(Verify normalization: (\int |\psi|^2 d\Omega = 1) indeed for the given coefficient.) Spin is an intrinsic degree of freedom. The spin operators (\hatS_x, \hatS_y, \hatS_z) obey the same commutation relations as orbital angular momentum:
Solution: First, (\langle S_x \rangle = \langle \psi | S_x | \psi \rangle = \frac\hbar2 \langle \psi | \sigma_x | \psi \rangle).
We also define ( \hatL^2 = \hatL_x^2 + \hatL_y^2 + \hatL_z^2 ), which commutes with each component: Quantum Mechanics Demystified 2nd Edition David McMahon
[ [\hatS_i, \hatS j] = i\hbar \epsilon ijk \hatS_k. ]
Solution: First, note that ( \sin\theta\cos\theta = \frac12\sin 2\theta ), and ( e^i\phi ) suggests ( m=1 ). But let’s check normalization and (L_z) action: ( \hatL_z = -i\hbar \frac\partial\partial\phi ). Applying to (\psi): ( -i\hbar \frac\partial\partial\phi \psi = -i\hbar (i) \psi = \hbar \psi ). Thus (\psi) is an eigenstate of (L_z) with eigenvalue ( \hbar ). So ( \langle L_z \rangle = \hbar ). (Verify normalization: (\int |\psi|^2 d\Omega = 1) indeed
[ \sigma_x = \beginpmatrix 0 & 1 \ 1 & 0 \endpmatrix,\quad \sigma_y = \beginpmatrix 0 & -i \ i & 0 \endpmatrix,\quad \sigma_z = \beginpmatrix 1 & 0 \ 0 & -1 \endpmatrix. ]
[ \hatL^2 |l,m\rangle = \hbar^2 l(l+1) |l,m\rangle, \quad l = 0, 1, 2, \dots ] [ \hatL_z |l,m\rangle = \hbar m |l,m\rangle, \quad m = -l, -l+1, \dots, l. ] ] Solution: First, note that ( \sin\theta\cos\theta =
An electron is in state (|\psi\rangle = \frac1\sqrt2 \beginpmatrix 1 \ i \endpmatrix). Find (\langle S_x \rangle) and (\langle S_y \rangle).