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The Hamiltonian of the Klein-Gordon Field may be written $$H=\int\frac{d^{3}p}{(2\pi)^{3}}\frac{1}{2\omega_{\mathbf{p}}}\omega_{\mathbf{p}}\left(a^{\dagger}(p)a(p)+\frac{1}{2}(2\pi)^{3}2\omega_{\mathbf{p}}\delta^{(3)}(0) \right)$$ where $a(p)$ and $a^{\dagger}(p)$ are the usual creation and annihilation operators and $\omega_{\mathbf{p}}=\sqrt{|\mathbf{p}|^{2}+m^{2}}$ is the oscillator frequency.

Act on a singly excited state $|q\rangle=a^{\dagger}(q)|0\rangle$ with this Hamiltonian. We find $$H|q\rangle=\left(\omega_{\mathbf{q}}+E_{0}\right)|q\rangle$$ where the (infinite) ground state energy $$E_{0}=V\int\frac{d^{3}p}{(2\pi)^{3}}\frac{\omega_{\mathbf{p}}}{2}$$ and the volume has been included using $$(2\pi)^{3}\delta^{(3)}(0)=\lim_{L \to \infty}\int_{-\frac{L}{2}}^{\frac{L}{2}}d^{3}xe^{i\mathbf{p}\cdot\mathbf{x}}|_{\mathbf{p}=0}=\lim_{L \to \infty}\int_{-\frac{L}{2}}^{\frac{L}{2}}d^{3}x=V$$

I am a bit confused because:

  1. In the ground state energy, the oscillator zero point energy seems to be contributing to an energy density (hence the volume factor).

  2. In the excitation energy, the oscillator excitation energy is not being treated as an energy density.

This appears to be a contradiction, since the oscillators should either contribute to the total energy or the energy density of the field, not both - can anybody explain this? Thanks :)

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  • $\begingroup$ You should consider the quantity $E_0/V$ i.e., the energy density which is finite. $\endgroup$ – SRS Apr 6 '17 at 11:04
  • $\begingroup$ @SRS Edited question. $\endgroup$ – klgklm Apr 6 '17 at 12:15
  • $\begingroup$ This is an interesting question. Electromagnetic zero point energy gives the universe a ludicrous total vacuum energy of $10^{123}$ Joules. The only difference here with the EM field is the energy dispersion relation. The divergency problem, only limited by the size of the universe, is the same. See also en.wikipedia.org/wiki/… $\endgroup$ – my2cts Oct 6 '18 at 11:42
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I'm not sure I understand exactly what you are asking. One intuitive way to see the calculation of the energy is $$E (\textrm{state } | q \rangle) = \sum\limits_{\textrm{Oscillators } \mathbf{p}} (n_{\mathbf{p}}+1/2) \omega_{\mathbf{p}} = \sum\limits_{\textrm{Oscillators } \mathbf{p} } n_{\mathbf{p}}\omega_{\mathbf{p}} + \sum\limits_{\textrm{Oscillators } \mathbf{p} } \frac{1}{2}\omega_{\mathbf{p}} =\omega_{\mathbf{q}} + E_0 $$ where we use that the excitation number $ n_{\mathbf{p}}$ is always zero, except when $\mathbf{p} = \mathbf{q}$. Does it answer the question?

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  • $\begingroup$ My integral form for $E_{0}$ has an extra factor of $V$, whereas yours doesn't... How does this arise? $\endgroup$ – klgklm Apr 6 '17 at 19:29
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"In the ground state energy, the oscillator zero point energy seems to be contributing to an energy density (hence the volume factor)"

It is not. For a single oscillator the zero point energy is finite and hence there is no problem with it. But for klien-gordon field, which is a collection of infinite number of oscillators, have infinite zero point energy. Therefore To deal with it we consider energy density instead which is $E_0 /V$. This also answered your (2) confusion.

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