Solution: First, (\langle S_x \rangle = \langle \psi | S_x | \psi \rangle = \frac\hbar2 \langle \psi | \sigma_x | \psi \rangle).
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 ).
For a particle (e.g., electron, proton, neutron), the eigenvalues of (\hatS^2) are (\hbar^2 s(s+1)) with (s = 1/2), and eigenvalues of (\hatS_z) are (\pm \hbar/2). Quantum Mechanics Demystified 2nd Edition David McMahon
[ \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. ]
[ \sigma_x |\psi\rangle = \beginpmatrix 0&1\1&0 \endpmatrix \frac1\sqrt2 \beginpmatrix 1\ i \endpmatrix = \frac1\sqrt2 \beginpmatrix i \ 1 \endpmatrix. ] [ \langle \psi | \sigma_x | \psi \rangle = \frac1\sqrt2 \beginpmatrix 1 & -i \endpmatrix \cdot \frac1\sqrt2 \beginpmatrix i \ 1 \endpmatrix = \frac12 (i - i) = 0. ] So (\langle S_x \rangle = 0). Solution: First, (\langle S_x \rangle = \langle \psi
[ [\hatS_i, \hatS j] = i\hbar \epsilon ijk \hatS_k. ]
(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: For a particle (e
7.1 Introduction In classical mechanics, angular momentum is a familiar concept: for a particle moving with momentum p at position r , the orbital angular momentum is L = r × p . In quantum mechanics, angular momentum becomes an operator, and its components do not commute. This leads to quantization, discrete eigenvalues, and the surprising property of spin – an intrinsic angular momentum with no classical analogue.