Pauli matrices There are 3 of these. They are\[ \sigma _{1}=\begin {pmatrix} 0 & 1\\ 1 & 0 \end {pmatrix} ,\sigma _{1}=\begin {pmatrix} 0 & -i\\ i & 0 \end {pmatrix} ,\sigma _{1}=\begin {pmatrix} 1 & 0\\ 0 & -1 \end {pmatrix} \] There are also sometimes called \(\alpha _{x},\alpha _{y},\alpha _{z}\). Not to be confused by component \(x,y,z\) of an ordinary vector. Important property is that \(\sigma _{i}^{2}=\begin {pmatrix} 1 & 0\\ 0 & 1 \end {pmatrix} =I\). Also they are all Hermitians (i.e. \(A^{\dag }=A\))\(.\) This is obvious for the first and last matrix, since there are symmetric and real (we know if a matrix is real and also symmetric, it is also Hermitian.). Another important property is that they are unitary. i.e. \(A^{\dag }=A^{-1}\). Also any two anticommute. This means \(\left [ M,N\right ] _{+}=MN+NM\).\[ \left [ \sigma _{x},\sigma _{y}\right ] =2i\sigma _{z}\] For Pauli matrices, \(\left [ \sigma _{i},\sigma _{j}\right ] =2i\sum \epsilon _{ijk}\sigma _{k}\). Hence\begin {align*} \left [ \sigma _{1},\sigma _{2}\right ] & =2i\sigma _{3}\\ \left [ \sigma _{2},\sigma _{1}\right ] & =-2i\sigma _{3}\\ \left [ \sigma _{1},\sigma _{3}\right ] & =-2i\sigma _{2}\\ \left [ \sigma _{3},\sigma _{1}\right ] & =2i\sigma _{2}\\ \left [ \sigma _{2},\sigma _{3}\right ] & =2i\sigma _{1}\\ \left [ \sigma _{3},\sigma _{2}\right ] & =-2i\sigma _{1} \end {align*}
Eigenvalues of Pauli matrices can be only \(1,-1.\)\[ \operatorname {Tr}\left (\sigma _{i}\right ) =0 \] And Pauli matrices do not commute. This means \(\sigma _{x}\sigma _{y}\neq \sigma _{y}\sigma _{x}\).
Electron \(\frac {1}{2}\) spin matrices
| Spin matrix | Eigenvalues | Eigenvectors |
| \(S_{x}=\frac {\hbar }{2}\begin {bmatrix} 0 & 1\\ 1 & 0 \end {bmatrix} \) | \(\frac {\hbar }{2},-\frac {\hbar }{2}\) | \(\frac {1}{\sqrt {2}}\begin {bmatrix} 1\\ 1 \end {bmatrix} \qquad \frac {1}{\sqrt {2}}\begin {bmatrix} -1\\ 1 \end {bmatrix} \) |
| \(S_{y}=\frac {\hbar }{2}\begin {bmatrix} 0 & -i\\ i & 0 \end {bmatrix} \) | \(\frac {\hbar }{2},-\frac {\hbar }{2}\) | \(\frac {1}{\sqrt {2}}\begin {bmatrix} -i\\ 1 \end {bmatrix} \qquad \frac {1}{\sqrt {2}}\begin {bmatrix} i\\ 1 \end {bmatrix} \) |
| \(S_{z}=\frac {\hbar }{2}\begin {bmatrix} 1 & 0\\ 0 & -1 \end {bmatrix} \) | \(\frac {\hbar }{2},-\frac {\hbar }{2}\) | \(\begin {bmatrix} 1\\ 0 \end {bmatrix} \qquad \begin {bmatrix} 0\\ 1 \end {bmatrix} \) |
And using \(\left [ S_{i},S_{j}\right ] =i\hbar \sum _{k}\epsilon _{ijk}S_{k}\). Hence \(\left [ S_{1},S_{2}\right ] =i\hbar S_{3}\) and \(\left [ S_{1},S_{3}\right ] =-i\hbar S_{2}\) and \(\left [ S_{2},S_{1}\right ] =-i\hbar S_{3}\) and \(\left [ S_{2},S_{3}\right ] =i\hbar S_{1}\) and \(\left [ S_{3},S_{1}\right ] =-i\hbar S_{2}\) and \(\left [ S_{3},S_{2}\right ] =-i\hbar S_{1}\). Hence\begin {align*} \left [ S_{x},S_{y}\right ] & =i\hbar S_{z}\\ \left [ S_{y},S_{x}\right ] & =-i\hbar S_{z}\\ \left [ S_{x},S_{z}\right ] & =-i\hbar S_{y}\\ \left [ S_{z},S_{x}\right ] & =i\hbar S_{y}\\ \left [ S_{y},S_{z}\right ] & =i\hbar S_{x}\\ \left [ S_{z},S_{y}\right ] & =-i\hbar S_{x} \end {align*}
And\[ S_{i}=\frac {\hbar }{2}\sigma _{i}\] And\[ \sigma _{i}^{2}=I \] And\begin {align*} S_{+}^{\dag }S_{+} & =S^{2}-S_{z}^{2}-\hbar S_{z}\\ & =\hbar ^{2}\\ S_{-}^{\dag }S_{-} & =S^{2}-S_{z}^{2}+\hbar S_{z}\\ & =\hbar ^{2} \end {align*}
Where \(S^{2}=\frac {3}{4}\hbar ^{2}I\).
Electron \(1\) spin matrices
| Spin matrix | Eigenvalues | Eigenvectors |
| \(S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \) | \(1,0,-1\) | \(\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \qquad \begin {bmatrix} -\frac {1}{\sqrt {2}}\\ 0\\ \frac {1}{\sqrt {2}}\end {bmatrix} \qquad \begin {bmatrix} \frac {1}{2}\\ \frac {-1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) |
| \(S_{y}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \) | \(1,0,-1\) | \(\begin {bmatrix} -\frac {1}{2}\\ -\frac {i}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \qquad \begin {bmatrix} \frac {1}{\sqrt {2}}\\ 0\\ \frac {1}{\sqrt {2}}\end {bmatrix} \qquad \begin {bmatrix} -\frac {1}{2}\\ \frac {i}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) |
| \(S_{z}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \) | \(\frac {1}{\sqrt {2}},0,\frac {-1}{\sqrt {2}}\) | \(\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \qquad \begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \qquad \begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \) |
And\begin {align*} S_{+}^{\dag }S_{+} & =S^{2}-S_{z}^{2}-\hbar S_{z}\\ & =\hbar ^{2}\\ S_{-}^{\dag }S_{-} & =S^{2}-S_{z}^{2}+\hbar S_{z}\\ & =\hbar ^{2} \end {align*}
Where \(S^{2}=2\hbar ^{2}I=\hbar ^{2}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} \).
If we are given state vector \(V\) and asked to find expectation value when measuring along \(x\) axis, then do \(\langle V|S_{x}|V\rangle \)