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This relates to page 20 of Peskin and Schroeder.

They state that the Fourier transform of the Klein-Gordon field satisfies the following:

$$\left[\frac{\partial^2}{\partial t^2}+(|\vec p|^2+m^2)\right]\phi(\vec p,t)=0 \tag{2.21},$$

which is the equation of motion of a simple harmonic oscillator with frequency:

$$\omega_\vec p=\sqrt{|\vec p|^2+m^2} \tag{2.22}.$$

This is fine, however their next equation is the Hamiltonian for the simple harmonic oscillator:

$$H_{SHO}=\frac{1}{2}p^2+\frac{1}{2}\omega^2\phi^2,$$

which, confusingly to me, does not have a mass $m$ in the denominator of the kinetic term. I have searched around a bit online and not found any reference to this, have I missed something?

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  • $\begingroup$ As you see from the expression you wrote for $\omega_{\vec{p}}$, $\omega$, $p$ and $m$ all have the same dimensions in Peskin units. $\endgroup$
    – secavara
    Jan 5, 2021 at 20:10
  • $\begingroup$ @secavara Ok I understand that, I don't see why it follows that they should omit the mass from the denominator? $\endgroup$
    – Charlie
    Jan 5, 2021 at 20:12
  • $\begingroup$ Nevermind, the next equation $(2.23)$ shows the actual relation between their dimensions. $\endgroup$
    – secavara
    Jan 5, 2021 at 20:14

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Neither does it have a mass in the numerator for the $\phi^2$ term! Peskin & Schroeder just do not bother with a constant $m$ is this context. As you can see, this part introduces you to the ladder operators, in order to apply the formalism to the Klein-Gordon hamiltonian. No need to worry about $m$'s, which are irrelevant to the commutation relations anyway, set it to 1 and work your way through the SHO properties.

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  • $\begingroup$ Ah, so it's just notational, no problem, thanks for your answer. $\endgroup$
    – Charlie
    Jan 5, 2021 at 20:14
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    $\begingroup$ Yes, there is no hidden magic trick other than notational shortcuts! You're welcome. $\endgroup$
    – gildran
    Jan 5, 2021 at 20:15

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