Spin 1 particle, spinor state

  • Thread starter Simp
  • Start date
  • #1
2
0
Hi there,
I have a question, something that is confusing me.
If a particle of spin 1 is measured to have m=1 along the x direction, would the spinor state just be a column vector with (1,0,0), which would also be the spinor if x was infact z. OR would the spinor be determined by multiplying the Sx matrix by a column vector of (a,b,c,d) and letting this equal the eigenvalue (hbar in this case) muliplied by column vector (a,b,c,d) and working out their values and normalising.?
Thanks!
 

Answers and Replies

  • #2
haushofer
Science Advisor
Insights Author
2,386
799
Well, I would write down the three-dimensional representation of SU(2) and look for the appropriate eigenvectors (which you call spinor states). An m=1 state along x is described by an eigenvector of S_x with eigenvalue 1, right?
 
  • #3
2
0
I am not sure what you mean by SU(2) ?
 
  • #4
Simp, welcome to PhysicsForums.
However, your thread is in the wrong forum, as this is the Quantum Mechanics forum and not the Physics Homework forum. If you need assistance with your homework problems, may I direct you to the appropriate forum: https://www.physicsforums.com/forumdisplay.php?f=152
Thank you.
 

Related Threads on Spin 1 particle, spinor state

Replies
5
Views
6K
Replies
3
Views
711
Replies
5
Views
2K
  • Last Post
Replies
7
Views
2K
Replies
2
Views
686
Replies
1
Views
1K
  • Last Post
Replies
7
Views
741
Replies
1
Views
1K
Replies
7
Views
4K
  • Last Post
Replies
6
Views
2K
Top