1

Suppose that I measure the spin of two electrons in the z direction, and that they have the same spin. Then we put both electrons in a box. After a long time, we know nothing about the spatial distribution of the electrons in the box, so their spatial distributions are equal. But their spins in the z direction are still equal. This seems to contradict the Pauli Exclusion Principle. How can this be resolved?

Riemann
  • 1,430
  • 1
    Why do you believe the two electrons will be in the same state (where Pauli exclusion would be a factor)? – Jon Custer Jan 11 '23 at 16:42
  • "After a long time... their spatial distributions are equal." Why? And why would "a long time" only affect the spatial distribution and not the spin distribution? – hft Jan 11 '23 at 18:52
  • "This seems to contradict the Pauli Exclusion Principle. How can this be resolved?" It is resolved because your premise is wrong. – hft Jan 11 '23 at 18:52

1 Answers1

3

Why would we know nothing? If you place the electrons in an antisymmetric spatial state, say $$ \psi_{12}(x_1,x_2)=\frac1{\sqrt{2}}\left(\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2)\right) $$ so the full spatial+spin state is antisymmetric as required by Pauli: $$ \psi_{12}(x_1,x_2)\vert \uparrow\rangle_1\vert\uparrow\rangle_2 $$ then the state will evolve $$ \Psi(x_1,x_2,t)=e^{-i t E_{12}/\hbar} \psi_{12}(x_1,x_2)\vert \uparrow\rangle_1\vert\uparrow\rangle_2 $$ so the distribution is well known.

Even if you use a linear combination of antisymmetric states: $$ \psi(x_1,x_2)=\sum_{ij}c_{jk} \frac1{\sqrt{2}}\left(\psi_k(x_1)\psi_j(x_2) -\psi_j(x_1)\psi_k(x_2)\right) $$ the time evolution will be given by $$ \sum_{ij}c_{jk} e^{-i E_{jk}t/\hbar}\frac{1}{\sqrt{2}}\left(\psi_k(x_1)\psi_j(x_2) -\psi_j(x_1)\psi_k(x_2)\right)\vert\uparrow\rangle_1\vert\uparrow\rangle_2 $$ which is still (legitimately) antisymmetric.

ZeroTheHero
  • 45,515