- #1

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$$\int_{-\infty}^{+\infty} \frac{d^3 k}{f(k)} \left[a(\textbf{k})e^{ikx}+a^*(\textbf{k})e^{-ikx}\right]$$

and proceeds to discuss the integration measure ##\frac{d^3 k}{f(k)}##. He chooses ##f(k)## such that the overall measure is Lorentz-invariant:

$$\frac{d^3 k}{f(k)} = d^4 k \delta(k^2 + m^2) \theta(k^0)$$

where ##k^2+m^2=\omega^2##, ##k^2 = k^{\mu}k_{\mu}##, and ##\theta(x)## is the Heaviside step function. If you integrate this measure over ##k^0##, then you get:

$$\int_{-\infty}^{+\infty} dk^0\delta(k^2+m^2)\theta(k^0) = \frac{1}{2\omega}$$

I understand all of the calculation behind this. I also understand that the choice of ##\delta(k^2+m^2)## is to enforce the on-shell condition (I don't quite understand why on-shell is important yet, but I'm sure that'll come later). What I don't understand is the presence of the Heaviside step function. I remember reading somewhere (I know PF hates when people say that, but I can't find the citation), that the reason for the step function is to choose the positive energy solutions (##k^0 > 0## implies that ##E >0##), but my question is this:

*Why are we allowed to arbitrarily throw out the negative energy solutions?*Is it just a choice we make?