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Problem: Spin-polarization principle

Any state of a single spin is an eigenvector of some component of the spin. In his book Quantum mechanics: The theoretical minimum, Leonard Susskind refers to this statement as the spin-polarization principle. Let us scrutinize the statement by considering the following problem.


Consider a spin-\(\tfrac{1}{2}\) particle in a generic spin-state given by \[\Psi = \begin{pmatrix} \psi_1 \\[0.1cm] \psi_2 \end{pmatrix} = \begin{pmatrix} e^{i \alpha} \cos \delta \\[0.1cm] e^{i \beta} \sin \delta \end{pmatrix} \,,\] where \(\alpha\), \(\beta\), \(\delta\) are some real numbers. The state is specified in the \(z\)-representation; that is, the states \[\begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 0 \\ 1 \end{pmatrix}\] have well-defined \(z\)-components of the spin equal to \(+\hbar/2\) and \(-\hbar/2\), respectively.

Find a unit vector \(\boldsymbol{n}\) (in three-dimensional Euclidean space) such that \(\Psi\) has a well-defined spin component along \(\boldsymbol{n}\) equal to \(+\hbar/2\).


For a solution, please see Issue #9 - Spin-polarization principle of Quantum Newsletter.