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Glenn Rowe

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*Quantum Field Theory for the Gifted Amateur*and have got to Chapter 17 on calculating propagataors. In their equation 17.23 they derive the expression for the free Feynman propagator for a scalar field to be $$\Delta\left(x,y\right)=\int\frac{d^{4}p}{\left(2\pi\right)^{4}}e^{-ip\cdot\left(x-y\right)}\frac{i}{\left(p^{0}\right)^{2}-E_{\boldsymbol{p}}^{2}+i\epsilon}$$ where ##p^0=E## represents an energy that is not on the mass shell, so that in general ##p^{0}\ne\sqrt{E_{\boldsymbol{p}}^{2}+m^{2}}##. I'm able to follow their derivation (I think), but then in Exercise 17.4, they ask us to show that the Feynman propagator for the quantum simple harmonic oscillator with spring constant ##m\omega_{0}^{2}## is given by $$\tilde{G}\left(\omega\right)=\frac{i}{m\left(\omega^{2}-\omega_{0}^{2}+i\epsilon\right)}$$

It seems to me that the energy of the harmonic oscillator in its "one-particle" state is ##\omega_0##, and the general energy (off the mass shell) is given by ##\omega## so that the position-space propagator would be given by $$G\left(x,y\right)=\int\frac{d^{4}p}{\left(2\pi\right)^{4}}e^{-ip\cdot\left(x-y\right)}\frac{i}{\omega^{2}-\omega_{0}^{2}+i\epsilon}$$

From there, we can read off the momentum-space Fourier component as $$\tilde{G}\left(\omega\right)=\frac{i}{\left(\omega^{2}-\omega_{0}^{2}+i\epsilon\right)}$$

I can't figure out where the extra factor of ##m## in the denominator comes from. Introducing the extra ##m## seems to mess up the units as well, since their general expression for the momentum-space propagator is $$\tilde{\Delta}\left(p\right)=\frac{i}{\left(p^{0}\right)^{2}-E_{\boldsymbol{p}}^{2}+i\epsilon}$$. I'm guessing I'm missing something simple (since pretty well all the exercises in the book aren't too complex once you understand the principles), but I just can't see it.