Pauli matrices and shared eigenvectors

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Discussion Overview

The discussion revolves around the properties of Pauli matrices and the conditions under which operators share eigenvectors, particularly focusing on the spin operators S2, Sz, and Sx. Participants explore the implications of commutation relations among these operators and the nature of their eigenspaces.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that since S2 commutes with Sz, they share their eigenspace, and similarly, since S2 also commutes with Sx, the eigenvectors of S2 and Sz should also be eigenvectors of Sx.
  • Another participant counters that while S2 and Sx share an eigenspace, it is distinct from the eigenspace shared by S2 and Sz, due to the non-commutation of Sx and Sz.
  • A further clarification is made that commutativity is not transitive, indicating that even if two pairs of operators commute, it does not guarantee that the first and last operators in the sequence also commute.

Areas of Agreement / Disagreement

Participants express disagreement regarding the implications of commutation on shared eigenspaces, with some asserting a misunderstanding of the relationships between the operators involved.

Contextual Notes

The discussion highlights the complexity of operator relationships in quantum mechanics, particularly the nuances of commutation and eigenspace sharing, without resolving the underlying assumptions or definitions involved.

Sunny Singh
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TL;DR
For a spin 1/2 particle, why does Sx, Sy and Sz don't share the complete eigenspace even though all of them commute with S^2
We know that S2 commutes with Sz and so they share their eigenspace. Now since S2 also commutes with Sx, as per my understanding, the eigenvectors of S2 and Sz should also be the eigenvectors of Sx. But since the paulic matrices σx and σy are not diagonlized in the eigenbasis of S2, it is clear that S2 and Sx don't share their eigenspace even though they commute with each other. How is that possible? what am i missing?
 
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To have common eigenspaces we need that all commute with each other. Every matrix commutes with ##I##, but that doesn't mean all matrices have the same eigenspace. And the ##S_{xyz}## do not commute.
 
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Sunny Singh said:
We know that S2 commutes with Sz and so they share their eigenspace.

Yes.

Sunny Singh said:
since S2 also commutes with Sx, as per my understanding, the eigenvectors of S2 and Sz should also be the eigenvectors of Sx.

Your understanding is incorrect. S2 and Sx share an eigenspace, but it's a different eigenspace from the one shared by S2 and Sz. The two eigenspaces must be different because Sx does not commute with Sz.
 
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To say it in another way, commutativity is not transitive. ##[A,B] = 0## and ##[B,C]=0## does not imply that ##[A,C]=0##.
 
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