Pauli spin matrices, operating on |+> with Sx

  • Thread starter L-x
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  • #1
L-x
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Homework Statement


What is the result of operating on the state |+> with the operator Sx?

here, |+> denotes the eigenstate of Sz with eigenvalue 1/2. I am working in units where h-bar is 1 (for simplicity, and because I don't know how to type it)

Homework Equations


[tex]S_i = \frac{1}{2} σ_i [/tex]


The Attempt at a Solution


My understanding of the physics of the problem is that after measurement the system will be in state where the x component of its spin is certain, as [Sx,Sz] != 0 this means it will be in some superposition of the states |+> and |-> My intuition tells me that the probability for the particle to be found in either state should be equal.

However, operating with the matrix representation of Sx :[tex]\frac{1}{2}
\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)
[/tex]
on |+> just gives (1/2)|->

What am i doing wrong?
 

Answers and Replies

  • #2
L-x
66
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I think I understand where I've gone wrong now. The problem is in my intuition, not in my appilcation of the pauli matrices.

as [tex]|+> = \frac{1}{\sqrt{2}}(|+_x> + |-_x>) [/tex]
(where [tex]|+_x> and |-_x>[/tex] are the eigenvectors of S_x with eigenvalues +/- 1/2)

when we operate with Sx we obtain [tex](2^{-\frac{1}{2}})(\frac{1}{2}|+_x> - \frac{1}{2}|-_x>) [/tex] which is equal to [tex]\frac{1}{2}|->[/tex]. The confusion I had was due to a faliure to distinguish between operation with a linear operator and measurement of an ovservable.
 
Last edited:
  • #3
G01
Homework Helper
Gold Member
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Yes, your reasoning is correct, and you have the correct result.
 

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