Can the rotation of an alpha particle be measured?

Question

Although we can measure the energy of an alpha particle

Measuring the energy of an alpha particle

as well as the position and momentum, I cannot find papers that explore the motion of an alpha particle about its own axis.

Two ways that I thought of to investigate this might be:

1. the billiard ball analogy - if an alpha particle has a spin and it hits a nucleus, then it should preferentially scatter in one direction

2. Lorentz force change - if an alpha particle is not spinning and it travels through a magnetic field, it will deviate a specific amount for a specific magnetic field strength. If an alpha particle is spinning, its magnetic field (as a result of Faraday's law that a spinning charge makes a magnetic field) will interact with the imposed magnetic field and cause a deviation from the expected path

Of course, generating a 'spinning' alpha particle is the trick. I could envision using conservation of angular momentum to do this. Get a bunch of alpha particles coming out of a cyclotron. Because these particles have been accelerated millions of times in a circle, they would have angular momentum. As they exit the cyclotron, they go straight so all that angular momentum goes from being a revolution in space to a rotation. Changing the polarity of the magnets in the cyclotron should give alpha particles spinning the opposite way. The paths of these two kinds of alpha particles should be different using a detector based one #1 or #2 above.

I am hoping a knowledgeable particle physicist may know of the experiments which explored this idea\ and give me a reference.

Answer 1

I cannot find papers that explore the motion of an alpha particle about its own axis.

That's because there isn't much to say about it. Alpha particles are quantum-mechanical objects and, as such, their rotational properties are governed by quantum mechanics. In QM, angular momentum is quantized, i.e. it can only take values from a discrete set of possibilities, which happens to be $\{0,\frac12,1,\frac32,2,\frac52,3,\ldots\}$, i.e. all integers and half-integers.

As it happens, alpha particles have angular momentum $J=0$. We know this from a huge base of evidence, both experimental (e.g. the lack of hyperfine structure in helium spectroscopy) and theoretical (where we have a solid understanding of nuclear structure). The implications of alpha particles having $J=0$ is that there is no sense in which they can be said to "rotate" internally. They hold no angular momentum beyond the 'orbital' angular momentum associated with the motion of their center of mass.

If you don't like this, tough. If you don't understand it, then you do need to start with an introductory quantum-physics textbook. The reason you can't find papers discussing the rotation of alpha particles it is that it's an easily solvable problem, at the level of an end-of-chapter exercise in an introductory nuclear physics textbook, or an obvious off-the-cuff remark (so see e.g. the start of §8.5 here).

But, just to emphasize, if you try to probe it experimentally, you won't find anything.

1. the billiard ball analogy - if an alpha particle has a spin and it hits a nucleus, then it should preferentially scatter in one direction.

Alpha particles don't have spin, and they don't scatter preferentially in any direction that would be spin-dependent.

2. Lorentz force change - if an alpha particle is not spinning and it travels through a magnetic field, it will deviate a specific amount for a specific magnetic field strength. If an alpha particle is spinning, its magnetic field (as a result of Faraday's law that a spinning charge makes a magnetic field) will interact with the imposed magnetic field and cause a deviation from the expected path.

For particles with spin, the correct understanding is not the Lorentz force, but rather the force on a magnetic dipole in an inhomogeneous magnetic field, which is given by $\mathbf F = \nabla (\boldsymbol \mu\cdot\mathbf B)$, i.e. the gradient of the projection of the magnetic field onto the magnetic dipole moment $\boldsymbol \mu$. For quantum particles, the magnetic dipole moment is proportional to the spin, so this effect can be used to separate particles according to their angular momentum; this configuration is known as a Stern-Gerlach device.

However, since alpha particles have no spin, they are not deflected by Stern-Gerlach devices.

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That said, there's an additional misconception in your post, specifically when you say

As they exit the cyclotron, they go straight so all that angular momentum goes from being a revolution in space to a rotation.

When they're circling around the cyclotron, the particles have orbital angular momentum $\mathbf L = \mathbf r\times\mathbf p$, which can be understood within classical mechanics, and which comes from the fact that the line of action of their velocity does not go through the origin (at the center of the cyclotron). If you release them so that they go straight out on a tangent to the circle, that property is still true, so they still have the same amount of (orbital) angular momentum.

Answer 2

Alpha particles are scalar particles: they are perfectly spherical, and hence, don't spin.

The alpha is a 4 particle state, which I am not going to write out, but the general idea is that the wave function is a product of a spatial wave function and a spin wave function:

$$ \psi = \psi(\vec x) \times |j, j_z\rangle $$

For a spin-$0$ S-state:

$$ \psi = f(r)Y_{l=0}^{m=0}(\theta, \phi) \times |0, 0\rangle $$

where $f(r)$ is the radial wave function.

$\psi$ is perfectly spherical, and by that, I mean it is invariant under a rotation by an angle $\phi$ about a unit vector $\hat n$. That rotation operates on the spatial wave function via:

$$ \psi(\vec x) \rightarrow e^{-i\phi\frac{\hat n\cdot \vec L}{\hbar}}\psi(\vec x) = \psi(\vec x)$$

where the last step uses $\vec L=0$.

Likewise, the spin part is mapped (using Wigner D-matrices to) $|0,0\rangle$.

What that means is that there is no way to even know if you have rotated the state. It is indistinguishable from the unrotated state.

This is analogous to translating a zero momentum plane-wave. A plane wave momentum eigenstate is:

$$ \psi(x) \propto e^{i(\vec p \cdot \vec x -E t)/\hbar}$$

which for $\vec p=0$ becomes:

$$ \psi(x) \propto 1 $$

which means that the state extends over all space with no phase variation. How do you translate it? It's just $1$ no matter what you do. Does it even mean anything to translate it? Likewise, rotating an alpha does nothing.

Answer 3

There is confusion in your question. Particles of the size of alpha particles, i,e. helium nuclei, are quantum mechanical entities and the modeling of their interactions depend on quantum mechanics , its equations and postulates.

In your question you are discussing the alpha particle as if it is a billiard ball. It is not. The term spin is taken from classical mechanics, but is defined differently: it is the value needed to be assigned to particles in particle interactions so that the law of angular momentum conservation holds absolutely at the level of particle physics .

The spin of particles is inferred from their interactions with other particles in specific experiments. The spin of the Higgs is found by the angular dependence of the Higgs production crossection, for example.

Try reading the wiki article.

Edit:

As you are thinking of the alpha particles classically like billiard balls and try to impose angular momentum conservation to the alpha ejected from the cyclotron , you are ignoring that the system is not closed when the alpha leave the cyclotron, and that all three conservation laws hold for closed isolated systems.

Take a simple classical example: Think of a string with a ball at the end and a man turning it around in the air. String breaks. What happens? Ball goes off straight on a tangent. What happens with the angular momentum it had? the man rotates the other way and balances it. The closed system was "man +string+ball".

In the cyclotron when the alpha particles leave, the system is open. The isolated system "cyclotron with alpha" will balance the angular momentum with the magnetic fields that were constraining the alpha tracks , the way the man+string constrains the ball track.

So even with classical physics the ball does not take rotation about itself when the forces acting upon it are removed. For a "spin rotation" to appear on an object, there should be a force applied whose vector direction does not point to the center of mass of the object.

For quantum mechanical objects,(as the alpha particle ), spins, i.e. rotations about self, are intrinsic and part of the definition of the particle.

Answer 4

I am posting my comment as an answer because it wont fit in a comment block. I marked Emilio's response as the answer because he was the only one who mentioned that data is available (the lack of hyperfine structure in helium spectroscopy). I will have to search more in that direction. Regarding the NUMEROUS answers that tell me that an alpha particle is a point charge, I appreciate not only your guidance but also the time you took to write down the QM model. However, I should explain my motivation for my question as it really boils down to the fact that the E field of alpha particle is not a spherically symmetric thing but more of an ellipsoid. How else could you arrange two protons except for being side by side. I am ignoring the neutrons as they do no contribute to the E field.

As these ellipsoid charges enter the cyclotron, the population is spinning in every which way. And since it is a positive charge, it will have a magnetic field. (Would I say magnetic moment? I am not sure of the correct term. I cannot imagine how schools expect students to learn online. Without being around other people, without being in a lab, without being able to talk in between classes the experience is bad. We are reducing school to an ersatz activity but that is another rant.)

Anyway this little football (american football not European, or should I say rugby ball?) will reach the end of a dee and start to experience the opposite electric field and it will start to accelerate. Since only one pointy end of the football will peek out into the external E field, that will start to get accelerated first. The football will start to tumble the same way that an off center kick will cause a real football to tumble. Or the same way that a discus thrower launches a discus with a spin. Imagine for clarity that the induced tumbling is in the plane of the cyclotron.

But once the whole football is in the homogeneous E field, the whole thing gets accelerated equally. Now its tumbling towards the other dee, gaining speed, faster and faster in a linear direction. but its still tumbling though the tumbling is not being accelerated. The tumbling of the football generates a magnetic field. If this field happens to be coincident with the externally applied magnetic field, it's fine. If however, the football is tumbling the other way, its magnetic field will be opposite to the externally applied field. And like what happens when two north poles are brought together, the football is repelled. The particle leaves the spiral orbit and crashes into the cyclotron.

Of course, there are details like what happens if the football happens to be tumbling in some other random direction, not in the plane of the cyclotron. Then I would say that its tumbling is a combination of velocities: the horizontal tumbling I talked about above and a vertical tumbling component. The horizontal tumbling component will force the particle out of the cyclotron plane in some cases as I described above. The vertical tumbling component will be in either of two directions (forward or backward?) This will generate a force towards or away from the the center of the cyclotron.

But ultimately, all footballs not spinning the proper way will get kicked out of the spiral and not make the whole journey.

That's why I wanted to see the data that said there is no spinning/tumbling/rotation of this little football of protons. Thanks again for your patience.