Angular momentum

An angular momentum operator \(\require{physics} \vb{T}\) is a vector operator whose Hermitian components satisfy the following commutation relations: \[\begin{align} & \comm{T^x}{T^y} = i T^z \\ & \comm{T^y}{T^z} = i T^x \\ & \comm{T^z}{T^x} = i T^y \thinspace , \end{align}\] which are the commutation relations for the \(\mathfrak{su}(2)\) Lie algebra.

We often introduce raising and lowering operators \(T^+\) and \(T^-\), as: \[\begin{align} & T^+ = T^x + i T^y \\ & T^- = T^x - i T^y \thinspace , \end{align}\] or, reversely as: \[\begin{align} & T^x = \frac{1}{2} \qty(T^+ + T^-) \\ & T^y = \frac{1}{2i} \qty(T^+ - T^-) \\ & T^z = T^z \thinspace , \end{align}\] with the commutators \[\begin{align} & \comm{T^+}{T^-} = 2 T^z \\ & \comm{T^z}{T^\pm} = \pm T^\pm \thinspace , \end{align}\] and \(T^+\) being the Hermitian adjoint of \(T^-\) \[\begin{equation} T^- = \qty(T^+)^\dagger \thinspace , \end{equation}\] and \(T^z\) being Hermitian: \[\begin{equation} \qty(T^z)^\dagger = T^z \thinspace . \end{equation}\]

Given the components of \(\vb{T}\), we introduce its squared norm as \[\begin{align} T^2 \equiv \vb{T} \vdot \vb{T} &= \qty(T^x)^2 + \qty(T^y)^2 + \qty(T^z)^2 \\ &= T^+ T^- - T^z + \qty(T^z)^2 \\ &= T^- T^+ + T^z + \qty(T^z)^2 \thinspace , \end{align}\] which obeys the following commutation relations: \[\begin{align} & \comm{T^2}{T^x} = \comm{T^2}{T^y} = \comm{T^2}{T^z} = 0 \\ % & \comm{T^2}{T^+} = \comm{T^2}{T^-} = \comm{T^2}{T^z} = 0 \thinspace . \end{align}\]