I'm writing a piece on the nuclear force, and I'm struggling with something. I always thought of the alpha particle as something with a tetrahedral disposition. If you search the internet on this there's plenty of hits. Ditto if you search for images:

enter image description here

The alpha particle is usually depicted as a tetrahedral arrangement of two protons and two neutrons. And not just in popscience pictures. Here it is again in a scholarpedia article clusters in nuclei by Professor Martin Freer. He says things like the alpha+alpha cluster structure is found in the ground state of 8Be, and gives this depiction showing the arrangement of four alpha particle clusters in the nucleus 16O:

enter image description here

However I'm struggling to find any hard scientific evidence of the tetrahedral disposition or configuration of the alpha particle. So my question is this:

Is there any hard scientific evidence that the alpha particle is tetrahedral?

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    $\begingroup$ @sammy gerbil : I suspect not. But there again, does anybody? That's really the crux of my question. Where's the evidence? $\endgroup$ Feb 4, 2017 at 17:34
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    $\begingroup$ The place to find the evidence is in polarized scattering experiments. The multiple stages of the $G_{EN}$ experiment scattered polarized electrons from a polarized helium-four target, which is exactly the configuration needed, though what you ask was not the focus of the experiment and would therefor be a side-analysis if they even if they collected all the right data. $\endgroup$ Feb 4, 2017 at 17:57
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    $\begingroup$ Your last image shows the clusters in a tetrahedral arrangement, but each cluster itself looks planar. $\endgroup$
    – Random832
    Feb 5, 2017 at 20:52
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    $\begingroup$ @JohnDuffield (However, I find Thomas' observation that all four particles are in the 1s nuclear orbital, and therefore the alpha should be "shaped" more like a sphere than a tetrahedron, entirely persuasive.) $\endgroup$
    – zwol
    Feb 6, 2017 at 13:43
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    $\begingroup$ @HolgerFiedler : see this: "Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole — the electromagnetic field." Maxwell unified electromagnetism. Unfortunately it sometimes seems like it never happened. Did you get my email? $\endgroup$ Feb 21, 2017 at 21:18

2 Answers 2


The alpha particle is a quantum mechanical system, and it is not clear what we might mean by drawing pictures of billiard balls arranged according to classical polyhedra.In particular, the alpha has quantum numbers $J^\pi=0^+$, so it has complete spherical symmetry. In a shell model picture, which provides a simple guide to the exact 4-body wave function, the alpha is a state in which all four particle (a neutron with spin up/down, and a proton with spin up/down) occupy the same 1s (spherically symmetric) orbital. This implies that the alpha should be drawn as a blob, with smeared out protons and neutrons.

The shell model wave function is not exact, and there are short range correlations, that means if I detect a spin up proton at the origin, then there is a slightly enhanced/reduced probability to find a spin up neutron/proton nearby, but these correlations do not in any sense favor tetrahedral configurations.

Larger nuclei (deformed nuclei, like plutonium) have (semi) classical shapes. The corresponding quantum mechanical wave function is a superposition of states with different orientations of the nucleus. The ground state is still isotropic, but excited states correspond to rotational bands. There is also a sense in which alpha particle cluster nuclei (like oxygen and carbon) involve large wave function components that favor certain geometric arrangements.

Postscript (experimental evidence): Entire text books (for example Bohr and Mottelson, Nuclear Structure) are devoted to explaining why the shell model provides an accurate guide to nuclear states. Modern variational (and exact numerical) wave functions can be found in http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.70.743.

Empirically, the simplest piece of evidence is the spectrum of excited states. A deformed nucleus has low-lying rotational and vibrational states. The alpha particle has a large gap (consistent with a closed shell), and the lowest excited state is $0^+$, consistent with a monopole vibration (see, for example, Fig. 3-2a in Bohr & Mottelson, vol I).

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    $\begingroup$ Thanks Thomas. I'm not fond of the billiard balls, and try to think of them as S-orbital wave things. $\endgroup$ Feb 4, 2017 at 18:14
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    $\begingroup$ This can't be the whole picture, since e.g. the ammonia molecule has the same properties, and there is plenty of evidence that it is pyramid-shaped. The angular-momentum argument says that there's a uniform probability amplitude for all the different orientations of the tetrahedron (if it exists), but it doesn't rule out the existence of inner structure in the body-fixed frame. However, I don't have a strong enough grasp of how body-fixed frames work in QM to provide an authoritative answer yet. $\endgroup$ Feb 5, 2017 at 21:47
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    $\begingroup$ @EmilioPisanty The main point is that a large object (a molecule) is more likely to have a classical shape than a small object (a nucleus). Indeed, we describe molecules using a Born-Oppenheimer approximation, which provides an effective potential for the nuclei. It makes sense to ask if the classical minimum of this potential corresponds to a geometric arrangement, like a tetrahedron. Of course, the actual QM wave function must be a superposition of all tetrahedral arrangements. $\endgroup$
    – Thomas
    Feb 6, 2017 at 0:22
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    $\begingroup$ @EmilioPisanty ... which is why I follow it up with point number two, the four-body wave function is to a good approximation given by the shell model, which is 4 particles in a 1-s orbital. The Born-Oppenheimer wave function of the 4 nuclei in the ammonium molecule is very different from that. $\endgroup$
    – Thomas
    Feb 6, 2017 at 0:50
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    $\begingroup$ @BillN: All even-even nuclei have 0+ ground states, but not all of them are spherically symmetric. The ground state of an even-even nucleus is spherically symmetric in the lab frame. It has to be, because that's how angular momentum works in quantum mechanics. It's deformed in the body-fixed frame, which is what we describe, for example, in calculations using the deformed shell model. The single-particle wavefunctions, as well as any many-particle wavefunctions, constructed from such a model are not states of good angular momentum, which is a defect in the model. $\endgroup$
    – user4552
    Oct 14, 2018 at 14:41

Thomas's answer is actually pretty nice, and I upvoted it. However, it doesn't seem to have satisfied everybody, and there are some aspects of it that I think are not quite right, or don't focus on quite the right things.

Is there any hard scientific evidence that the alpha particle is tetrahedral?

The most straightforward response to this is that the notion of a tetrahedral cluster of grapes is clearly a cartoon inspired by classical intuition, and it would be absurd on the face of it to imagine that it was an accurate model of the actual quantum-mechanical system. It's not really of interest to discuss the tetrahedral cartoon in its most literal sense, because it's silly. What is at least somewhat interesting, in principle, is to ask whether the correlations among the neutrons and protons have any properties that resemble at all the kinds of correlations we would imagine from the tetrahedral cartoon.

The discussion of correlations between nucleons seems to have caused a lot of confusion in the long comment thread under Thomas's answer, so let's discuss a more straightforward example. Consider positronium in its ground state. A standard textbook treatment would start by writing down the wavefunction in separable form as something like $\Psi(x_0)\Phi(x_1)$, where $x_0$ is the vector indicating the position of the center of mass, and $x_1$ is the position of the positron relative to the electron (or relative to the c.m.). The correlations are described by the fact that $\Phi$ really tells us the wavefunction of both particles, and these correlations are perfect due to conservation of momentum. If we wish, we can completely ignore $\Psi(x_0)$, or if we care, we can let it be a state of good momentum.

But for many-body systems, this approach becomes difficult, and the classic method of attack is to instead write down a single-particle potential and populate it with particles, using occupation numbers that obey the relevant statistics. This is much more tractable for $N>2$ particles, but it has the disadvantage that the states we construct are not states of good momentum. If we apply it to positronium, then the correlations between the electron and positron are sort of there, because they both tend to live in the same region of space, but these correlations are not described exactly. There are spurious fluctuations in the total momentum, which violates conservation of momentum.

Emilio Pisanty wrote in a comment:

However, I don't have a strong enough grasp of how body-fixed frames work in QM

When we talk about body-fixed frames in nuclear physics, it's basically a way of talking about correlations between nucleons, but using a model in a specific way. Let's make an analogy with the example of the broken translational symmetry in the case of positronium.

In nuclear physics, we often violate several good symmetries at once in the same way that I described above for positronium. For a deformed rare earth nucleus, for example, we would probably use a single-particle potential with a prolate ellipsoidal shape, and we would also introduce pairing as described by the Bogoliubov approximation. The resulting many-body wavefunctions have unphysical fluctuations in momentum $\textbf{p}$, total angular momentum $J$, neutron number $N$, and proton number $Z$. For a nucleus with mass number (i.e., particle number) $A$, the relative sizes of these fluctuations all scale down with $A$, so for many heavy nuclei, for many observables, this basically produces no problems.

The ground state of an even-even nucleus such as an alpha particle is spherically symmetric in the lab frame. It has to be, because that's how angular momentum works in quantum mechanics. An even-even nucleus can be deformed in the body-fixed frame, which is what we describe, for example, in calculations using the deformed shell model. So the fact that the helium nucleus has a $0^+$ ground state really tells us nothing about whether it has a particular deformed shape such as a tetrahedron.

So when we want to tell whether a particular nucleus is deformed in its ground state, we do not get that information from its ground-state spin. We get it from other observables. If an even-nucleus is a prolate ellipsoid (which is the shape that essentially all stably deformed nuclei have), there is a rotational band built on the ground state, with spin-parity going like $0^+$, $2^+$, $4^+$, ... The energies go like $J(J+1)$. The half-life for gamma decay down this band by E2 transitions is quite short, indicating collective motion. Semi-classically, this band is interpreted as end-over-end rotation, since a quantum rotor can't rotate about an axis of symmetry. Angular momentum can be generated about the symmetry axis only by particle-hole excitations, which display none of the observational hallmarks described above.

If helium were really configured in the kind of tetrahedral configuration shown in the cartoons, then it would have some of these rotational characteristics, but not all of them. It would certainly have low-energy rotational bands built on the ground state, but we don't observe any such bands. The ground state would lack parity symmetry in the body-fixed frame, and if we were to take the cartoons totally literally, then it would also have a large electric dipole moment. This dipole moment would vanish in the true ground state (similar to the ammonia molecule, which is a classic example described, e.g., in the Feynman Lectures). However, there would be negative-parity rotational states interlaced with the positive-parity states, and there would be strong E1 transitions between these positive- and negative-parity states. We don't observe anything like this. There is evidence that a few nuclei do have reflection-asymmetric shapes, so this is not just speculative. The alpha particle's properties look nothing like the properties we would expect for the reflection-asymmetric shape.

So there is very direct observational evidence that the structure of the alpha particle is nothing like the cartoon, not even in some vaguely semi-classical way.

There are also clear theoretical reasons why we would not expect such a structure for helium. It's doubly magic, and doubly magic nuclei never have any stable deformation in their ground state.


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