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Spin Annhilation and Creator Operators Matrix Representation

  • #1
55
2

Homework Statement


Given the expression
[tex] s_{\pm}|s,m> = \hbar \sqrt{s(s+1)-m(m\pm 1)}|s,m \pm 1>[/tex]
obtain the matrix representations of s+/- for spin 1/2 in the usual basis of eigenstates of sz

Homework Equations


[tex] s_{\pm}|s,m> = \hbar \sqrt{s(s+1)-m(m\pm 1)}|s,m \pm 1>[/tex]
[tex] S_{+} = \hbar
\begin{bmatrix}
0 &1 \\
0 & 0
\end{bmatrix}
[/tex]
[tex] S_{-} = \hbar
\begin{bmatrix}
0 &0 \\
1 & 0
\end{bmatrix}
[/tex]

The Attempt at a Solution


So I've gotten the first part. You just sub into s and m for spin up or spin down yielding
[tex] s_{+}|\downarrow> = \hbar |\uparrow>[/tex]
[tex] s_{-}|\uparrow> = \hbar |\downarrow>[/tex]
In most textbooks I've checked, they just skip from what I've gotten above straight to the matrix representations. But I'm totally confused as to how the matrix elements of the matrices are found as you go from one to the other.
 

Answers and Replies

  • #2
Orodruin
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The matrix element in position ij describes the matrix element between the states i and j, ie,
$$\langle i| A | j\rangle = A_{ij}$$
In your case you only have two states (although the formula given can be used to express the action of ##s_\pm## on a state with higher total spin as well)

Edit: Also note that > is a LaTeX relation whereas \rangle is a LaTeX delimeter. Compare ##|a>## to ##|a\rangle##
 
  • #3
55
2
The matrix element in position ij describes the matrix element between the states i and j, ie,
$$\langle i| A | j\rangle = A_{ij}$$
In your case you only have two states (although the formula given can be used to express the action of ##s_\pm## on a state with higher total spin as well)

Edit: Also note that > is a LaTeX relation whereas \rangle is a LaTeX delimeter. Compare ##|a>## to ##|a\rangle##
Thank you so much for explaining. And thanks for the LaTex help. Tried using \ket but that didn't work so had to improvise
 

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