# Continuous plane wave solutions to Klein Gordon Field Equation

The continuous plane wave solution to the Klein Gordon Field Equation can be written as

$$\phi(x) = \int\frac{d^3\vec{k}}{\sqrt{2(2\pi)^3w_\vec{k}}} a(\vec k) e^{-ikx} + \int\frac{d^3\vec{k}}{\sqrt{2(2\pi)^3w_\vec{k}}} b(\vec k) e^{ikx}$$

where $$x$$ and $$k$$ are 4-vectors.

The LHS is a function of 4-vector $$x$$, i.e. it is a function of the 4 components $$x^0,x^1,x^2$$ and $$x^3$$.

Similarly, the RHS is a function of 8 components $$k^0,k^1,k^2,k^3$$ and $$x^0,x^1,x^2,x^3$$, but integrated over $$k^1,k^2,k^3$$, meaning it reduces to a function of 5 components ($$k^0$$, $$x^0,x^1,x^2,x^3$$).

So apparently the LHS and RHS don't agree with each other. One side is a function of 4 components and the other is a function of 5 components.

What am I seeing wrong here?

• Do you remember where the $w_{\vec{k}}$ came from? Jan 27 '20 at 11:25
• @jacob1729 $w_\vec{k}$ is the component $k^0$ which is the energy in natural units Jan 27 '20 at 11:32

These solutions are on-shell, meaning that they maintain $${\bf{k}}^2=m^2$$ which gives $$k_0$$ as function of the 3-vector $$k$$. This is why we need to integrate only over $$k_1,k_2,k_3$$ (and also the reason for the measure of $$\omega_{\vec{k}}^{-1/2}$$).
• does that mean by integrating over $k^1,k^2,k^3$, it is implicitly implied that $k^0$ is also integrated and eliminated from the equation? Jan 27 '20 at 11:36
• Yes. We start with $\int\! d^4k \delta({\bf{k}}^2-m^2)$ and then using the delta function we integrate over $k_0$, getting the factor of $1/\sqrt{\omega_{\vec{k}}}$ in the integration measure.