Let me be sort of specific. We consider a Weyl particles on $S^{1}$, with the following Hamiltonian $$\mathcal{H}=v\int_{0}^{2\pi}\psi^{\dagger}(x)(-i\partial_{x})\psi(x)dx$$ Such particles have the property that $$\rho(x)=\psi^{\dagger}(x)\psi(x)$$ is ill defined, in the sense that the expectation value of $\rho(x)$ diverges. For this reson we change the definition of normal ordering to $$\rho(x)=:\psi^{\dagger}(x)\psi(x):=\psi^{\dagger}(x)\psi(x)-\bar{\rho}(x)$$ Where the over bar is the expectation value. With this symbolism, we have that $$[\rho(x), \rho(x')]=\frac{i}{2\pi}\delta'(x-x')$$ Istead of usual $[\rho(x), \rho(x')]=0$. We also know that for chiral bosons one has $$[\varphi(x), \varphi(x')]=-i\pi~\mbox{sign(x-x')}$$ So by differentiating the bosonic commutator twice with respect to $x$ and then with respect to $x'$, we get precisely the commutator of densities, provided that $$\rho(x)=\frac{1}{2\pi}\partial_{x}\varphi(x)$$ So far so good. Than one sees that $$[\rho(x), \psi(x')]=-\psi(x)\delta(x-x')$$ In order to satisfy this commutation relation we have to make a clever guess. That is, if one asumes $\psi(x)=e^{i\varphi(x)}$, it is easy to show that indeed the algebra is satisfied. I'm pretty much satisfied with everithing by this point, it seems that we've found a canonical transformation and bla-bla-bla. What normally is then claimed in the literature that $$\mathcal{H}=v\int_{0}^{2\pi}\psi^{\dagger}(x)(-i\partial_{x})\psi(x)dx=\frac{v}{4\pi}\int_{0}^{2\pi}(\partial_{x}\varphi(x))^{2}dx$$ And this is a place where I get stuck. So $$\psi^{\dagger}(x)(-i\partial_{x})\psi(x)=e^{-i\varphi(x)}(-i\partial_{x})e^{i\varphi(x)}=e^{-i\varphi(x)}(\partial_{x}\varphi(x))e^{i\varphi(x)}=e^{-i[\varphi(x), \cdot]}\partial_{x}\varphi(x)=?$$ I.e. what is the commutator $[\partial_{x}\varphi(x), \varphi(x)]$? Does it even make sense? In this expression may also write $$(\partial_{x}\varphi(x))e^{i\varphi(x)}=2\pi\rho(x)\psi(x)=2\pi\rho(x)\psi(x')\big{|}_{x'=x}=2\pi\psi(x')\rho(x)\Big{|}_{x'=x}-2\pi\psi(x)\delta(x-x')\Big{|}_{x'=x}=2\pi\psi(x)\rho(x)+?=e^{i\varphi(x)}\partial_{x}\varphi(x)+?$$ What am I doing wrong?

  • $\begingroup$ And, if one wonders about the divergence of the expectation value of the density, it is singular in the sense $\bar{\rho(x)}=<\psi^{\dagger}(x)\psi(x)>=\lim_{\delta\rightarrow{0}}<\psi^{\dagger}(x)\psi(x+\delta)>=\lim_{\delta\rightarrow{0}}\Big(\frac{1}{2\pi{i}\delta}+\mathcal{O}(1)\Big)$. So the rigurous redefinition of normal ordering is $:\psi^{\dagger}(x)\psi(x):=\lim_{\delta\rightarrow0}(\psi^{\dagger}(x)\psi(x+\delta)-\frac{1}{2\pi{i}\delta})$ $\endgroup$ Sep 11, 2017 at 9:19
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    $\begingroup$ You can have a look at chapter 7 of this nice paper: arxiv.org/abs/cond-mat/9805275 I'm also quiet sure that the book of Gianmarchi on Bosonization will cover this point also in depth... $\endgroup$
    – tired
    Sep 11, 2017 at 9:33

1 Answer 1


I am basically following the first chapter of Bosonization by Michael Stone.

Starting from the bosonization formula: $$\psi(x) = :e^{i\phi(x)}:$$ And the Hamiltonian $$H = i \int dx \psi^{\dagger}_R(x) \partial_x \psi_R(x) $$

In the bosonization of the composite terms, we need to point split and take care of the normal ordering according to:

$$ :e^{ia\phi(x_1)}::e^{ib\phi(x_2)}: =:e^{ia\phi(x_1)+ib\phi(x_2)}:e^{ab\log{(x_1-x_2)}} $$

Thus we have: $$\begin{align*} \psi^{\dagger}_R(x) \partial_x \psi_R(x) & = \lim_{x' \to x} :e^{-i \phi(x')}: \partial_x :e^{i \phi(x)}:\\ &= \lim_{x' \to x} :e^{-i \phi(x')}: :e^{i \phi(x)}:i \partial_x \phi(x)\\ &=\lim_{x' \to x} (x'-x)^{-1}:e^{-i \phi(x') + i \phi(x)}: i \partial_x \phi(x)\\ &= \lim_{x' \to x} (x'-x)^{-1} \Big(1 - i (x'-x) \partial_x \phi(x) + O((x'-x)^2)\Big)i \partial_x \phi(x)\\ & = \lim_{x' \to x}\frac{i}{x-x_1}+ \partial_x \phi(x) \partial_x \phi(x) + O(x'-x) \end{align*} $$ The singular term is taken care of by the normal ordering of the right hand side: Thus we get: $$H = i \int dx (\psi^{\dagger}_R(x) \partial_x \psi_R(x) = i \int dx :( \partial_x \phi(x))^2 :$$

Update: Proof of the normal ordering identity:

$$:e^{ia\phi(x_1)}::e^{ib\phi(x_2)}: = e^{ia\phi_{-}(x_1)+ia \phi_{+}(x_1)} e^{ib\phi_{-}(x_2)+ib \phi_{+}(x_2)}$$

where $\phi_{+}$ contains only creation operators and $\phi_{-}$ annihilation operators. In order to normal order the expression we need to commute between the second and third terms.

Using the Campbell–Baker–Hausdorff (CBH)

$$e^{A} e^{B} = e^{A+B} e^{[A, B]/2}$$

(for $[A, B] = const.$)

The relevant modes from the mode expansion

$$\phi_{-}(x)= \sum_{n>0} \sqrt{\frac{2}{n}} a_n e^{-\frac{nx}{L}}$$ $$\phi_{+}(x)= \sum_{n>0} \sqrt{\frac{2}{n}} a_n^{\dagger}e^{\frac{nx}{L}}$$

$L$ is the quantization box length which will be taken eventually to infinity

Using $$[a_n, a_m^{\dagger}] = \delta_{mn}$$ We get: $$[\phi_{-}(x_1), \phi_{+}(x_2)] = 2\sum_{n>0}\frac{1}{n}e^{-\frac{ n(x_1-x_2)} {L} } = 2 \log(1-e^{-\frac{ (x_1-x_2)} {L} })\rightarrow_{L\to \infty} 2\log(x_1-x_2) + \mathrm{const.}$$

  • $\begingroup$ That makes a lot of sense right now, thank you very much for your answer and your reference!!! $\endgroup$ Sep 15, 2017 at 7:08
  • $\begingroup$ Can you please give a proof of the statement $:e^{ia\varphi(x)}::e^{ib\varphi(x')}:=:e^{ia\varphi(x)+ib\varphi(x')}:e^{iab\ln(x-x')}$? $\endgroup$ Sep 18, 2017 at 11:30
  • $\begingroup$ @Kiryl Pesotski I have added an explanation $\endgroup$ Sep 18, 2017 at 13:34
  • $\begingroup$ Could you please elaborate on the last line of your derivation? Is $x_1 \equiv x'$? Why does $(x'-x)^{-1} \partial_x \phi = 1/(x-x')$? $\endgroup$
    – Faser
    Oct 12, 2021 at 18:14

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