Wightman quantum field - Interpretation

Question

I have a question regarding the interpretation of the Wightman quantum field in mathematical quantum field theory.

A quantum field $\phi$ is a operator-valued distribution. This means that $\phi$ is a linear function

$$\phi:\mathcal{S}(\mathbb{R}^{n})\to L(D,\mathcal{H}),$$

where $S(\mathbb{R}^{4})$ denotes the Schwartz space, $\mathcal{H}$ denotes a Hilbert space and $D$ denotes a dense subset of $\mathcal{H}$, such that $\forall\Psi_{1}, \Psi_{2}\in D$

$$\langle\Psi_{1}\mid\phi(\cdot)\Psi_{2}\rangle:\mathcal{S}(\mathbb{R}^{n})\to\mathbb{C}$$

is a tempered distribution. From the distribution theory we know that for regular distributions $T(f)$ there exists a function $T(x)$ such that

$$T(f)=\int\mathcal{d}^{4}x\, T(x)f(x).$$

If a distribution is irregular, such a function does not exist, like for the delta distribution. Nevertheless, the notation $T(x)$ is often used in physics, as for the delta-distribution $\delta(x)$ and also for the quantum field $\phi(x)$.

Now to my question:

If we write $\phi(x)$, the intepretation is quite clear: the value of the quantum field at the space-time point $x$, but if we write $\phi(f)$ as a distribution, how can the function $f$ be interpreted? Does it have a physical meaning, or is it just a relic of the mathematical description?

Thank you!

Answer 1

I like to think of it this way: We can only measure something with finite spatial resolution and for a finite time. So any experiment only measures an average over a small spacetime region. This is basically
$$ \phi(f)=\int \phi(x)f(x) d^4x $$ for some compactly supported smooth function $f$. This basically is what the distribution definition is doing. For technical reasons (we like Fourier transforms) people prefer Schwartz class to compactly supported test functions, but I doubt that it makes much difference to the physics.

Answer 2

The smearing (test-) function is not a relic of the mathematical description, but a key ingredient of the theory. To quote from the fathers Wightman and Streater (PCT, Statistics, and all That):

It was recognized early in the analysis of field measurements for the electromagnetic field in quantum electrodynamics that, in their dependence on a space-time point, the components of fields are in general more singular than ordinary functions. This suggests that only smeared fields be required to yield well-defined operators. For example, in the case of the electric field, $\mathcal{E}(x,t)$ is not a well-defined operator, while $\int dx ~ dt ~ f(x) \mathcal{E}(x,t) = \mathcal{E}(f) $ is.

Another quote comes from BLT (Introduction to Aximatic Field Theory, 1975):

We define a quantum (or quantized) field as an operator- valued tensor distribution. Such a definition corresponds better to the real physical situation than the more familiar notion of a field as a quantity defined at each point of space- time. Indeed, in experiments the field strength is always measured not at a mathematical point $x$ but in some region of space and in a finite interval of time. Such a measurement is naturally described by the expectation value of the field as a distribution applied to a test function with support in the given space-time region.

It is also worth noting that classical fields are also distributions.

Answer 3

These smearing functions are closely related to the single particle wave functions.

Fix a decomposition of spacetime into space times time. Fourier transform the space coordinate. In the special case $f(t,p) = \delta_0(t) \psi(p)$, which can be reached by taking a family of gaussian approximations, the operator $\phi^\dagger(f)$ creates a particle with momentum-space wavefunction $\psi(p)$ at time $t=0$.