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I have read this question:

Elementary particle (electron) and non-elementary (proton) spagettification

and the comments where it says:

But no real elementary particle can be confined in a region smaller than its compton wavelength.

An these questions:

What is the physical significance of Compton wavelength?

Confining a particle into a region shorter than its Compton wavelength

https://en.wikipedia.org/wiki/Compton_wavelength

Where Veritas' answer says:

Yes, this will happen. But you cannot confine particle in the vacuum. To confine a particle, you must have some potential. The energy to produce pairs must come exactly from this binding potential. For example, you can confine electron using a very strong electric field. To confine an electron in a region smaller than its Compton wavelength you need a field with enough energy to create electron position pairs. Particle in a vacuum will never be confined.

So which one is right?

Question:

  1. Can elementary particles be confined into a smaller region then their Compton wavelength?
Árpád Szendrei
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  • I think the answer is already derivable from the links you posted. Trying to confine a particle below its Compton wavelength will cause pair creation around it, essentially defeating the meaning of confinement of a single particle, due to the presence of a cloud of new particles around it. – Avantgarde Aug 06 '18 at 16:52
  • @Avantgarde, how do you know pair creation will happen, if we do not know the details of how the confinement is accomplished? And even if pair creation occurs, how does that prevent the original particle to disappear within known region of dimensions smaller than the Compton length? This kind of argument with pair creation seems very thin. – Ján Lalinský Aug 06 '18 at 20:59
  • @JánLalinský Trying to probe distances smaller than the Compton wavelength ($\lambda_c$) implies that probing energies are greater than the rest mass of the involved particle. This is the regime where QFT (and therefore, particle creation) becomes essential for the system's description. Also, I didn't say that the original particle will disappear: probing distances smaller than $\lambda_c$ will impart energies $\geq$ rest mass of the original particle. – Avantgarde Aug 06 '18 at 21:16
  • @Avantgarde you seem to describe a high-energy collision and think that is neccessary to confine the particle. But to confine means to limit the particle's position to some bounded available space, not necessarily to collide it with another particle. Confinement can be achieved by bounding the particle to an atom, or molecular system. Or inside a static field trap. – Ján Lalinský Aug 07 '18 at 18:05
  • @JánLalinský No, that's not what I mean. You can confine however way you want. After that, we want to measure the particle's position. Now, reducing the position measurement's uncertainty below $\lambda_c$ necessarily means $E \geq m$. This is just Heisenberg's uncertainty principle. – Avantgarde Aug 07 '18 at 20:06
  • @Avantgarde well the question is about confinement, not about measurement of position. What is E? Energy of the light used to look at the particle? – Ján Lalinský Aug 08 '18 at 12:10
  • @JánLalinský That's the point of https://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics. Only after you perform a measurement on a particle to find its position will you get information about the extent to which it is 'confined'. If confinement was perfect, then $\Delta x =0$ and there would be no quantum mechanics. The measurement of position within $\lambda_c$ will induce a momentum uncertainty (of the particle being measured) so large as to induce a $\Delta E$ enough for particle production. (I'm speaking roughly here, so ignore $\Delta$, factors of $2$, etc.) – Avantgarde Aug 08 '18 at 13:58
  • @Avantgarde I think you are jumping to conclusions. Confinement does not require measuring position - see Michale Momayezi's answer. And even if such measurement is done, it is not clear how lack of knowledge of momentum "induces" pair creation. Such lack of knowledge of momentum is always present, even when the result of position measurement has uncertainty greater than $\lambda_c$, so I do not see why passing this imaginary limit should change anything. – Ján Lalinský Aug 08 '18 at 17:33
  • @JánLalinský Confinement does not make objective sense without measurement, which is a defining aspect of quantum mechanics! Also, the answer below is based on the Bohr model which is rife with shortcomings, in particular, it: is not relativistic (as needed for confinement within a distance $\lambda_c$) and violates Heisenberg's uncertainty principle. So you cannot use the obsolete Bohr model to explain the concerned phenomena. – Avantgarde Aug 08 '18 at 18:38

1 Answers1

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I think it is possible to confine an electron mostly into a region smaller than its Compton wavelength ($\lambda_C=2.4\mathrm{pm}$).

First consider Bohr's model for a single electron in the field of a nucleus of charge Ze. The ground state has a normalized probability density proportional to $\exp\left(-{Z r}/{\mathrm{a}_1}\right)$. Here $\mathrm{a}_1$ is the hydrogen Bohr radius of 53pm. So that Bohr diameter is 44 times $\lambda_C$.

Next consider a fully ionized uranium nucleus an add one electron. Its Bohr radius will be 92 times smaller and hence $\mathrm{a}_{92}=0.58\mathrm{pm}$. The diameter is 1.06pm, which is still considerably smaller than $\lambda_C$.

When we add a second electron to fill the s-shell, the orbital increases in diameter. For example in helium the radius of the full s-shell 31pm instead of the scaled Bohr radius of 53pm/2; ie a 17% increase.

If this approximation holds inside a uranium atom, we can estimate that the s-shell is confined mostly to a sphere smaller than $\lambda_C$.

This can be extended to heavier nuclei as long as they are stable enough to allow for the formation of an electron shell before they decay.

Michael Momayezi
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  • I do not think it is correct to use the Bohr model to explain phenomena like that asked in the question. The model has many problems, particularly, being non-relativistic and violating Heisenberg's uncertainty principle. – Avantgarde Aug 08 '18 at 18:45
  • @Avantgarde, I agree in principle, but in this case, the Bohr model, the nonrelativistic Schroedinger's equation and also common relativistic models based on the Dirac equation give the same prediction for hydrogenic atoms; the typical radius is given by the Bohr radius divided by $Z$. If $Z$ is big enough (greater than 43), the radius is smaller than $\lambda_C/2$. The relativistic models predict even slightly smaller radius that the non-relativistic ones. – Ján Lalinský Aug 09 '18 at 15:48
  • @JánLalinský It would be good if you could put the relevant calculations and the explanation into an answer so that everyone can have a look. The moment you say that a particle is at a certain radius, you're effectively using old quantum theory concepts (like those of Bohr model) which assume infallibility of particle position. This is prohibited in modern quantum theory. – Avantgarde Aug 09 '18 at 16:38
  • I did not say that; I said something about typical radius, which is the value $r_0$ in $\psi = Ce^{-r/r_0}$. – Ján Lalinský Aug 09 '18 at 17:30
  • @JánLalinský Yes, I know. And such conclusions about radial position preclude modern quantum measurement theory in the context of OP's question. I am not aware of any paper/book that discusses typical radii in QFT. Feel free to share them if you know, however. – Avantgarde Aug 09 '18 at 18:28
  • @Avantgarde, the question did not require answer in the framework of "modern quantum measurement theory" or QFT. Confinement evokes particle in a potential wall, or bound states of two attracting particles; QFT has little to say about them, the equations are intractable. Most bound state calculations are approximate models based on the Dirac equation. So I think Michael's answer is valid. – Ján Lalinský Aug 09 '18 at 18:37
  • @JánLalinský No, when you're considering length scales $\leq \lambda_c$, you cannot avoid QFT, let alone use the old Bohr model. Measurement is the cornerstone of quantum mechanics without which you will be doing (deterministic) classical mechanics, so I'm not sure why you want to ignore it. And surely, bound states can be handled in QFT, see https://physics.stackexchange.com/questions/217575/how-are-bound-states-handled-in-qft – Avantgarde Aug 09 '18 at 19:03