# Peskin and Schroeder, where is the mass in the denominator of the simple harmonic oscillator Hamiltonian?

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?

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

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.