3
$\begingroup$

If two non-entangled free fermions with the same spatial wavefunctions (which do not yet overlap) throughout time approach each other, at what point during the overlapping of the two spatial wavefunctions will the two electrons have related spins (because of Pauli's Exclusion Principle), which were unrelated before the overlap?

I know, the free fermions have no quantized positions (though you can argue that the wavefunctions are superpositions of an infinite number of space "basis states", with infinitely small weight factors, while the wavefunctions grow larger through time, ending up smeared out all over space, which implies a well-defined energy), and thus no quantized energy and no energy quantum number (while their spins do have quantum numbers).

But nevertheless, I have the feeling (god forbid!) that, during the overlap of the two fermion wavefunctions (let's assume they overlap completely at a certain time), their spins must become opposite at some point (which wasn't the case before the start of the overlapping).

$\endgroup$
  • $\begingroup$ You seem to have neglected the basic scattering aspect of this arrangement: the potential between them dominates the dynamics unless their relative momentum is very high, indeed, in which case they are not in the same state even when their location overlap is very strong. $\endgroup$ – dmckee Dec 9 '17 at 2:43
  • $\begingroup$ @dmckee indeed, for the purposes of the OP it would probably be more interesting to ask about the hypothetical outcome of two identical non-interacting fermions sent on a collision course. $\endgroup$ – Rococo Dec 9 '17 at 3:27
  • $\begingroup$ I changed my question because I indeed forgot about the interaction aspect! $\endgroup$ – descheleschilder Dec 9 '17 at 8:37
2
$\begingroup$

Assuming non-interacting $\frac{1}{2}$-spin dirac fermions, assuming that you are asuming $\psi_1(x,t)=\psi_2(-x,t)$ for the spatial part of the first-quantized wavefunctions in some reference frame, being $\psi_1$ and $\psi_2$ spinors. Given a basis $\psi_I(x,t)$ of solutions of the Dirac equation ($I$ not necessarily numerable), $\Psi(x,t)=\sum_I \psi_I(x,t) b^\dagger_I$, for $b^\dagger_I$ behave as "creation operators", $\{b^\dagger_I,b^\dagger_J\}=\{b_I,b_J\}=0$, $\{b^\dagger_I,b^\dagger_J\}=\delta_{IJ}$, and $\{\Psi(x,t),\Psi(x',t')\}=0$, $\{\overline{\Psi(x,t)},\overline{\Psi(x',t')}\}=0$, $\{\overline{\Psi(x,t)},\Psi(x',t')\}=\delta^3(x-x')\delta(t-t')$. You can proof easily that this means that $\{\psi_I\}$ forms a orthonormal basis. This will be demanded to the solutions of the classical equation of motion, now immerse ourselves in the Dirac Equation. Now, we say that $\psi_1$ satisfies Dirac equation, and $\psi_2$ too, and that means that $P\psi_2(x,t)=P\psi_1(-x,t)=\gamma^0\psi_1(x,t)$ (I am using here the parity invariance of the Dirac equation) also too, so $(i\partial\!\!\!/-m)\gamma^0\psi_1(x,t)=\gamma^0(i\gamma^0\partial_0-i\gamma^i \partial_i-m)\psi_1(x,t)$, so $\partial_0\psi_1(x,t)=-im\gamma^0\psi_1(x,t)$, so, wavefunctions with same quantum numbers and parity reversed spatial non-spinor part can't approach each other, because they need to be static.

Was this a little useful on answering your question? Maybe I misunderstood the question.

$\endgroup$
  • $\begingroup$ It is surely useful! Especially the part of decomposing the wavefunctions in a continuous orthonormal basis. But don't obey the two wavefunctions the following conditions: $\psi_{1}(x,t)=-\psi_{1}(-x,t)$ and $\psi_{2}(x',t)=-\psi_{2}(x',t)$ before they start to overlap? $\endgroup$ – descheleschilder Dec 9 '17 at 16:47
  • $\begingroup$ After some thinking, the symmetry of the spatial wavefunction applies of course only to the two particles if they overlap, and not to a single fermion wavefunction, so $\psi_{12}(x,t)=\psi_{12}(-x,t)$, so I'm not assuming $\psi_{1}(x,t)=\psi_{2}(-x,t)$, but I consider (before the overlap) two spatially separated spinors $\psi_{1}$ and $\psi_{2}$, which of course don't obey the conditions I wrote in the preceding comment. I don't understand why in your answer $\psi_{1}(x,t)=\psi_{2}(-x,t)$. $\endgroup$ – descheleschilder Dec 9 '17 at 18:32
  • $\begingroup$ With $\psi_{12}(x,t)$ is possibly that you mean $\psi_{12}(x_1,x_2,t)$? If you think in the second it'll be more evidente why I assumed what i assumed. $\endgroup$ – Iliado Odiseo Dec 11 '17 at 0:54
0
$\begingroup$

... at what point during the overlapping of the two spatial wavefunctions will the two electrons have related spins (because of Pauli's Exclusion Principle), which were unrelated before the overlap?

The (intrinsic) spin of the electron (and other subatomic particles) named two phenomena of the behavior of electrons. The first is the deflection of the moving electron in an external magnetic field. The second is the Pauli Exclusion Principle. How such different phenomena could be imagined to have the same cause?

It is without doubt, that the intrinsic spin is related one by one with the (intrinsic) magnetic dipole moment of the electron. Now try to connect both phenomena with this moment. For the deflection see How does the electrons' magnetic dipole moment get influenced when electrons are moving through a magnetic field?.

For the Pauli principle it is easy, simply two electrons would align each other like two bar magnets, north to south and south to north pole. This is the lowest energetically state together.


A little digression: I’m wondering why for the electrons and protons in atomic bonds it is assumed that their charges staying unchanged. It seems more practicable to say that their charges neutralize each other to some extend AND their magnetic dipole moments determine their arrangement in atoms and molecules. Following this idea I just find it easy to understand chemical bonds and the arrangement of electrons in atomic shells.


About the distance of the Pauli principle. It comes into play when the magnetic dipole moments are stronger the electric repulsion forces. And this happens in atoms and molecules only. To calculate it one has to find an equation which take in account the neutralization of the electric electron-proton interaction in atoms. I’m not educated in quantum theories, but perhaps this is what they are doing?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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