5 It is clear that integrating over final state energies simply produces a sum over vibrational transitions such that the n ’th term in Eq. (A4) represents T$_{0}$$_{n}$ . Calculating T$_{1}$ ( ε$_{i}$, ε$_{f}$ ) is a bit more involved since the integrand in (A1) now includes the first Laguerre polynomial L$_{1}$ ( x ) = 1 − x . We start by writing where T$_{1}$ ( ε$_{i}$, ε$_{f}$ ) = T$_{0}$ ( ε$_{i}$, ε$_{f}$ ) + ˜ T ( ε$_{i}$, ε$_{f}$ ) , (A5) ˜ T ( ε$_{i}$, ε$_{f}$ ) = Γ$_{L}$Γ$_{R}$ e − 2 $^{g}$g ∫ ∞ 0 dse i ( ε 0 − ε$_{i}$ − gω 0 + i Γ / 2) s/ ℏ exp( ge iω 0 $^{s}$) × ∞ 0 dte − i ( ε 0 − ε$_{i}$ − gω 0 − i Γ / 2) t/ $^{ℏ}$exp( ge − iω 0 $^{t}$) e − i ( ε$_{i}$ − ε$_{f}$ ) t/ ℏ × ∫ ∞ −∞ dτ 2 π ℏ $_{3}$e i ( ε$_{i}$ − ε$_{f}$ ) τ/ $^{ℏ}$exp [ ge − iω 0 $^{τ}$eiω 0 $^{t}$(1 − e − iω 0 $^{t}$)(1 − e iω 0 $^{s}$) ] × [ (1 − e iω 0 $^{t}$)( e − iω 0 t − 1) + (1 − e iω 0 $^{s}$)( e − iω 0 s − 1) − e − iω 0 $^{τ}$(1 − e iω 0 $^{t}$)(1 − e iω 0 $^{s}$) − e iω 0 $^{τ}$(1 − e − iω 0 $^{t}$)(1 − e − iω 0 $^{s}$) ] . Taylor expanding the second exponential in the τ integral and performing the integration gives ˜ T ( ε$_{i}$, ε$_{f}$ ) = Γ$_{L}$Γ$_{R}$ e − 2 g g ℏ 2 ∫ ∞ 0 dse i ( ε 0 − ε$_{i}$ − gω 0 + i Γ / 2) s/ $^{ℏ}$exp( ge iω 0 $^{s}$) × ∞ 0 dte − i ( ε 0 − ε$_{i}$ − gω 0 − i Γ / 2) t/ ℏ exp( ge − iω 0 $^{t}$) $_{×}$[ ∞ ∑ m =0 g m m ! e iω 0 $^{t}$(1 − e − iω 0 $^{t}$)m $^{+2}$(1 − e iω 0 $^{s}$)$^{m}$δ ( ε$_{i}$ − ε$_{f}$ − m ℏ ω$_{0}$ ) + ∞ ∑ m =0 g m m ! e − iω 0 $^{s}$(1 − e − iω 0 $^{t}$)$^{m}$(1 − e iω 0 $^{s}$)m $^{+2}$δ ( ε$_{i}$ − ε$_{f}$ − m ℏ ω$_{0}$ ) + ∞ ∑ m =0 g m m ! (1 − e − iω 0 $^{t}$)m $^{+1}$(1 − e iω 0 $^{s}$)m $^{+1}$δ ( ε$_{i}$ − ε$_{f}$ − ( m + 1) ℏ ω$_{0}$ ) + ∞ ∑ m =0 g m m ! e − iω 0 ( s − t $^{)}$(1 − e − iω 0 $^{t}$)m $^{+1}$(1 − e iω 0 $^{s}$)m $^{+1}$δ ( ε$_{i}$ − ε$_{f}$ − ( m − 1) ℏ ω$_{0}$ ) ] . The first two terms are each others complex conjugated and the last two terms factorizes (s and t integrals) into complex conjugated and the integrals can then be performed. The final result is rather complicated but consist of an infinite number of terms, each of which involves a delta function δ ( ε$_{i}$ − ε$_{f}$ − m ℏ ω$_{0}$ ), where m runs from -1 to infinity. We can thus obtain T$_{10}$ and T$_{11}$ by collecting terms involving m = 1 and m = 0 respectively. The results are and T$_{10}$ ( ε$_{i}$ ) = Γ$_{L}$Γ$_{R}$ e − 2 $^{g}$g ∣ ∞ ∑ l =0 g l l ! 1 ε$_{i}$ − ε$_{0}$ − ( l − 1 − g ) ℏ ω$_{0}$ + i Γ / 2 − ∞ ∑ l =0 g l l ! 1 ε$_{i}$ − ε$_{0}$ − ( l − g ) ℏ ω$_{0}$ + i Γ / $_{2}$∣ ∣ ∣$^{2}$ , T$_{11}$ ( ε$_{i}$ ) =Γ$_{L}$Γ$_{R}$ e − 2 g ∣ ∞ ∑ l =0 g l l ! 1 ε$_{i}$ − ε$_{0}$ − ( l − g ) ℏ ω$_{0}$ + i Γ / $_{2}$∣ ∣ ∣$^{2}$ +2 g Γ$_{L}$Γ$_{R}$ e − 2 $^{g}$Re ( 2 ∑ j =0 ( − 1) j ( 2 j ) ∞ ∑ l =0 g l l ! 1 ε$_{i}$ − ε$_{0}$ − ( j + l − 1 − g ) ℏ ω$_{0}$ + i Γ / 2 ∞ ∑ l =0 g l l ! 1 ε$_{i}$ − ε$_{0}$ − ( l − g ) ℏ ω$_{0}$ − i Γ / 2 ) + g $^{2}$Γ$_{L}$ Γ$_{R}$ e − 2 g ∣ 2 ∑ j =0 ( − 1) j ( 2 j ) ∞ ∑ l =0 g l l ! 1 ε$_{i}$ − ε$_{0}$ − ( j + l − 1 − g ) ℏ ω$_{0}$ + i Γ / $_{2}$∣ ∣ ∣$^{2}$ . ∗ Electronic address: tolsen@fysik.dtu.dk 1 H. Park, J. Park, A. K. L. Lim, E. H. Anderson, A. P.