- #1
bananabandana
- 113
- 5
If we have the normal KG scalar field expansion:
$$ \hat{\phi}(x^{\mu}) = \int \frac{d^{3}p}{(2\pi)^{3}\omega(\mathbf{p})} \big( \hat{a}(p)e^{-ip_{\mu}x^{\mu}}+\hat{a}^{\dagger}(p)e^{ip_{\mu}x^{\mu}} \big) $$
With ## \omega(\mathbf{p}) = \sqrt{|\mathbf{p}^{2}|+m^{2}}##
Then why do we associate positive energy states with ##\hat{a}(p)e^{-ip_{\mu}x^{\mu}}## and negative energy states with ##a^{\dagger}(p)e^{ip_{\mu}x^{\mu}}##?
For some reason I thought this was the wrong way round (just because of the sign of exponential, the fact ##p_{0} = \omega(\mathbf{p}) = E_{\mathbf{p}}##, and using metric sign ##(+,-,-,-)##?
$$ \hat{\phi}(x^{\mu}) = \int \frac{d^{3}p}{(2\pi)^{3}\omega(\mathbf{p})} \big( \hat{a}(p)e^{-ip_{\mu}x^{\mu}}+\hat{a}^{\dagger}(p)e^{ip_{\mu}x^{\mu}} \big) $$
With ## \omega(\mathbf{p}) = \sqrt{|\mathbf{p}^{2}|+m^{2}}##
Then why do we associate positive energy states with ##\hat{a}(p)e^{-ip_{\mu}x^{\mu}}## and negative energy states with ##a^{\dagger}(p)e^{ip_{\mu}x^{\mu}}##?
For some reason I thought this was the wrong way round (just because of the sign of exponential, the fact ##p_{0} = \omega(\mathbf{p}) = E_{\mathbf{p}}##, and using metric sign ##(+,-,-,-)##?