Negative and Positive energy modes of KG equation

  • #1

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 ##(+,-,-,-)##?
 

Answers and Replies

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

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.
 

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