# Simple question on computing commutation relation

In bosonization, one faces with the following commutator: $$[\phi(x_1), \theta(x_2)]=\sum_{q\neq 0} \frac{\pi}{Lq} e^{iq(x_2-x_1)-\alpha |q|}\tag{1}$$ where $$q$$ is an non-zero integer multiple of $$2\pi/L$$, $$L$$ is the system size, and $$\alpha>0$$ is an infinitesimal regulator. If one takes the limit $$L \to \infty$$ and then take $$\alpha \to 0$$, then one obtains $$[\phi(x_1), \theta(x_2)] \to i\int_0^\infty \frac{dq}{q} \sin(q(x_2-x_1)) e^{-\alpha |q|} \to i \frac{\pi}{2} {\rm sgn}(x_2-x_1).\tag{2}$$ So far is what the book says. However, I noticed that the series can be evaluated exactly, using $$-\ln (1-x) = \sum_{n=1}^\infty \frac{x^n}{n}.\tag{3}$$ Using this, we compute $$\sum_{q> 0} \frac{\pi}{Lq} e^{iq(x_2-x_1)-\alpha |q|}=-\frac{1}{2} \ln \left( 1-e^{-\frac{2\pi\alpha}{L} +i\frac{2\pi}{L}(x_2-x_1)} \right)\tag{4}$$ and similar expression for summation over $$q<0$$. Then the final exact result is $$[\phi(x_1), \theta(x_2)]= -\frac12 \left[ \ln \left( 1-e^{-\frac{2\pi\alpha}{L} +i\frac{2\pi}{L}(x_2-x_1)}\right) - \ln \left( 1-e^{-\frac{2\pi\alpha}{L} -i\frac{2\pi}{L}(x_2-x_1)} \right)\right].\tag{5}$$ I should recover the expression in the book by taking $$L\to\infty$$ and $$\alpha\to 0$$, but it seems that the limit is different than the book. Where is the problem?

• Which book? Which page? – Qmechanic Oct 8 '19 at 8:05

FWIW, OP's expression (5) does become the RHS of eq. (2) if we first take the limit $$\alpha\to 0$$ and then take the limit $$L\to\infty$$. In contrast, if we take the limits of OP's expression (5) in the opposite order we get zero.