The mere formal expression of an equation is not very useful, unless complemented by a more or less intuitive understanding. Different people may have different intuitions of a given formula or different mental images of one physical systems (more on that later).

The interesting part is that putting together two such different intuitions of a relation can yield non-trivial results with almost no algebraic manipulation, as I'll show below. What is the meaning of the following formula ?

\[\frac{1}{\sqrt{2\pi} \sigma} \int_{-\infty}^{\infty} \text{d}x \exp (i q x) \exp \left (- \frac{x^2}{2 \sigma ^2} \right ) \tag{1}\]- It can be seen as the Fourier transform of a Gaussian (or normal) distribution with zero mean and standard deviation \(\sigma\): \( \displaystyle \underbrace{\int_{-\infty}^{\infty} \text{d}x \exp (i q x)}_{\cal{F}(\cdot)} \frac{1}{\sqrt{2\pi} \sigma} \exp \left (- \frac{x^2}{2 \sigma ^2} \right ) = \cal{F} \left ( \cal{N} (0,\sigma) \right ) \tag{1a} \)
- ... or as the expectation of a phase factor over said Gaussian distribution: \( \displaystyle \underbrace{\int_{-\infty}^{\infty} \text{d}x \frac{1}{\sqrt{2\pi} \sigma} \exp \left (- \frac{x^2}{2 \sigma ^2} \right )}_{\left \langle \cdot \right \rangle} \exp (i q x) = \left \langle \exp (i q x) \right \rangle \tag{1b} \)

Equating the rhs expressions in the two equations above (which are simply two different readings of (1)!) and recalling that the Fourier transform of a Gaussian is itself a Gaussian yields the very useful identity:

\[\left \langle \exp (i q x) \right \rangle = \cal{N} \left (0,\frac{1}{\sigma} \right ) \tag{2} \]employed, for instance, in proving the Siegert relation.Note that both sides of (2) are now functions of \(q\), since we have already integrated over \(x\).

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