# A question about verifying the transverse of electric field

I came accross a question about verifying the transverse of electric field in Peskin and Schroeder's QFT p179.

Given

$$\mathcal{A}^{\mu}(\mathbf{k}) = \frac{ -e}{| \mathbf{k} | } \left( \frac{p'^{\mu}}{k \cdot p'} - \frac{p^{\mu}}{k \cdot p} \right) \tag{6.7}$$

and

$$\mathcal{E} = -i \mathbf{k} \mathcal{A}^0 (\mathbf{k}) + i k^0 \mathcal{A}(\mathbf{k}) \tag{6.9}$$

it is said

you can easily check that the electric field is transverse: $\mathbf{k} \cdot \mathcal{E}(\mathbf{k}) =0$

I tried to vertify this almost trivial relation and I didn't get it... Here is my attempt $$\mathbf{k} \cdot \mathcal{E}(\mathbf{k}) = \left( \frac{ -e}{| \mathbf{k} | } \right) \left[ - i \mathbf{k} \frac{ \mathbf{k} p'^0 - k^0 \mathbf{p}'}{ k \cdot p'} + i \mathbf{k} \frac{ \mathbf{k} p^0 - k^0 \mathbf{p}}{ k \cdot p} \right] \tag{1}$$

I may use $p^{\mu}=(p^0, \mathbf{0})$ (since the particle is initially at rest) and $k^0=|\mathbf{k}|$. It seems that Eq. (1) is still not necessarily zero. Would you provide, at least, any hint how to show Eq. (1) is zero?

Hint: You should find something like: $$\propto \frac{k_\mu p'^\mu}{k_\mu p'^\mu} - \frac{k_\mu p^\mu}{k_\mu p^\mu} = 0$$
Spoiler: I omit the $-e/|\vec k|$ factor for brevity. We have $$A^\mu = \frac{p'^{\mu}}{k^\mu p'_\mu} - \frac{p^{\mu}}{k^\mu p_\mu}$$ and $$k^0=|\vec k|.$$ We can begin by rewritting $$\vec \epsilon = -i \vec k A^0 + i k^0 \vec A \\ =-i\vec k \left(\frac{p'^0}{k^\mu p'_\mu} - \frac{p^0}{k^\mu p_\mu}\right) +ik^0 \left(\frac{\vec p'}{k^\mu p'_\mu} - \frac{\vec p}{k^\mu p_\mu}\right)$$ Now taking the dot product, $$\vec k \cdot \vec \epsilon\\ =-i|\vec k|^2 \left(\frac{p'^0}{k^\mu p'_\mu} - \frac{p^0}{k^\mu p_\mu}\right) +ik^0 \left(\frac{\vec k \cdot\vec p'}{k^\mu p'_\mu} - \frac{\vec k \cdot\vec p}{k^\mu p_\mu}\right)$$ Now utilise $k^0=|\vec k|$; $$=-i k_0 \left(\frac{k_0 p'^0}{k^\mu p'_\mu} - \frac{k_0 p^0}{k^\mu p_\mu}\right) +ik^0 \left(\frac{\vec k \cdot\vec p'}{k^\mu p'_\mu} - \frac{\vec k \cdot\vec p}{k^\mu p_\mu}\right)\\ \propto \frac{k_\mu p'^\mu}{k_\mu p'^\mu} - \frac{k_\mu p^\mu}{k_\mu p^\mu} = 0$$