Hamiltonian of the Dirac Field

Question

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 $(***)$?

Answer 1

The point is that for the mode expansion you have positive and negative frequency modes, which manifest as the signs in the exponential. When using the integral (A), you will get $\delta^{(3)}(\vec{p} + \vec{q})$, which gives after further integral $\vec{p} = - \vec{q}$. So calling it "relabeling" might be inaccurate, IMO. The relabeling, i think, is meant to be $\vec{k} \rightarrow - \vec{k}$ before using the equation (A).