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:

  1. 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?
  2. 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.
  • 1
    $\begingroup$ But there is no mass appearing in the free Hamiltonian, right? I am looking directly at the Lagrangian density in eq. (1.1), there is only a kinematic term plus interaction (the cosine). If there is a mass, it is due to the quadratic term in the cosine. Could this be the reason that the field has not mass $m$ but also is not really massless? $\endgroup$
    – Cream
    Commented Feb 18, 2021 at 20:27
  • $\begingroup$ As far as I know, there is no mass. However, if you consider the configuration $\varphi\approx 0$ and expand the cosine term to fourth order, then it is possible to interpret $\alpha_0=m^2$. $\endgroup$ Commented Feb 18, 2021 at 21:59

1 Answer 1


I think I have an idea whats going on now:

As I see it, the problem with the mass is that Coleman is considering the full, interacting theory. The mass in this theory would be changed from the "bare" mass $\alpha_0$ by renormalization. Therefore, he defines a general field $\varphi_m$ with time-ordering $N_m$ at first for arbitrary $m$.
So the answer would be: There is mass, but we do not know it yet. Therefore, it remains unspecified for now.

The computation of the normal ordering is straight-forward (you were probably missing the correct normalization of the commutator $[a,a^\dagger]$): \begin{align} &\mathcal{H_0} = \frac{\pi^2}{2} + \frac{(\partial_x \varphi)^2}{2} ~,~~ \mathcal{H_0} = N_m(\mathcal{H_0}) + E_0(m) \\ \Rightarrow& E_0(m) = \frac{1}{2} \int \frac{dk dk'}{4\pi^2} \frac{\sqrt{\omega_{k,m}\omega_{k',m}}}{2} e^{i(k-k')x} \underbrace{[a_{k',m}, a^\dagger_{k,m}]}_{=2\pi\delta(k-k')} + \frac{1}{2} \int \frac{dk dk'}{(2\pi)^2} \frac{k^2}{2\sqrt{\omega_{k,m}\omega_{k',m}}} e^{i(k-k')x} \underbrace{[a_{k',m}, a^\dagger_{k,m}]}_{=2\pi\delta(k-k')} \\ =& \frac{2\pi}{2\times 8\pi^2} \int dk \Big( \omega_{k,m} + \frac{k^2}{\omega_{k,m}} \Big) = \int \frac{dk}{8\pi} \frac{2k^2 + m^2}{\omega_{k,m}}. \end{align}

Does that clarify things?


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