# How to get to negative-momentum solutions from negative-energy solutions of the Klein-Gordon equation?

I'm trying to derive the general solution to the (complex, classical) Klein-Gordon equation:

$$\phi = \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{k}}} \left(B(k)e^{ikx} + C(k)e^{-ikx}\right).$$

I started with a four-integral over the ansatz solution:

$$\int d^4k\ A(k^{\mu})e^{-ik_{\mu}x^{\mu}},$$

and then applied the on-shell constraint; i.e., multiplying by

$$\delta(\omega^2 - \omega_{k}^2) = \frac{1}{2\omega_{k}}\left[\delta(\omega+\omega_{k})+\delta(\omega-\omega_{k})\right],$$

to get:

$$\phi = \int \frac{d^3k}{2\omega_{k}} A(k) \left(e^{-i(\omega_{k}t - kx)} + e^{i(\omega_{k}t + kx)}\right).$$

But I have absolutely no idea where to go from here. I presume that $A(k^{\mu})$ becomes $B(k)$ and $C(k)$ somehow during the process of enforcing the on-shell constraint and I just missed it, but there's also the problem of what to do with the negative energy solutions. Also there's the $2\omega_{k}$ vs $(2\pi)^3 \sqrt{2\omega_{k}}$ thing, but I'm not sure if that's just a choice in normalisation.

Okay, I've just been a bit daft. Clearly 2am wasn't the best time to be doing this.

Following application of the on-shell condition you get:

$$\phi = \int \frac{d^3k}{2\omega_{k}} \left(B(k)e^{-i(\omega_{k}t - kx)} + B(k)e^{i(\omega_{k}t + kx)}\right),$$

where $B(k)$ is just what you get when you integrate the $A(k^{\mu})$ over $\omega$.

Then in the $\omega_{k}t + kx$ part of the integral you make the variable change:

$$k \to k' = -k,$$

which doesn't affect $\omega_{k}$ (i.e., $\omega_{k} = \omega_{k'}$) since $\omega_{k}$ is defined from the square of $k$. It does affect $d^3k$:

$$d^3k \to d^3k' = -d^3k$$

but this is just absorbed into the new coefficient, $C(k')$. Then since $k'$ is just a dummy variable you can rename it to be $k$ again, and you get:

$$\phi = \int \frac{d^3k}{2\omega_{k}} \left(B(k)e^{-i(\omega_{k}t - kx)} + C(k)e^{i(\omega_{k}t - kx)}\right) \Box$$

And then if the field is real you impose the $\phi = \phi^{*}$ condition which tells you $C(k) = B^{*}(k)$.