I am new in PSE.
I am studying the $1+1$ sine-Gordon theory from Sidney Coleman's article of 1975 (pdf). The Hamiltonian is \begin{align} H=\int dx\;\left(\mathcal{H}_0-\frac{\alpha_0^2}{\beta}\cos\left(\beta\varphi\right)-\gamma_0\right),\quad \mathcal{H}_0=\frac{1}{2}\pi^2+\frac{1}{2}\varphi'^2 \end{align} where $x$ is a spatial coordinate, primes are derivatives with respect to $x$, $\alpha_0$ and $\beta$ are real positive numbers and $\gamma_0$ is a factor to adjust the minimum energy to zero. The quantization of the theory is analogous to the Klein-Gordon field theory, that is, we can write (in the Schrödinger picture) \begin{align} \varphi(x)&=\int \frac{dk}{2\pi}\;\frac{1}{\sqrt{2\omega_{k,m}}}\left(a_{k,m} e^{-ikx}+a^\dagger_{k,m} e^{ikx}\right),\\ \pi(x)&=i\int \frac{dk}{2\pi}\;\sqrt{\frac{\omega_{k,m}}{2}}\left(a_{k,m} e^{-ikx}-a^\dagger_{k,m} e^{ikx}\right), \end{align} with \begin{align} \omega_{k,m}=\sqrt{k^2+m^2}. \end{align} Here are my questions:
- Sidney says: "The normal ordered prescription, rearringing all the $a$'s to the right and all the $a^\dagger$'s on the left is ambiguous, because does not tell us what $m$ is. Of course, if we are really doing interaction-picture perturbation theory, it would be senseless to choose $m$ to be other than that mass which occurs in the free Hamiltonian. However, for the sG equation, it is not clear what is the most profitable way to divide the Hamiltonian into a free and an interaction part, or, indeed, whether any such division is profitable. Therefore, for the moment at least, we will not specify $m$, and we will denote by $N_m$ the normal-ordering operation defined by the mass $m$". He will not specify $m$, but anyway he is defining a normal-ordering with mass $m$? There is mass or there is not?
- In equation (2.17) of his article, he replaces in $N_m\left(H_0\right)$ the field expansions $\varphi$ and $\pi$, obtaining a vacuum energy \begin{align} E_0(m)=\int \frac{dk}{8\pi}\;\frac{2k^2+m^2}{\omega_{k,m}}. \end{align} How on earth I can obtain this expression? I computed it five times "just replacing the field expansion" in the free Hamiltonian $H_0$ but always I obtain a similar (but no the same) expression. Maybe this helps to answer, but I believe that my two questions reduces to answer "what is $N_m(\mathcal{O})?$", where $\mathcal{O}$ is an arbitrary operator.