Electron as an example has a de Broglie wavelength and could diffract. If it has a single wavelength the time extent of the particle's pulse duration would be infinite .. If it carries a broadband will be finite in time .. my question is how is it determined the duration in time of the electron or it is assumed to be the classical diameter and what is its spectral content ?
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If a pulse is broadband, i.e. if its Fourier transform is spread out over a wide range of frequencies, then its time spread tends to get narrower. This idea is made more precise in three common ways:
- The theorem that a function $f:\mathbb{C}\to\mathbb{C}$ and its Fourier transform cannot both have a compact support in $\mathbb{R}$ (a compactly supported function is one which is nonzero only within a compact set such as a closed interval);
- The Payley-Wiener theorem (see Wiki page with this name), which describes growth rates of Fourier transforms, in particular, there is a version that shows that $F(\omega) = \mathfrak{F} f(t)$ where $\mathfrak{F}$ is the Fourier transform is bounded by $|F(\omega)| \leq C e^{A\omega}$ for some $C>0$ and is also $\mathbf{L}^2$ integrable over $\omega \in \mathbb{R}$ if $f(t)$ is nonzero only in the interval $t\in[-A,\,A]$. The Payley-Wiener causality criterion (see this criterion in my answer here) also holds here because compactly supported functions can be thought of as special cases of causal one;
- The uncertainty principle for Fourier transforms: if $f(t)\in \mathbf{L}^2(\mathbb{R})$ and $F(\omega)$ its Fourier transform, then the product of the root mean square spreads of both functions is bounded as follows. Without loss of generality, assume that $f(t)$ is real and $\int_{-\infty}^\infty x\,f(x)\,\mathrm{d}\,x = \int_{-\infty}^\infty \omega\,F(\omega)\,\mathrm{d}\,\omega = 0$, then:
$$\sqrt{\frac{\int_{-\infty}^\infty t^2\,|f(t)|^2\,\mathrm{d}\,x}{\int_{-\infty}^\infty |f(t)|^2\,\mathrm{d}\,x}}\;\sqrt{\frac{\int_{-\infty}^\infty \omega^2\,|F(\omega)|^2\,\mathrm{d}\,\omega}{\int_{-\infty}^\infty |F(\omega)|^2\,\mathrm{d}\,\omega}} \leq \frac{1}{2}\quad\quad\quad\quad(1)$$
- and moreover the inequality is saturated by Gaussian $f(t) \propto \exp\left(-\frac{t^2}{2\,\sigma^2}\right)\,e^{-i\,\omega_0\,t}$ for some real constants $\sigma$ and $\omega_0>0$, i.e. such functions (their Fourier transforms are also Gaussian) achieve equality in the above bound.
This last inequality is the main reason why the "breadth" or "extent" of a pulse is measured by its root mean square spread, because the uncertainty principle bounds the products of these measures of spread.
What is its spectral content? If you mean "what wavelengths are present", then that is defined by the "broadbandedness" of the pulse that you postulate in the first place, i.e. this question is implicitly answered by its first part.
This should answer all your questions, since the you have worded it in "signal processing" language: broadbandedness, confinement and pulse diameter / duration. However, your quantum optics and quantum mechanics tags bespeak your wanting to understand all this in a quantum mechanics framework, so let's look at how all the above fits into quantum mechanics. The link begins and ends with the canonical commutation relationship (CCR):
$$[\hat{\mathbf{X}},\,\hat{\mathbf{P}}]=\hat{\mathbf{X}}\,\hat{\mathbf{P}} - \hat{\mathbf{P}}\,\hat{\mathbf{X}} = i\,\hbar\,\mathbf{I}\quad\quad\quad\quad(2)$$
where $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ are respectively the position and momentum observables. Let's simplify this to a one-dimensional particle. It can be shown that, given the CCR, $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ must have continuous spectrums - see my answer here and especially the link therein for more information. The Merzbacher "Quantum Mechanics" reference therein also shows how to derive the uncertainty principle from the CCR, but in a way that is a little different from my favourite "signal processing" approach I am about to show you. The eigenvalues of $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ can be any real values, so let's assume that our co-ordinates in quantum state space have been chosen so that $\hat{\mathbf{X}}$ is a simple multiplication operator $\hat{\mathbf{X}} f(x) = x\,f(x)$ and we write the quantum state as a wavefunction $\psi(x)\in\mathbf{L}^2(\mathbb{R})$ of the position observable's eigenvalue $x$, so that now $|\psi(x)|^2$ is the probability density of the position eigenvalue given the quantum state $\psi(x)$. So now let's think of $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ as operators belonging to some linear space $\mathcal{L}$ of suitably well behaved operators on $\mathbf{L}^2(\mathbb{R})$. We need to assume that $\mathcal{L}$ is a space of diagonalizable (i.e. spectrally factorisable) operators with continuous spectrums. Then we can think of the Lie bracket $\hat{\mathbf{P}}\mapsto[\hat{\mathbf{X}},\,\hat{\mathbf{P}}]$ as linear operator $\operatorname{ad}_{\hat{\mathbf{X}}}:\mathcal{L}\to\mathcal{L}$ on the space $\mathcal{L}$ of our operators. This mapping's kernel is the linear space of operators mapped to nought by $\operatorname{ad}_{\hat{\mathbf{X}}}$; these are precisely the linear space of generalised multiplication operators $f(x)\mapsto g(x) f(x)$ for some fixed $g(x) \in \mathbf{L}^2(\mathbb{R})$ defining each such operator. The reason we know this is the whole kernel is that operators commute if and only if those operators have the same eigenvectors and those commuting with $\hat{\mathbf{X}}$ are the kernel members we seek. So a kernel member must be a multiplication operator. Now a particular solution to the CCR can readily be verified to be $f(x)\mapsto -i\,\hbar\,\mathrm{d}_x\,f(x)$, and the coset of all operators $\mathbf{P}$ fulfilling the CCR is the kernel displaced by any particular solution, so the most general $\hat{\mathbf{P}}$ we can have in this particular co-ordinate system is:
$$f(x) \mapsto (-i\,\hbar\,\mathrm{d}_x + g(x))\,f(x) = -i\,\hbar\,\mathbf{Q}^{-1} \mathbf{D} \mathbf{Q} f(x)\quad\quad\quad\quad(3)$$
where $\mathbf{D} f(x) = \mathrm{d}_x f(x)$ and $\mathbf{Q} f(x) = \exp\left(\frac{i}{\hbar} h(x) \right) f(x)$ where $h(x) = \int g(x)\mathrm{d}x$. So now we can rotate our state space co-ordinates so that $f(x) \mapsto \mathbf{Q} f(x)$ and the $\hat{\mathbf{X}}$ and $\hat{\mathbf{P}}$ observables transform in this space as $\hat{\mathbf{X}}\mapsto \mathbf{Q}\hat{\mathbf{X}}\mathbf{Q}^{-1} = \hat{\mathbf{X}}$ and $\hat{\mathbf{P}}\mapsto \mathbf{Q}\hat{\mathbf{P}}\mathbf{Q}^{-1} = -i\,\hbar\,\mathbf{D}$. Therefore:
The CCR alone implies there is an orthogonal co-ordinate system for the quantum state space wherein:
$$\begin{array}{lcl}\hat{\mathbf{X}} f(x) &=& x\,f(x)\\ \hat{\mathbf{P}} f(x) &=& -i\,\hbar\,\mathrm{d}_x\,f(x)\end{array}\quad\quad\quad\quad(4)$$
So naturally, for the simplest description, we shall write our wavefunction $\psi(x)$ in these co-ordinates. So, now, what are the eigenfunctions of $\hat{\mathbf{P}}$ in our co-ordinate system? They are precisely the eigenfunctions of the derivative operator $\mathrm{d}_x$ i.e. the functions of the form $\psi_k(x) = \exp(i\,k\,x)$ for some real $k$. So the Fourier transform is the unitary co-ordinate transform that we must impart to our co-ordinate system to express our wavefunctions as functions $\Psi(p)$ of the momentum eigenvalue, where $p = -\hbar\,k$ (equivalently: as linear superpositions of he eigenfunctions $\exp(i\,k\,x)$ in the position eigenspace). Momentum space and position space are unitarily related by the Fourier transform (the Fourier transform in unitary by Parseval's / Plancherel's theorem). So now we can apply (1) directly to the rms spreads of the wavefunction in these two, Fourier-transformed co-ordinate systems; we get:
$$\sigma_\hat{\mathbf{X}} \sigma_\hat{\mathbf{P}} \geq \frac{\hbar}{2}\quad\quad\quad\quad(5)$$
where $\sigma_\hat{\mathbf{X}}$ is the rms spread of a wavefunction pulse $\psi(x)$ in the position eigenspace and $\sigma_\hat{\mathbf{P}}$ is the rms spread of the same state $\Psi(p)$ written in the momentum eigenspace.
Lastly, you speak of spreads in time for wavefunctions. Strictly speaking, the often quoted uncertainty principle $\Delta E\,\Delta t \geq \frac{\hbar}{2}$ is not general because there is no time observable in nonrelativitistic quantum mechanics; time is simply a parameter. But if we have a free nonrelativistic particle then $E = p^2/(2\,m)$ and therefore the spread in energy and momentum are related by $\Delta E = p\,\Delta p/m$. The time of arrival of a pulse is uncertain by the amount $\Delta t = \Delta x / v$, so that $\Delta E\,\Delta t = \Delta p\,\Delta x = \hbar/2$. A more general derivation of the energy-time uncertainty relationship is to be found in Joshphysic's answer here.
Selene Routley
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