# Quantization of free real scalar massless field in 2d

Is there a reference to literature where one explicitly constructs quantization of the free real scalar massless field in the 2-dimensional space-time? In particular, how the propagator looks like?

The 4d case is treated in many standard textbooks in QFT, but the 2d case seems to be different due to extra divergences which do not exist in higher dimensions.

For example, using the equation $\Box_x \langle T\phi(x)\phi(y)\rangle=\delta^{(2)}(x-y)$ and the analogy with the 4d case (the Feynman rule) one might expect that the propagator should be proportional to $$\frac{1}{p^2+i\varepsilon}=\frac{1}{(p_0)^2-(p_1)^2+i\varepsilon}.$$ However in 2d this generalized function is not well defined, i.e. it diverges as $\varepsilon \to +0$ (even its imaginary part diverges).

ADD: Let's prove the above claim that the imaginary part diverges. We have $$Im\left(\frac{1}{p^2+i\varepsilon}\right)=\frac{1}{2i}\left(\frac{1}{p^2+i\varepsilon}-\frac{1}{p^2-i\varepsilon}\right)=-\frac{\varepsilon}{(p^2)^2+\varepsilon^2}.$$ Let $\phi(p)$ be a smooth non-negative function which equals to 1 in the unit ball and vanishes outside some larger ball. Then \begin{eqnarray*} -\int dp^2Im\left(\frac{1}{p^2+i\varepsilon}\right)\cdot \phi(p)=\int d^2p \frac{\varepsilon}{(p^2)^2+\varepsilon^2}\cdot \phi(p)=\int d^2p \frac{1}{(p^2)^2+1}\cdot \phi(\sqrt{\varepsilon}p), \end{eqnarray*} where the last equality is obtained by the change of variables $p\mapsto \sqrt{\varepsilon}p$. As $\varepsilon\to +0$, the last integral becomes at least

\begin{eqnarray} \int d^2p \frac{1}{(p^2)^2+1}. \end{eqnarray} Let us show that this is infinite. Let's make change of variables $x=p_0-p^1,\, y=p_0+p_1$. Then the last integral is $$\frac{1}{2}\int dxdy\frac{1}{(xy)^2+1}=\frac{1}{2} \int dy\int \frac{dx}{(xy)^2+1}=\frac{1}{2} \int dy \frac{1}{|y|}\left(\int \frac{dz}{z^2+1}\right)=\infty,$$ where the second equality is obtained by the change of variables x=z/|y|$. The result is proven. • Can you be more specific about the divergences that you have in mind? – SRS Mar 30 '16 at 14:01 • @SRS: I added a comment about it. However I think that there is also another way leading to divergences, which is based on computations rather than analogies, but it would require much more space to explain that. – MKO Mar 30 '16 at 14:25 • Pick up any book on string theory or 2d CFTs (but mostly the former) The first or second nontrivial chapter will be about this. – Prahar Mar 31 '16 at 5:42 • Secondly, I don't understand what is the problem with the propagator you wrote down. It's perfectly fine (in momentum space). What is this divergence you are talking about?? – Prahar Mar 31 '16 at 5:43 • @MKO: you've just rediscovered the Dirac delta function... – Adam Mar 31 '16 at 11:36 ## 1 Answer I was able to find an answer to my question in literature. The reference is: A.S. Wightman, "Introduction to some aspects of quantizes fields", in "Lectures notes, Cargese Summer School, 1964". At the bottom of p. 204 Wightman writes ".. there is no such mathematical object as a free scalar field of mass zero in two-dimensional space-time unless one of the usual assumptions is abandoned." Than he shows that one may abandon the assumption of positivity of scalar product in Hilbert space, and he constructs quantization of the free massless scalar field in a "Hilbert" space with indefinite metric. Let me repeat the argument explaining the above quotation. Let$\phi(x)$be such a field. Consider the function$\langle 0|\phi(x)\phi(y)|0\rangle \equiv F(x-y)$. It satisfies$\Box F=0$. Its Fourier trnsform$\tilde F$is a Lorentz invariant distribution satisfying$p^2\tilde F(p)=0$and supported on$p^0\geq 0$. Hence$\tilde F$is supported on the union of two half-lines$\{p_0=p_1, p_0\geq 0\}$and$\{p_0=-p_1, p_0\geq 0\}$. Moreover if the scalar product is positive definite then$\tilde F$is a non-negative measure. However one can show that any non-negative Lorentz invariant measure which is supported on the above two half-lines must be proportional to$\delta^{(2)}(p)$. That means that$F(x-y)=const\$.

Finally let me add that I have found yet another source (which I have not studied in detail) where the authors apparently claim that one can quantize the free massless scalar field in 2d if one does keep the positive definiteness of the scalar product, but abandons the assumption that the vacuum vector has finite norm. See Bogolyubov, Logunov, Oksak, Todorov "General principles of QFT", Section 11.1.