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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 (pages 108, 109); specifically, I do not see how to derive $(5.12)$

Let's go step by step. Given the Hamiltonian operator

$$H= \int d^3 x \psi^{\dagger} \gamma^0 (-i \gamma^i \partial_i + m) \psi \tag{*}$$

Given the following mode expansion

$$(-i\gamma^i \partial_i + m)\psi=\sum_{s=1}^2\int \frac{d^3 k}{(2\pi)^3} \sqrt{\frac{\omega_{\vec k}}{2}} \gamma^0 \Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x} - c_{\vec k}^{s \dagger} v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big] \tag{5.10}$$

Given the mode expansions for $\psi$ and $\psi^{\dagger}$ operators

$$\psi(\vec x) = \sum_{r=1}^2 \int \frac{d^3 q}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec q}}}\Big[b_{\vec q}^r u^r (\vec q) e^{i \vec q \cdot \vec x}+c_{\vec q}^{r \dagger}v^r(\vec q) e^{-i\vec q \cdot \vec x}\Big] \tag{5.4a}$$

$$\psi^{\dagger}(\vec x) = \sum_{r=1}^2 \int \frac{d^3 q}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec q}}}\Big[b_{\vec q}^{r \dagger} u^{r \dagger} (\vec q) e^{-i \vec q \cdot \vec x}+c_{\vec q}^r v^{r \dagger}(\vec q) e^{i\vec q \cdot \vec x}\Big] \tag{5.4b}$$

I understand that by applying $(5.4b)$ and (5.10) to $(*)$ we get

$$H=\int \frac{d^3 x d^3 p d^3 q}{(2\pi)^6} \sqrt{\frac{\omega_{\vec k}}{4 \omega_{\vec q}}}\Big[b_{\vec q}^{r \dagger} u^{r \dagger} (\vec q) e^{-i \vec q \cdot \vec x}+c_{\vec q}^r v^{r \dagger} (\vec q) e^{i \vec q \cdot \vec x}\Big] \cdot \Big[b_{\vec k}^s u^s (\vec k) e^{i \vec k \cdot \vec x}-c_{\vec k}^{s \dagger} v^s (\vec k) e^{-i \vec k \cdot \vec x}\Big] \tag{**}$$

OK so far. But now he carries on asserting the following

$$H=\frac 1 2 \int \frac{d^3 k}{(2\pi)^3} \Big[ b_{\vec k}^{r \dagger} b_{\vec k}^s[u^{r \dagger} (\vec k) \cdot u^s (\vec k)]-c_{\vec k}^r c_{\vec k}^{s \dagger} [v^{r \dagger} (\vec k) \cdot v^s(\vec k)]-b_{\vec k}^{r \dagger} c_{-\vec k}^{s \dagger}[u^{r \dagger}(\vec p) \cdot v^s (-\vec k)]+c_{\vec k}^r b_{-\vec k}^s [v^{r \dagger} (\vec k) \cdot u^s (-\vec k)]\Big] \tag{***}$$

Where we've done the following integral

$$\int \frac{d^3 x}{(2\pi)^3} e^{i \vec x \cdot (\vec k - \vec q)}= \delta^{(3)} (\vec k - \vec q) \tag{A}$$

I am OK with everything on $(***)$ but the fact that he applies $\vec k \rightarrow -\vec k$ only to the last two terms just to make the last two terms vanish based on the following inner-product identities (which we assume to be true in this post):

$$u^{r \dagger} (\vec k) \cdot u^s (\vec k)=v^{r \dagger} (\vec k) \cdot v^s (\vec k)=2 p_0 \delta^{r s}, \ \ \ \ u^{r\dagger} (\vec k) \cdot v^s (-\vec k)=v^{r \dagger} (\vec k) \cdot u^s (-\vec k)=0 \tag{****}$$

Then we finally get

$$H = \int \frac{d^3 k}{(2\pi)^3} \omega_{\vec k} \Big( b_{\vec k}^{s \dagger}b_{\vec k}^s - c_{\vec k}^s c_{\vec k}^{s \dagger}\Big) = \int \frac{d^3 k}{(2\pi)^3} \omega_{\vec k} \Big( b_{\vec k}^{s \dagger} b_{\vec k}^s-c_{\vec k}^{s \dagger}c_{\vec k}^s+(2\pi)^3\delta^{(3)}(0) \Big) \tag{5.12}$$

Where we've used the commutation relation:

$$[c_{\vec k}^r, c_{\vec q}^{s \dagger}] = -(2 \pi)^3 \delta^{rs} \delta^{(3)} (\vec k - \vec q) \tag{5.5}$$

So my question is:

  1. Why is it allowed to simply apply $\vec k \rightarrow -\vec k$ only to the last two terms in $(***)$?
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