Undergrad Pauli matrices and shared eigenvectors

Click For Summary
SUMMARY

The discussion centers on the commutation relations between the spin operators S2, Sz, and Sx, specifically regarding their shared eigenspaces. It is established that while S2 commutes with both Sz and Sx, they do not share the same eigenspace due to the non-commutativity of Sx and Sz. The key takeaway is that commutativity is not transitive; just because two operators commute with a third does not mean they commute with each other.

PREREQUISITES
  • Understanding of quantum mechanics and spin operators
  • Familiarity with the Pauli matrices (σx, σy, σz)
  • Knowledge of commutation relations in linear algebra
  • Concept of eigenspaces and eigenvectors
NEXT STEPS
  • Study the implications of non-commuting operators in quantum mechanics
  • Explore the properties of Pauli matrices in detail
  • Learn about the mathematical framework of eigenvalues and eigenvectors
  • Investigate the significance of commutation relations in quantum systems
USEFUL FOR

Students and professionals in quantum mechanics, physicists studying spin systems, and anyone interested in the mathematical foundations of quantum theory.

Sunny Singh
Messages
19
Reaction score
1
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?
 
Physics news on Phys.org
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.
 
  • Like
Likes vanhees71 and Sunny Singh
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.
 
  • Like
Likes vanhees71 and Sunny Singh
To say it in another way, commutativity is not transitive. ##[A,B] = 0## and ##[B,C]=0## does not imply that ##[A,C]=0##.
 
  • Like
Likes vanhees71
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 16 ·
Replies
16
Views
23K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
11K
  • · Replies 19 ·
Replies
19
Views
3K