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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?

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    $\begingroup$ Do you remember where the $w_{\vec{k}}$ came from? $\endgroup$
    – jacob1729
    Jan 27 '20 at 11:25
  • $\begingroup$ @jacob1729 $w_\vec{k}$ is the component $k^0$ which is the energy in natural units $\endgroup$
    – TaeNyFan
    Jan 27 '20 at 11:32
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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 $\omega_{\vec{k}}^{-1/2}$).

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  • $\begingroup$ 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? $\endgroup$
    – TaeNyFan
    Jan 27 '20 at 11:36
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    $\begingroup$ 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. $\endgroup$
    – user245141
    Jan 27 '20 at 11:46

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