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There is a analog between harmonic oscillator $x=\frac{1}{\sqrt{2\omega}}(a+a^\dagger)$ and quantum field $\phi=\int dp^3\frac{1}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p e^{ipx}+a^\dagger e^{-ipx})$, which is used to quantize the field operator.

However, one thing confuse me is about the coefficient $\frac{1}{\sqrt{2\omega}}$. For field operator, this comes from Lorentz invariance, just because we have integrated time t. However, for harmonic oscillator, there seems no apparent Lorentz symmetry give me this. Is there any hidden symmetry behind the harmonic oscillator?

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2 Answers

For the standard harmonic oscillator ($m \equiv 1$) with

$$ \tag{1} H = \frac{1}{2}\left(p^2 + \omega^2 x^2 \right)$$

you need this factor to get it into the ‘nicer’ form using the ladder operators:

$$ \tag{2} H = \omega \left( a^\dagger a + \frac{1}{2} \right) \quad. $$

That is, if you substitute

$$ x = \frac{1}{\sqrt{2\omega}} \left( a + a^\dagger \right) \quad ; \quad p = i \sqrt{\frac{\omega}{2}} \left( a - a^\dagger \right) $$

into the original equation (1), you get

$$ H = \frac{\omega}{2} \left( a^\dagger a + a a^\dagger \right) $$

which, after normal ordering by making use of

$$ [ a , a^\dagger ] = 1 \Leftrightarrow [ x , p ] = i \hbar $$

is equivalent to the second equation (2).

To answer your question: The factor of $\frac{1}{\sqrt{2\omega}}$ arises not from Lorentz invariance but from the frequency $\omega$ of the harmonic oscillator, which is related to the energy between two excited states.

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I am truly confused what the question is actually asking and in what sense this answer is trying to answer it. The right dependence of the frequency of the QFT oscillators surely is dictated by the Lorentz invariance because it must be the frequency associated with the energy of one excitation in a given mode and the energy for a given mode is calculated via $E=\sqrt{p^2+m_0^2}$ from Lorentz inv. A simple single harmonic oscillator has no Lorentz symmetry but $\omega$ is its free parameter and this simple "building" block appears in QFT and elsewhere, with one value of $\omega$ or another. –  Luboš Motl Dec 9 '12 at 6:34
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What's the problem or question here? It's hard to see. The Solar System has an elliptical orbit of the Earth with periodicity 1 year which is related to seasons, calendars, and other things. Kepler's problem predicts elliptical orbits with some parameters but it knows nothing about the Mayan or Jewish calendar. Is that a problem? Why should it? In some contexts, parameters may be determined by some input/assumptions, in others by other inputs, in more general ones, they can't be determined at all. Different situations have different assumptions which lead to different detailed consequences. –  Luboš Motl Dec 9 '12 at 6:35
    
@LubošMotl I was trying to show that the factor of $\frac{1}{\sqrt{2\omega}}$ in $x$ is not a result of implied Lorentz invariance but a result of us wanting to diagonalise the original HO Hamiltonian. You are right that it is, in this sense, a free parametre of the original Hamiltonian. –  Claudius Dec 9 '12 at 11:04
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Dear Claudius, the physically invariant, correct statement you made is that $\hbar\omega$ is the spacing between adjacent levels. But what you call $x,p,H$ depends on conventions. For example, you could rescale $x,p$ by factors $K,L$ and change the coefficients of $x^2$ and $p^2$ in the Hamiltonian $H$ correspondingly so that the physics (spectrum of $H$ etc.) doesn't change at all. –  Luboš Motl Dec 10 '12 at 8:21
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The normalization in front of $a$ and $a^\dagger$ is totally arbitrary until you specify what their commutation relation is. Fixing what multiple of identity $[a,a^\dagger]$ you like fixes the proper normalization to be used (i.e. in order to agree with the canonical commutation relation $[q,p]=i$ among conjugate variable). This is true in QM as much as in QFT (that is in fact a QM theory). The factor $1/\sqrt{2\omega(p)}$ is not coming from Lorentz (Lorentz says only that $\omega(p)=\sqrt{p^2+m^2}$) but rather from your choice of normalization of the states such as $a^\dagger(p)|0\rangle$. Some people like for instance to impose a Lorentz preserving normalization, some don't. It's a free choice with no physical content. You could have an arbitrary function in front of $a$ and $a^\dagger$ but still end up with the same Hamiltonian just by picking the proper $[a,a^\dagger]$ normalization.

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