In massless 2d scalar field theory, the expression $\langle \phi(x)\phi(y) \rangle = -\ln|x-y|$ is best regarded as a mnemonic. (A first hint of trouble: $|x-y|$ has dimensions of length, so you can't just stick it into a logarithm.) Likewise, the expression $e^{-S}$ for the measure. The Euclidean measure is usually supported on functions for which $S$ isn't defined.

The idea behind this shorthand is that any Euclidean correlation function in the massless 2d scalar model can be written as a limit of a correlation function in the massive model. The massive 2d theory is Gaussian and completely characterized by the 2-point function $\langle \phi(x)\phi(y)\rangle$, which blows up like $-\ln(m|x-y|)$ as $m \to 0$. This means that some correlation functions will not remain finite in this limit, which reduces the size of the Hilbert space and the collection of field operators.

Concretely, this works as follows..

Recall the Wightman/GNS construction of the Hilbert space and field operators: The collection of Euclidean correlation functions specifies a linear functional $E$ on the algebra $\mathcal{A}_+$ of Euclidean observables which are supported at positive time. Reflection about $t=0$ gives a map $\theta: \mathcal{A}_+ \to \mathcal{A}_-$, which defines a (degenerate) sesquilinear form $s: \mathcal{A}_+ \times \mathcal{A}_+ \to [0,\infty]$. Call an element $a$ of $\mathcal{A}_+$ $L^2$-finite if $E(\theta(a)a) < \infty$, by analogy with the case where $E$ is the expectation value of a normed functional measure. If $s$ is reflection positive, we get an inner product space by restricting attention to the space $\mathcal{V}$ of $L^2$-finite vectors and quotienting out by the null ideal of $s$. Completing gives a Hilbert space $\mathcal{H}$. The elements of $\mathcal{V}$ act on a space of smooth vectors within $\mathcal{H}$ via right multiplication.

When we take the massless limit, we reduce the size of $\mathcal{V}$, as the evaluation observables $ev_x(\phi) = \phi(x)$ cease to $L^2$-finite. (Note that $E(\theta(ev_x) ev_y) = \langle \phi(x)\phi(y)$.) This is why people say $\phi$ isn't a quantum field in the massless case. However, many local observables do survive this limit, such as the famous vertex operators. (And various idiocies occur, such as zero becoming part of the Hamiltonian's continuous spectrum, thus 'no vacuum vector'.)

It's harmless abuse of notation to write expressions in terms of $\phi$ (or its Fourier components, which you've written $a_n$), as long as you remember that the computation has to commute with the construction above.

This post imported from StackExchange MathOverflow at 2014-12-07 12:42 (UTC), posted by SE-user user1504