# Negative and Positive energy modes of KG equation

## Main Question or Discussion Point

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 $(+,-,-,-)$?

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king vitamin
Gold Member
The operators $\hat{a}$ and $\hat{a}^{\dagger}$ satisfy
$$[\hat{a}(p),\hat{a}^{\dagger}(p')] = (2 \pi)^3 \delta^3(p - p')$$
and
$$\hat{a}(p)|0\rangle = 0$$
The physical states of Klein-Gordon theory all have positive* energy, and are given by (up to normalization)
$$\hat{a}^{\dagger}(p_1)\hat{a}^{\dagger}(p_2) \cdots \hat{a}^{\dagger}(p_n) |0 \rangle$$
and energy
$$E = \omega(p_1) + \omega(p_2) + \cdots + \omega(p_n)$$

It's a little imprecise to associate energy to the operators $\hat{a}$ and $\hat{a}^{\dagger}$ themselves. But the operator $\hat{a}^{\dagger}(p)$ raises the energy by $\omega(p)$, and the operator $\hat{a}(p)$ either annihilates the state or takes you to a state with energy lower by $\omega(p)$, so I suppose you can heuristically think of them "carrying energy" $\omega(p)$ and $-\omega(p)$ respectively. But when you want to be precise, you should go back to the above statements.

* To ignore issues regarding the ground state energy, I'll take the ground state energy to be zero, aka taking the normal-ordered Klein-Gordon Hamiltonian.

• vanhees71