I was studying Tong's lecture notes and there's a specific mathematical step I do not see how to derive (page 108); specifically, I do not see how to derive $(5.9)$ and $(5.10)$
Let's assume the following equation holds (which is a sum of plane wave solutions for $(-i\gamma^i \partial_i + m)\psi$)
$$(-i\gamma^i \partial_i + m)\psi=\sum_{s=1}^2\int \frac{d^3 k}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec k}}}\Big[b_{\vec k}^s(-\gamma^i k_i +m) u^s (\vec k) e^{i \vec k \cdot \vec x}+c_{\vec k}^{s \dagger}(\gamma^i k_i +m)v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big] \tag{*}$$
Given the defining equations for the spinors (which we also assume to hold)
$$(\gamma^{\mu} k_{\mu} - m)u^s(\vec k)= \begin{pmatrix} -m & k_{\mu}\sigma^{\mu} \\ k_{\mu} \bar \sigma^{\mu} & -m \\ \end{pmatrix} u^s(\vec k)=0 \tag{4.105}$$
$$(\gamma^{\mu} k_{\mu} + m)v^s(\vec k)= \begin{pmatrix} m & k_{\mu}\sigma^{\mu} \\ k_{\mu} \bar \sigma^{\mu} & m \\ \end{pmatrix} v^s(\vec k)=0 \tag{4.111}$$
Then Tong asserts that using $(4.105)$ and $(4.111)$ we get
$$(-\gamma^i k_i +m) u^s (\vec k)=\gamma^0 k_0 u^s (\vec k), \ \ \ \ (\gamma^i k_i +m) v^s (\vec k)=-\gamma^0 k_0 v^s (\vec k) \tag{5.9}$$
And then using $(5.9)$ we get
$$(-i\gamma^i \partial_i + m)\psi=\sum_{s=1}^2\int \frac{d^3 k}{(2\pi)^3} \sqrt{\frac{\omega_{\vec k}}{2}} \gamma^0 \Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x} - c_{\vec k}^{s \dagger} v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big] \tag{5.10}$$
But I am completely lost in how to get $(5.9)$ and $(5.10)$
Could you please explain how to derive them?
Maybe for $(5.10)$ we could start from the plane wave solutions $\psi(x)$ and $\psi^{\dagger}(x)$
$$\psi(\vec x) = \sum_{s=1}^2 \int \frac{d^3 k}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec k}}}\Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x}+c_{\vec k}^{s \dagger}v^s(\vec k) e^{-i\vec k \cdot \vec x}\Big]$$
$$\psi^{\dagger}(\vec x) = \sum_{s=1}^2 \int \frac{d^3 k}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec k}}}\Big[b_{\vec k}^{s \dagger} u^{s \dagger} (\vec k) e^{-i \vec k \cdot \vec x}+c_{\vec k}^s v^{s \dagger}(\vec k) e^{i\vec k \cdot \vec x}\Big] \tag{5.4}$$
But no idea how to proceed (Contour integration, Cauchy residue theorem...?)
PS: please let me know if more details need to be included. Thanks.
EDIT
We get (5.9) from $\gamma^\mu k_\mu=\gamma^0k_0+\gamma^ik_i$.
Alright, let's expand out (4.105) and (4.111)
$$(\gamma^0 k_0+\gamma^i k_i - m)u^s(\vec k)= \begin{pmatrix} -m & k_0 + k_i\sigma^i \\ k_0 - k_i\sigma^i & -m \\ \end{pmatrix} u^s(\vec k)=0 \tag{4.105}$$
$$(\gamma^0 k_0+\gamma^i k_i - m)v^s(\vec k)= \begin{pmatrix} m & k_0 + k_i\sigma^i \\ k_0 - k_i\sigma^i & m \\ \end{pmatrix} v^s(\vec k)=0 \tag{4.111}$$
Where I have used:
$$\sigma^{\mu}=(1,\sigma^i), \ \ \ \ \bar\sigma^{\mu}=(1,-\sigma^i) \tag{4.63}$$
But ,unfortunately, I still do not see why this leads to (5.9)
We get (5.10) by substituting the two halves of (5.9) in (*) to remove the $\pm\gamma^ik_i+m$ operators.
By simply plugging (5.9) into (5.10) I get
$$(-i\gamma^i \partial_i + m)\psi=\sum_{s=1}^2\int \frac{d^3 k}{(2\pi)^3} \sqrt{\frac{\omega_{\vec k}}{2}} \gamma^0 k_0\Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x} - c_{\vec k}^{s \dagger} v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big]$$
Note I get $k_0$, which shouldn't be there. What am I missing? Thanks.
EDIT 1
$$\gamma^0k_0v^s+\gamma^ik_iv^s=\gamma^\mu k_\mu v^s=-mv^s\implies(\gamma^ik_i+m)v^s=-\gamma^0k_0v^s.$$
We get (5.9) from $\gamma^\mu k_\mu=\gamma^0k_0+\gamma^ik_i$. We get (5.10) by substituting the two halves of (5.9) in (*) to remove the $\pm\gamma^ik_i+m$ operators.
In particular, from (4.105)$$\gamma^0k_0u^s+\gamma^ik_iu^s=\gamma^\mu k_\mu u^s=mu^s\implies(-\gamma^ik_i+m)u^s=\gamma^0k_0u^s.$$The proof from (4.111) of the second half of (5.9) is similar.