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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

$$\gamma^0k_0v^s+\gamma^ik_iv^s=\gamma^\mu k_\mu v^s=-mv^s\implies(\gamma^ik_i+m)v^s=-\gamma^0k_0v^s.$$

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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.

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  • $\begingroup$ J.G. Thanks for your answer. Could you please check my edit? $\endgroup$ – JD_PM Jun 21 at 20:31
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    $\begingroup$ @JD_PM You get (5.9) by rearranging the parts of (4.105), (4.111) that equate something not explicitly written as a matrix to $0$; there's no need to work with the $\sigma^\mu$ etc. The $k^0$ in (5.10) is $\omega_{\vec{k}}$, so when multiplied by $\frac{1}{\sqrt{2\omega_{\vec{k}}}}$ you get $\sqrt{\frac{\omega_{\vec{k}}}{2}}$ as required. $\endgroup$ – J.G. Jun 21 at 20:45
  • $\begingroup$ J.G. Thanks I see now how to get (5.10). However I am afraid I still do not understand how to get (5.9); could you please explain how to rearrange (4.105), (4.111) so that equate something not explicitly written as a matrix to $0$? $\endgroup$ – JD_PM Jun 22 at 8:44
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    $\begingroup$ @JD_PM See edit. $\endgroup$ – J.G. Jun 22 at 8:53
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    $\begingroup$ J.G. Ahhh I got it. I really appreciate your patience and quick responses. I really hope to come across you again. $\endgroup$ – JD_PM Jun 22 at 9:21

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