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In a simple atom say hydrogen, there is an electron cloud which is spherical in shape. What about a free electron, how big or small will that cloud be? I think the term cloud here means the likelihood the electron can be found, as for a free electron in absent of other forces does it stay as a point-like particle or a standing wave spreading out across the universe?

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"What about a free electron, how big or small will that cloud be?"

As big or as small as you want it to be, in principle – that is, in an infinite vacuum at zero Kelvin. The most common example of a normalized wave packet is the Gaussian (because it's rotationally symmetric and computationally simple): \begin{align} \psi_G(\mathbf{x}) &= \frac{1}{(2\pi)^{3/4} \sigma^{3/2}} \exp\left(-\frac{[\mathbf{x}-\mathbf{x}_0]^2}{4\sigma^2} + i\frac{\mathbf{p}_0\cdot\mathbf{x}}{\hbar}\right), \end{align} where the packet is centered at $\mathbf{x}_0$ and has mean momentum $\mathbf{p}_0$ with real space "width" (standard deviation) $\sigma$. $\sigma$ can, in principle, be as large or as small as you like. The only question is whether you have the energy needed to make it small because the mean energy in the electron is given by: $$\langle \psi_G| H_{\mathrm{free}} |\psi_G\rangle = \frac{\mathbf{p}_0^2}{2m} + \frac{\hbar^2}{8m\sigma^2},$$ which diverges as $\sigma \rightarrow 0$.

Pick any form for the electron wave packet you like, and you'll get similar results. Just be careful to keep the packet normalizable, otherwise you'll end up with annoyances caused by non-physical states like the plane waves $\psi_{\mathrm{pw}} = \frac{1}{\sqrt{2\pi}} \exp\left(i \frac{\mathbf{p}_0\cdot \mathbf{x}}{\hbar}\right)$.

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  • $\begingroup$ what would we obtain for $\sigma$ at room temperature (with $k_b T$=0.026 eV available)? or, what would be the required energy to keep the electron "small"? $\endgroup$
    – khaverim
    Aug 25 '17 at 1:53
  • $\begingroup$ @khaverim your question is somewhat vague. At room temperature in what situation? In contact with a photon gas? Are there more electrons around, or is it a thermodynamic ensemble of one particle states? Is the temperature high enough that pair production occurs ($kT \sim m_e c^2 \approx 500 \operatorname{keV}$)? $\endgroup$ Aug 25 '17 at 2:18
  • $\begingroup$ I am curious, per my answer which derives from something a professor said to the class, and the pretty negative response, what it would take to constrain a gaussian electron to $\sigma=$ ~50 Angstroms (a "small" volume). I would say then, at room temperature in a human-like situation. Perhaps not immediately near gaseous molecules (as to interact with them), but with the same energy available as ambient air. $\endgroup$
    – khaverim
    Aug 25 '17 at 2:23
  • $\begingroup$ Specifically, I want to have something to ask/say to my professor when I next see him re: this question $\endgroup$
    – khaverim
    Aug 25 '17 at 2:23
  • $\begingroup$ Well, for one thing the thermal wavelength you calculated has nothing to do with $\sigma$ and everything to do with $\sqrt{\mathbf{p}_0}$, in my answer's notation. See, the thermodynamic ensemble uses the expectation of the Hamiltonian I calculated in my answer. The "thermal wavelength" comes from finding the thermal mean energy and solving for the wavelength ($\lambda = h/p$) in the large $\sigma$ limit (i.e. the gas is diffuse). $\endgroup$ Aug 25 '17 at 2:46
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The surrounding cloud is just a picture of a zone where the electron has a very high probability of being there. Note that we are dealing with probabilities, so the electron has the possibility (a very small one) of being in the other side of the universe. Questions like "where is exactly the electron?" or "what is the shape and size of an electron?" are, in a certain sense, "forbidden" by the Heisenberg's uncertainity principle, it doesn't matter if the electron is bounden to an atom or if it's a free electron.

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Since it's isolated (and therefore has no potential energy), the quick answer is that it resembles (but is not exactly) a Gaussian distribution on the order of 10 - 50 angstroms wide. You can calculate it yourself from the equation for thermal wavelength:

$$ \Lambda = \sqrt{\frac{h^2}{2\pi m k_BT}} $$

At room temperature, using the mass of an electron we have $\Lambda = 4.3\times10^{-9}$ meters, or 43 Angstroms.

So it would "look" something like this, where red indicates a "denser" part of the "cloud" (though this 2D gaussian doesn't do justice to a 3D function): enter image description here

This is a 3D gaussian. We have to use cross sections to see the probability density though:

enter image description here

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    $\begingroup$ On what basis are you saying it will be Gaussian? $\endgroup$
    – CDCM
    Aug 24 '17 at 2:14
  • $\begingroup$ As it happens my professor for QM and Stat Mech. just told us this today. If I recall correctly it's because the Fourier transform of a gaussian (the wavefunction of an electron can be defined as an integral of plane waves weighted by complex exponentional probability functions [that's an FT itself]) is just another gaussian. --- then you get the "shape" of it by the square modulus of the wavefunction at each position in space, which gives you these kinds of images -- red is a higher quantity (probability) $\endgroup$
    – khaverim
    Aug 24 '17 at 2:17
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    $\begingroup$ This answer is wrong without more qualifications. A free electron can in principle have any normalizable wavefunction at all, gaussians are just convenient models. Furthermore, you seem to assume that your electron is part of a gas, which wasn't stated in the question. $\endgroup$
    – Javier
    Aug 24 '17 at 2:21
  • $\begingroup$ Yes but my question is why should it be Gaussian to begin with, the electron could be in a different state. Further, Gaussian wave packets disperse - they become more broad in time. $\endgroup$
    – CDCM
    Aug 24 '17 at 2:24
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    $\begingroup$ No, we use Gaussians because they are convenient. Mathematically, Gaussians are a measure zero set within $L^2$, so actually the probability of a wavefunction being exactly a Gaussian is zero. And the concept of temperature simply doesn't apply to a lone electron. You need a whole bunch of them to talk about thermodynamics. $\endgroup$
    – Javier
    Aug 24 '17 at 3:04

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