Fermion creation operator in boson basis

I've been reading Giamarchi, Quantum Physics in One Dimension, Chapter 2 on 1d bosonization, and in appendix B.1, he derives equation B.2, which represents the fermion creation operator $$\psi_r (x)$$ with bosonic operators and the ladder operator $$U_r$$, but I don't understand how he did this, especially where the $$\epsilon$$ came from. \begin{align} \psi_r^{\dagger}(x) &= \frac{1}{\sqrt{L}} e^{-irk_Fx}e^{i\phi_r^\dagger(x)}U_r^\dagger e^{i\phi_r(x)}\\ &=\lim_{\epsilon\to0}\frac1{\sqrt{2\epsilon L}}e^{-i r(k_F-\pi/L)x}e^{i\phi_r^\dagger(x)+i\phi_r(x)}U_r^\dagger \tag{B.9} \end{align}

If anyone has gone over the derivation, it would be appreciated if you could elaborate.

EDIT: I found this paper by Jan von Delft, which explains this pretty well, especially pages 16-17 and the appendix.

• I added MathJax, please use it in the future for typesetting math. See guide and links therein. Jun 28 '19 at 5:30
• This was first derived by just noticing that combinations of these operators had the correct commutation relations to represent the creation (and with a hermitian conjugate, destruction) of fermion modes. Rigorous derivations were first given in F. D. M. Haldane, J. Phys. C 14 2585 (1981) and R. Heidenreich, R. Seiler, D. A. Uhlenbrock, J. Stat. Phys. 22, 27 (1980).
– Buzz
Jul 2 '19 at 18:16
• @Buzz I've read Haldane before, but he only wrote down the equation in (3.37) and claiming it can be trivially obtained, which I'm not quite sure why. I also don't see where the $\epsilon$ comes from. I haven't read the other paper though. Jul 4 '19 at 0:16