnarray}&&{\| {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau {X}^{\dagger }(t,\tau ){A}_{\beta }{\rm{\Theta }}(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\| }_{\infty },{\| {\displaystyle \int }_{0}^{\infty }{\rm{d}}\tau X{{\rm{\Theta }}}^{\dagger }(t,\tau ){A}_{\beta }X(t,\tau ){\tilde{\rho }}_{S}(t){A}_{\alpha }(t){{\mathcal{B}}}_{\alpha \beta }(\tau )\| }_{\infty }\\ &&\quad \leqslant {\displaystyle \int }_{0}^{{\tau }_{\mathrm{tr}}}{\rm{d}}\tau \parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty }| {{\mathcal{B}}}_{\alpha \beta }(\tau )| +{\displaystyle \int }_{{\tau }_{\mathrm{tr}}}^{\infty }{\rm{d}}\tau \ \}\parallel {\rm{\Theta }}(t,\tau ){\parallel }_{\infty }| {{\mathcal{B}}}_{\alpha \beta }(\tau )| \\ &&\quad \lesssim \mathrm{min}\left\{2{\tau }_{B},\displaystyle \frac{{\tau }_{B}h}{{{\rm{\Delta }}}^{2}{t}_{f}}+\displaystyle \frac{{\tau }_{B}^{3}\tilde{h}}{2{t}_{f}}\right\}+2\displaystyle \frac{{\tau }_{M}^{2}}{{\tau }_{\mathrm{tr}}},\end{eqnarray} \tag{ F8 }$$

in place of (F7b). A similar modification can be computed for the term in (F6c). To compute the order of $\frac{{\tau }_{M}^{2}}{{\tau }_{\mathrm{tr}}},$ we recall that ${\tau }_{\mathrm{tr}}\sim \beta \mathrm{ln}(\beta {\omega }_{c})\gg \beta$ (equation (75)), and that ${\tau }_{M}=\sqrt{2\beta /{\omega }_{c}}.$ Thus $\frac{{\tau }_{M}^{2}}{{\tau }_{\mathrm{tr}}}\sim {[{\omega }_{c}\mathrm{ln}(\beta {\omega }_{c})]}^{-1}.$ It follows that we can safely ignore the $\frac{{\tau }_{M}^{2}}{{\tau }_{\mathrm{tr}}}$ term in equation (F8) provided equation (77) is satisfied. The analysis of the term in (F6c) does not change this conclusion.