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I Negative and Positive energy modes of KG equation

  1. Dec 18, 2017 #1
    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 ##(+,-,-,-)##?
     
  2. jcsd
  3. Dec 19, 2017 #2

    king vitamin

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    Gold Member

    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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