In Peskin & Schroeder, the Momentum operator of the Dirac field is given as: $$ {\bf P}=\int d^3x \psi^\dagger \left(-i\nabla\right) \psi=\int \frac{d^3p}{(2\pi)^3}\sum_s {\bf p}(a_p^{s\dagger}a_p^s+b_p^{s\dagger}b_p^s).\tag{3.105} $$ I am having trouble deriving the right-hand side of this equation from the left hand, since I tried plugging in $\psi,\bar{\psi}$ in the Schrödinger picture but got stuck : $$ P^i=\int d^3x\psi^\dagger (-i\partial_i)\psi= $$ $$\int d^3x\int \frac{d^3pd^3q}{(2\pi)^6}\frac{1}{\sqrt{2E_p}}\sum_s (b_p^s\bar{v}^s(p)e^{-ip\cdot x}+a_p^{s\dagger}\bar{u}^s(p)e^{ip\cdot x} )(-i\partial_i)(\frac{1}{\sqrt{2E_q}}\sum_r (a_q^s u^s(q)e^{-iq\cdot x}+b_q^{r\dagger}v^s(q)e^{iq\cdot x} )) $$ $$=\int d^3x\int \frac{d^3pd^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_pE_q}}\sum_s (b_p^s\bar{v}^s(p)e^{-ip\cdot x}+a_p^{s\dagger}\bar{u}^s(p)e^{ip\cdot x} )\sum_r q^i(-a_q^s u^s(q)e^{-iq\cdot x}+b_q^{r\dagger}v^s(q)e^{iq\cdot x} )) $$ $$ =\int d^3x\int \frac{d^3pd^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_pE_q}}q^i\sum_{s,r} (b_p^s\bar{v}^s(p)e^{-ip\cdot x}+a_p^{s\dagger}\bar{u}^s(p)e^{ip\cdot x} )(-a_q^s u^s(q)e^{-iq\cdot x}+b_q^{r\dagger}v^s(q)e^{iq\cdot x} )) $$
Now I would want to use the definition of the delta function to turn q to p and get rid of the integration over x, and the anticommuting (?) terms to vanish but can't quite think of a reason why they should. Am I going about this the wrong way? Edit: thanks to Hyperon's hints I was able to get to: $$ P^i=\int d^3x\psi^\dagger (-i\partial_i)\psi= $$ $$\int d^3x\int \frac{d^3pd^3q}{(2\pi)^6}e^{ip\cdot x}e^{-iq\cdot x}\frac{1}{\sqrt{2E_p}}\sum_s (b_{-p}^sv^{s\dagger}(-p)+a_p^{s\dagger}u^{s\dagger}(p) )(-i\partial_i)\left(\frac{1}{\sqrt{2E_q}}\sum_r (a_q^r u^r(q)+b_{-q}^{r\dagger}v^r(-q)\right) $$ $$=\int d^3x\int \frac{d^3pd^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_pE_q}}e^{ip\cdot x}e^{-iq\cdot x}\sum_{s,r} (b_{-p}^sv^{s\dagger}(-p)+a_p^{s\dagger}u^{s\dagger}(p) ) q^i(-a_q^r u^r(q)+b_{-q}^{r\dagger}v^r(-q) )) $$ $$ =\int \frac{d^3pd^3q}{(2\pi)^3}\frac{1}{2\sqrt{E_pE_q}}q^i\delta^{(3)}(p-q)\sum_{s,r} (b_{-p}^sv^{s\dagger}(-p)+a_p^{s\dagger}u^{s\dagger}(p) )(-a_q^r u^r(q)+b_{-q}^{r\dagger}v^r(-q) ) $$ $$ =\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}p^i\sum_{s,r} (b_{-p}^sv^{s\dagger}(-p)+a_p^{s\dagger}u^{s\dagger}(p) )(-a_p^r u^r(p)+b_{-p}^{r\dagger}v^r(-p) ) $$
But I still fail to see how the terms of the type $b_{-p}^sv^{s\dagger}(-p)a_p^r u^r(p)$ vanish from the final result.
