1
$\begingroup$

Ernest Rutherford performed the gold foil experiment; alpha particles were fired at a gold foil and the alpha particles were scattered. This result disproved Thomson's plum pudding model of atoms.

This got me wondering, how does QM fit into this picture? How do we use the wave nature of alpha particles to explain what is going on in this experiment? When does the wavefunction collapse? Did the wavefunction spread out across the entire space prior to measurement? If so, doesn't it mean that the particles were not "fired" but rather just under the influence of the potentials of the gold atoms?

P.S. My understanding of the physical interpretation of QM is all over the place so my question might not even be valid in the first place. If so, sorry for that.

$\endgroup$
9
  • $\begingroup$ QM is not needed to interpret Rutherford scattering. It is purely classical. $\endgroup$
    – Jon Custer
    Commented Oct 2, 2021 at 9:13
  • $\begingroup$ @JonCuster I think OP is asking "how are the results consisted with QM". Of course you can interpret the experiment in a classical manner, but because we're dealing here with subatomic particles, one should also get the same results when trying to use QM (which, in a way, is a more accurate description of what is going on). $\endgroup$ Commented Oct 2, 2021 at 9:30
  • $\begingroup$ @OfekGillon I concur. The Rutherford experiment can be viewed as an emitter-with-low-luminosity experiment. The assistent would first sit half an hour in total darkness, to which in response the eyes make the threshold to detect light lower and lower, allowing the flashes to be seen at all. Flashes occurred one at a time, allowing counting. Question: quantummechanically, when an Alpha particle scatters of a gold nucleus, is it subsequently described as existing in a superposition of being scattered in any direction? $\endgroup$
    – Cleonis
    Commented Oct 2, 2021 at 10:38
  • $\begingroup$ @Cleonis is alluding to this concept. $\endgroup$
    – J.G.
    Commented Oct 2, 2021 at 12:41
  • 1
    $\begingroup$ Given that Rutherford could describe the scattering (including the differential cross section) just fine in 1908, well before quantum, it is clear that QM is not required to analyze it. $\endgroup$
    – Jon Custer
    Commented Oct 2, 2021 at 15:07

3 Answers 3

1
$\begingroup$

There are two parts in the question. First, given potential, how to find the scattering amplitudes as a function of scattering angle. Second, how to obtain the scattering potential in the first place.

The answer to the first question is twofold again. Normally, yes you need QM to compute scattering amplitudes correctly. However Rutherford did not do that and just did the calculation using classical scattering theory for a Coulombic potential. But there he was helped by one of those rare lucky coinsidences that helped researchers when they needed to leap the gap. Namely, the scattering amplitudes for Coulombic field coinside for classical and quantum theories. This is a feature of Coulombic potential that makes it special in this sense.

Second, the scattering potential. This is where QM was necessary. Classical theory could not (and still cannot) accommodate for a point-like concentration of positive charge in the center of an atom without compromising its long-term stability. Rutherford reached this conclusion with his classical scattering calculation, which he was lucky is still valid in QM domain, and the rest is history.

$\endgroup$
0
$\begingroup$

The Rutherford experiment

Rutherford scattering is the elastic scattering of charged particles by the Coulomb interaction. It is a physical phenomenon explained by Ernest Rutherford in 19111 that led to the development of the planetary Rutherford model of the atom and eventually the Bohr model. Rutherford scattering was first referred to as Coulomb scattering because it relies only upon the static electric (Coulomb) potential, and the minimum distance between particles is set entirely by this potential. The classical Rutherford scattering process of alpha particles against gold nuclei is an example of "elastic scattering" because neither the alpha particles nor the gold nuclei are internally excited

This can be mathematically model with classical mechanics , no need of quantum mechanics.

The experimental data that showed the need for a new theory for the atomic dimensions were the photoelectric effect, the black body radiation and the spectra of atoms. These lead to the semiclassical Bohr model and finally to quantum mechanics for describing with accuracy interactions at the atomic level.

Quantum mechanics uses the solutions of the differential wave equations with extra postulates that pick up those solutions that describe the measurements possible in those dimensions.

A main postulate is that the solutions for a given system are not solutions for individual particles but for a statistical accumulation, predicting the probability of measurement . I.e. there should be an accumulation of same boundary condition events to see how the quantum theory fits the data.

Expectation Value Postulate For a physical system described by a wavefunction $Ψ $, the expectation value of any physical observable q can be expressed in terms of the corresponding operator Q as follows:

wavf

It is a probabilistic theory.

How do we use the wave nature of alpha particles to explain what is going on in this experiment?

The so called "wave nature" is an envelope that can roughly describe this probabilistic expectation, the wave functions are sinusoidal solutions of the wave equation, but is not needed to model the experiment.

When does the wavefunction collapse?

Wave function collapse is a confusing way of stating that a measurement or an interaction picks up one of the probable states that the wavefunction models. After interaction/measurement a differen wave function appears due to the different boundary conditions that result with an interaction.

Did the wavefunction spread out across the entire space prior to measurement?

The wavefunction is defined over the space of the problem by construction of the theory. The measurement picks up one probable point, ( like throwing 1 and 6 in a two die throw)

If so, doesn't it mean that the particles were not "fired" but rather just under the influence of the potentials of the gold atoms?

Certainly the wavefunction is the proper quantum mechanical mathematical solution for the available Coulomb potentials of the experiment :"alpha particle scattering off gold nucleus".

$\endgroup$
2
  • $\begingroup$ While I understand that QM is not necessary to explain this phenomenon, but it does not mean that QM cannot be invoked to explain. Since we are in the atomic scale, I would think that the effects of QM is significant enough to require a QM explanation to this scattering phenomenon. Perhaps a more refine way of posing my question would be "how does one reconcile the QM picture and classical picture in this experiment?" $\endgroup$
    – Tham
    Commented Oct 2, 2021 at 13:35
  • $\begingroup$ The R, scattering has been examined in many ways quantum mechanically, and of course it agrees with the classical calculation see arxiv.org/pdf/2108.04388.pdf , also for detailed study arxiv.org/pdf/quant-ph/0403050.pdf for particular cas $\endgroup$
    – anna v
    Commented Oct 2, 2021 at 14:02
0
$\begingroup$

Yes the wavefunction spreads out across the entire space prior to measurement.

But don't forget the wavefunction is itself a mathematical tool.

Consider the following example from ordinary probability. Suppose you have a piece of paper and two envelopes. The paper is put into one of the envelopes but you don't know which. So the probability distribution describing your knowledge is 50:50 between the envelopes. If the envelopes are located on a table, separated by 10 cm, then the probability distribution as function of position has two bumps separated by 10 cm. Now send one of the envelopes to Australia. Now the probability distribution has two bumps separated by many thousands of miles. Now open the envelope near you. At this stage the probability distribution suddenly collapses to either a bump near you (if the paper was in your envelope) or a bump in Australia.

The above facts concerning classical probability distributions do not capture all that one can say about quantum wavefunctions. One difference is that the distribution I described was about your state of knowledge rather than the physical situation itself, whereas the wavefunction is about the physical situation itself (or, if you prefer, the maximum amount of knowledge that anyone could have). Nonetheless you should not think of wavefunctions as just like waves; they are subtle tools and what they do is give the distribution of quantum amplitude. Quantum amplitude is much debated in interpretations of quantum theory. It is the thing whose modulus-squared give a probability for some event or scenario. In the Rutherford scattering experiment, there is a quantum amplitude for "alpha particle left source and arrived at detector located at $x$" for each $x$. The quantum amplitude, then, is a complex number $\psi$ which depends on $x$, so we write it as $\psi(x)$ and thus you have a wavefunction. And in the end it is complicated permanent things such as an electrical discharge involving billions of atoms (e.g. in a particle detector), or dots on a photographic film made from billions of chemical reactions which are observed. It is the probabilities of those complicated permanent events (called "measurements") which the mathematical formalism of quantum theory ultimately describes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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