What are the limitations of magnetisation for S=1/2 systems?

[Petar Mali]

If we have case

\sigma=\frac{1}{2}-\frac{1}{N}\sum_{\bf{k}}\langle\hat{S}^-\hat{S}^+\rangle_{\bf{k}}

where \sigma is magnetisation. How we know that \sigma must be less than \frac{1}{2}. Or why is

\frac{1}{N}\sum_{\bf{k}}\langle\hat{S}^-\hat{S}^+\rangle_{\bf{k}}>0

Thanks for your answer.

 

[daveyrocket]

Just look at the matrix elements of the spin raising and lowering operators.
Or alternatively, multiply out the spin operators to get

S^+ S^- = S^2 \sigma^+ \sigma^- = S^2 (\sigma_x^2 + \sigma_y^2 + 2\sigma_z)

The \sigma_i^2 matrix is the identity, so its expectation value is 1. \sigma_z has matrix elements of +1 and -1, so the quantity in the parentheses has to be between 0 and 4. S^2 is 1/4, so the result is between 0 and 1.

 

[Petar Mali]

Just look at the matrix elements of the spin raising and lowering operators.
Or alternatively, multiply out the spin operators to get

S^+ S^- = S^2 \sigma^+ \sigma^- = S^2 (\sigma_x^2 + \sigma_y^2 + 2\sigma_z)

The \sigma_i^2 matrix is the identity, so its expectation value is 1. \sigma_z has matrix elements of +1 and -1, so the quantity in the parentheses has to be between 0 and 4. S^2 is 1/4, so the result is between 0 and 1.

You use if I see well

\hat{S}^+=S\hat{\sigma}^+

\hat{S}^-=S\hat{\sigma}^-

and how you define \hat{\sigma}^+ and \hat{\sigma}^-?

 

[daveyrocket]

\sigma^{\pm} = \sigma_x \pm i\sigma_y