transverse momentum expansion of hard partonic scattering amplitudes, in particular, from the on-shell condition for the unobserved particle k$_{2}$ in Fig. 1. This momentum can be written as k$_{2}$ = p$_{1}$ + p$_{2}$ − k$_{1}$ , where we can parameterize k$_{1}$ as k$_{1}$ = P$_{h}$/z + k$_{1}$$_{T}$ . Thus, the on-shell condition will lead to the following expansion, where the second term will result into a derivative term contribution to the transverse spin dependent differential cross section. As discussed above, for convenience, we use a light-like vector n$_{h}$ to constrain the transverse momentum expansion which is per- pendicular to the final state hadron’s momentum. Therefore, k$_{1}$$_{T}$ can be expressed as k µ 1 T = k$_{1}$ − k$_{1}$ · P$_{h}$n µ h − k$_{1}$ · n$_{h}$P µ $_{h}$. Substituting this expression into the above equation, we will obtain the contribution from the matrix element associated with the operator ∂$_{T}$ψ which will result into the contribution of ˆ H ( z ). The derivative on the Delta function will be translated into the derivative on the twist-three fragmentation function ˆ H ( z ) by partial integral. This contribution is similar to what has been calculated for the twist-three distri- bution functions [17]. In particular, the hard part is proportional to that from the relevant twist-two Born diagrams. In our case, it is the transverse spin transfer coefficients calculated in Ref. [36]. However, there exist major difference between these two contributions. The twist-three distribution contributions come from the pole contributions from the initial/final state interactions, and the derivative terms have contributions from the expansion of the on-shell condition for the unobserved particle and the double pole contribution from the final state interaction diagrams [17]. In the twist-three fragmentation function contributions, in contrast, we do not take pole contributions, and the derivative terms only come from the expansion of the on-shell condition of the unobserved particle in the above 2 → 2 processes. The final result for the derivative contributions of ˆ H ( z ) takes the following form, where 7 x ′ min = − T/z S + U/z , z$_{min}$ = − T + U S , (9) E$_{h}$ d $^{3}$Δ σ ( s$_{⊥}$ ) d $^{3}$P$_{h}$ = ϵ$_{⊥}$$_{αβ}$S α ⊥ 2 α 2 s S ∑ a,b,c ∫ 1 x ′ min dx ′ x ′ f$_{b}$ ( x $^{′}$) 1 x h$_{a}$ ( x ) ∫ 1 z$_{min}$ dz z [ − z ∂ ∂z ( ˆ H ( z ) z 2 )] × 1 x $^{′}$S + T/z { P β h + z ( p$_{2}$ · P$_{h}$n β h − p$_{2}$ · n$_{h}$P β $_{h}$) − z ˆ u } H$_{ab}$$_{→}$$_{c}$ (ˆ s, ˆ t, ˆ u ) , (8) δ (( p$_{1}$ + p$_{2}$ − k$_{1}$ ) $^{2}$) = δ (( p$_{1}$ + p$_{2}$ − P$_{h}$/z ) $^{2}$) − 2( p$_{1}$ + p$_{2}$ ) · k$_{1}$$_{T}$δ $^{′}$(( p$_{1}$ + p$_{2}$ − P$_{h}$/z ) $^{2}$) , (7)