Correctly interpretting distributions

distribution-theoryfunctional-analysismathematical physicspartial differential equationsquantum-field-theory

In Folland's Quantum Field Theory, he gives the free field as
$$ \phi(t, x) = \int \frac{1}{\sqrt{2\omega_p}} \big(e^{-ip \cdot x – i\omega_p t} a(p) + e^{-ip \cdot x + i\omega_p t}a^\dagger(p)\big) \frac{d^3p}{(2\pi)^3}.$$
Here $\omega_p \in \mathbb{R}$, $p, x \in \mathbb{R}^3$, and $a$ is an operator-valued distribution. Folland uses the physics convention of denoting the adjoints by $\dagger$, so $a^\dagger$ is the adjoint of $a$.

$\phi$ is actually an operator-valued distribution, and before introducing the above formula Folland says "maintaining the notational fiction that distributions are functions…". I am wondering if one is to get rid of this notational fiction, what is the proper way to interpret $\phi$ or ($a$ and $a^\dagger)$ in this case. Would it be that for some test functions $f \in C_c^\infty(\mathbb{R}^3)$ and $g \in C_c^\infty(\mathbb{R})$ we have that $\phi(f, g)$ is the operator such that
$$ \phi(f, g) = \iint g(t)f(x)\phi(t,x) dtdx^3 \\= \iint g(t)f(x)\int \frac{1}{\sqrt{2\omega_p}} \big(e^{-ip \cdot x – i\omega_p t} a(p) + e^{-ip \cdot x + i\omega_p t}a^\dagger(p)\big) \frac{d^3p}{(2\pi)^3}dtdx^3.$$

Is this the proper way to interpret such a distribution? What confuses me is in the above $a$ and $a^\dagger$ are also distributions, so do they need test functions of their own?

Similarly, in the same section, Folland says

Suppose $\phi$ is a tempered distribution on $\mathbb{R}^4$ that satisfies the Kliein-Gordon equation $$(\partial^2 + m^2)\phi = 0.$$

When we say a (tempered) distribution satisfies a PDE does mean it satisfies the PDE in the weak sense (i.e. with weak derivatives)?

Best Answer

First of all, let me review some generalities: A "distribution" $T\in\mathcal{D}^{\prime}(\mathbb{R}^{4})$ on $\mathbb{R}^{4}$ is defined to be a linear and continuous functional of the form

$$T:\mathcal{D}(\mathbb{R}^{4})\to\mathbb{C}$$

where $\mathcal{D}(\mathbb{R}^{4})$ is some space of test functions. Usual choices are $C_{c}^{\infty}(\mathbb{R}^{4})$, $C^{\infty}(\mathbb{R}^{4})$ and $\mathcal{S}(\mathbb{R}^{4})$ equipped with some specific topologies, where in the latter case these distributions are usually called "tempered distributions".

Now, for every (nice enough, i.e. locally intergrable) function $f$, you can define a distribution via

$$T_{f}(\varphi):=\int_{\mathbb{R}^{4}}\varphi(x)f(x).$$

However, not every distribution is of this form. A distribution $T$ for which there is a $f$ such that $T=T_{f}$ is called "regular".

Now to your questions:

  1. If the quantum field $\phi$ would be regular, then the correct way of writing it would be $$\Phi(\varphi)=\int_{\mathbb{R}^{4}}\varphi(x)\phi(x),$$ where $x=(t,\vec{x})$ denotes a 4-vector, i.e. an element of $\mathbb{R}^{4}$. So, you don't have to take two test functions one for the space and one for the time component. (not every function $f\in\mathcal{D}(\mathbb{R}^{4})$ can be factorized as $f(x)=f_{1}(t)f_{2}(\vec{x})$.)
  2. However, the operator-valued distribution $\phi$, i.e. the quantum field, is in general$^{1}$ not regular, i.e. it can not be written as a function. What Folland means with "maintaining the notational fiction that distributions are functions" is that you write all the equations as if it the distribution $\Phi$ was regular, in order to simplify notation. The situation is actually very similar to the delta function: The delta distribution is defined by $$\delta(\varphi)=\varphi(0)$$ Now, one can show that this distribution is not regular, i.e. there does not exist a function $\delta$ (denoted by the same symbol for simplicity) such that $$\delta(\varphi)=\int_{\mathbb{R}^{4}}\delta(x)\varphi(x)\,\mathrm{d}x=\varphi(0)$$ However, still, it is often very useful to write down equation in terms of the fictional function $\delta$, in order to keep the notation simple. These equations make then only sense if we integrate over them. In other words, one views the expression $$\int_{\mathbb{R}^{4}}\delta(x)\varphi(x)\,\mathrm{d}x$$ just as a notation for the actual distribution $\delta$. The same is meant in the context of quantum field theory. The quantum field is a operator-valued distribution $\Phi$, which is not regular, i.e. there does not exist a function $\phi$. However, to simplify notation, we write down all equation in terms of this "fictional" function $\phi$. These equation then have to be interpreted in the sense that we have to integrate against a test function to recover the actual well-defined object.

As to your confusion with the creation and anihilation operator: Yes, $a,a^{\dagger}$ are also distributions. What you really do is to define a quantum field as

$$\Phi(f)\propto a(f)+a^{\dagger}(f)$$

for all test functions $f$ (This formalism is sometimes called "Segal quantization"). Then, you introduce the fictional field $\phi$, i.e. you formally write

$$\Phi(f)=\int\,f(x)\varphi(x)$$

where, as explained above, $\varphi$ does not exist as an actual function. This, in turn, leads to the formula you used, i.e. using ficitional functions $a$ and $a^{\dagger}$ for the distributions denoted with the same symbol.

$^{1}$ More formally, one can show, that there does not exist an operator-valued function satisfying the Wightman axioms of quantum field theory. From the physical point of view, this could have been expected and is basically a consequence of the uncertainty principle: If you measure your field at some point, then this would cause very large fluctuations of momentum, which in turn cannot lead to a well-defined operator. This is also the reason why one more formally considers distributions functions in order to defined quantum fields.

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