Let $M$ be a smooth manifold and denote $C^\infty_0(M)$ the space of smooth functions with compact support. In Mathematics a distribution is defined to be a continuous linear functional $\phi : C^\infty_0(M)\to \mathbb{R}$. The space of distributions is usually denoted $\mathfrak{D}'(M)$.

So a distribution is a map that takes a function and outputs a number in a linear and continuous way. The Delta distribution centered at $x\in M$ is for example


Another way to create distributions is to pick $f\in C^\infty_0(M)$ and define

$${f}^\diamond[g]=\int_M fg.$$

That much is fine. The problem is the following: in Physics one often forgets all this and treats distributions as functions. So a Physicst will almost never bother writing $\phi[f]$ or just $\phi$. They write $\phi(x)$ which is not really correct, since $\phi$ isn't a function on $M$ at all.

The issue though is that there is a terminology around which makes me quite confused. One often talks about "smeared" fields written as


and talks about the field in "unsmeared form" writting it just $\varphi(x)$. This confuses further, because it is known that it is not true that given $\varphi$ there is $f$ such that $\varphi = f^\diamond$.

This terminology may be found for example in Fewster's notes on QFT on curved spacetime, but I've seem it elsewhere.

This seems to imply that when one picks $\phi\in C^\infty_0(M)$ it is unsmeared and when one picks $\phi^\diamond\in \mathfrak{D}'(M)$ and apply it to a function it is smeared (but notice that $\phi^\diamond[f]$ is a real number, not even a field anymore after applying to $f$).

So what really is this smeared and unsmeared terminology about and how does this makes contact with distribution theory from mathematics?

  • $\begingroup$ The notation $u(x)$ for a distribution $u$ is a nice way of reminding everyone "which" space of test functions you act on, and is especially useful when working with product space/tensor products. $\endgroup$ – Ryan Unger Sep 29 '17 at 22:38
  • $\begingroup$ It's just an abuse of notation, exactly like pretending $\delta(x)$ is a function. The "smeared field" talk is for intuition: it explains why we use distributions. $\endgroup$ – Javier Sep 29 '17 at 23:28

Quantum (unsmeared) fields are often referred to as operator-valued distributions in mathematical physics.

They surely are linear maps from some symplectic linear space of test functions to the self-adjoint operators affiliated to a W*-algebra. However, they may fail in general to be continuous maps so they are not technically distributions.

Classical fields instead are usually standard "distributions", or more precisely elements of the continuous dual of the symplectic space of test functions (that may however not necessarily be $C_0^{\infty}(M)$, or $\mathscr{S}(M)$ if it is possible to define it).

The notation $\varphi(x)$, often used by physicists, is an abuse and it could be avoided. Let me remark however that analysts use often the notation $f(x)$ for distributions in $\mathscr{S}'(\mathbb{R}^d)$, especially if they are in a "function subspace", like $L^p(\mathbb{R}^d)$ or $W^{r,p}(\mathbb{R}^d)$ (and so for classical fields, that are usually assumed to be in some Sobolev space, the notation is still mathematically rather acceptable).


It is true that many distributions are not represented by functions, but it is also true that any distribution may be approximated arbitrarily well by functions, which are very nice. (An example: $\delta_0[f] = \lim_{\epsilon \to 0} \int f(x) \frac{1}{\sqrt{2\pi\epsilon}} exp(-\frac{x^2}{2\epsilon})$.) This is really a quite general fact about nuclear spaces. The nuclear vector space $\mathcal{S}$ of test functions is dense in the dual space $\mathcal{S}'$ of distributions.

Much the same thing is going on when physicists use the abusive notation $\phi(x)$. They're implicitly approximating operator-valued distributions $\phi$ by operator-valued functions, mollifying the highest frequency modes. So, they're really using $\phi_\epsilon(x) = \int \phi(y) w_\epsilon(x-y) dy$, where $w_\epsilon$ is some regulating bump function of width $\epsilon$. This is OK, as long as you're careful not to draw conclusions about short distances faster than you shrink $\epsilon$.


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