3
$\begingroup$

In "Mathematical Methods in Physics" by Blanchard and Brüning the introduction of regular distributions and then general distributions is motivated by the following idea:

When we think about the fact that the values of measurements of physical quantities are obtained by an average process, then the interpretation appears reasonable that many physical quantities can be described mathematically only by objects of the type

$$ I_g(\phi) = \int_{\mathbb{R}^n} g(x) \phi(x) \,\mathrm{d} x $$

(This formula defines a regular distribution on the space of test functions $\mathcal{C}_0(\mathbb{R}^n)$, i.e. the continous functions on $\mathbb{R}^n$ with compact support.)

I am looking for concrete examples which illustrate this motivation in more detail.

$\endgroup$
3
  • 1
    $\begingroup$ What "motivation" are you searching, and what are the quantities in the integral? This said, the ultimate example is QFT, where fields are "operator valued distributions" that have to be smoothed against some functions to become actual operators (but simpler are examples coming from statistical mechanics or electromagnetism). $\endgroup$
    – gented
    Jul 15 '17 at 11:38
  • $\begingroup$ @GennaroTedesco: That's part of my question: I want some examples which in particular show what could the quantities in the integral be. Would be great to have examples from very simple and more complex settings as well to get the picture. $\endgroup$
    – Julia
    Jul 15 '17 at 13:27
  • $\begingroup$ The uncertainty from quantum physics also motivates having distributions: instead of just a function that assigns a value to a point in (position or momentum) space, we better do so to test functions with support spread around that point. In QFT this is why quantum fields should really be thought of as (operator-valued) distributions. $\endgroup$ Jul 16 '17 at 14:04
1
$\begingroup$

In probability theory the expectation value of an random variable $X$ is defined as $$E[X] = \int dx\; x \cdot g(X=x) $$ where $g(X=x)$ is the probability density function of $X$ evaluated at $x$. The integral becomes a sum, if the distribution is discrete instead of being continuous. In physics, a measurement is associated with errors. E.g, if we measure the current in a circuit, the measured value $I_0$ might be accompanied by a Gaussian error distribution.

If we generalize the idea of a random variable to a function of a random variable $\phi(X)$, we get $$E[\phi] = \int dx\; \phi(x) \cdot g(X=x)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.