Collection of Solved Problems in Physics

Spin-1/2 projection along a direction in the xz-plane

Task number: 4413

• Hint 1

  Recall or look up the procedure for determining the eigenvalues and eigenvectors of matrices.

• Solution to hint 1

  The eigenvalues \(\lambda\) of a matrix \(A\) are obtained by solving the characteristic equation

  \[ \det \left ( A - \lambda \mathbb{E} \right ) = 0 \, , \]

  where \(\mathbb{E}\) denotes the identity matrix of the appropriate dimension.

  The eigenvector \(\vec u\) corresponding to the eigenvalue \(\lambda\) of the matrix \(A\) is obtained by solving the equation

  \[ \left ( A - \lambda \mathbb{E} \right ) \cdot \vec u = \vec o \, , \]

  where \(\vec o\) denotes the zero vector.

• Solution a)

  The eigenvalues \(\lambda\) are determined by substituting into the characteristic equation \(\det \left ( P - \lambda \mathbb{E} \right ) = 0\) and simplifying

  \[ \det \left ( P - \lambda \mathbb{E} \right ) = \det \left ( \frac{\hbar}{4} \begin{pmatrix} -\sqrt{3} & 1 \\ 1 & \sqrt{3} \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) = \det \begin{pmatrix} - \frac{\sqrt{3}}{4} \hbar - \lambda & \frac{\hbar}{4} \\ \frac{\hbar}{4} & \frac{\sqrt{3}}{4} \hbar - \lambda \end{pmatrix} = \] \[ = \lambda^2 - \frac{3}{16} \hbar^2 - \frac{\hbar^2}{16} = \lambda^2 - \frac{\hbar^2}{4} = \left (\lambda - \frac{\hbar}{2} \right ) \left (\lambda + \frac{\hbar}{2} \right ) = 0 \, . \]

  The eigenvalues are therefore \(\lambda_{1{,}2} = \pm \frac{\hbar}{2}\).

  We now focus on the eigenvector \(\vec u\) corresponding to the eigenvalue \(\lambda = + \frac{\hbar}{2}\). It is determined by substituting into the equation \(\left ( P - \lambda \mathbb{E} \right ) \cdot \vec u = \vec o\) and simplifying

  \[ \left ( P - \lambda \mathbb{E} \right ) \cdot \vec u = \left ( \frac{\hbar}{4} \begin{pmatrix} -\sqrt{3} & 1 \\ 1 & \sqrt{3} \end{pmatrix} - \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \] \[ = \frac{\hbar}{4} \begin{pmatrix} -\sqrt{3} - 2 & 1 \\ 1 & \sqrt{3} - 2 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equation by the factor \(\frac{\hbar}{4}\) and write it out component‑wise. This yields a system of two linear equations for the two unknowns

  \[ \left ( -\sqrt{3} - 2 \right ) u_1 + u_2 = 0 \, , \] \[ u_1 + \left ( \sqrt{3} - 2 \right ) u_2 = 0 \, . \]

  These equations are linearly dependent. From the second equation we therefore immediately see that one solution of the system is \(u_1 = 2 - \sqrt{3}, \, u_2 = 1\). In this way we have determined the direction of the eigenvector corresponding to the eigenvalue \(+ \frac{\hbar}{2}\). We now normalise it

  \[ \frac{1}{\sqrt{|u_1|^2 + |u_2|^2}} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \frac{1}{\sqrt{7 - 4 \sqrt{3} + 1}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} = \] \[ = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} \, . \]

  We thus obtain the unit eigenvector of the matrix \(P\) corresponding to the eigenvalue \(+ \frac{\hbar}{2}\)

  \[ \vec u = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} \, . \]

  For the eigenvalue \(- \frac{\hbar}{2}\), the procedure is analogous. The explicit calculation of the corresponding eigenvector can be found below in the section Comment. Here we proceed more directly. Since the two eigenvectors must be orthogonal to each other, the unit eigenvector corresponding to the eigenvalue \(- \frac{\hbar}{2}\) is

  \[ \vec v = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} - 1 \\ 2 - \sqrt{3} \end{pmatrix} \, . \]

• Comment – calculation of the second eigenvector

  We now focus on the eigenvector \(\vec v\) corresponding to the eigenvalue \(\lambda = - \frac{\hbar}{2}\). It is determined by substituting into the characteristic equation \(\left ( P - \lambda \mathbb{E} \right ) \cdot \vec v = \vec o\) and simplifying

  \[ \left ( P - \lambda \mathbb{E} \right ) \cdot \vec u = \left ( \frac{\hbar}{4} \begin{pmatrix} -\sqrt{3} & 1 \\ 1 & \sqrt{3} \end{pmatrix} + \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \] \[ = \frac{\hbar}{4} \begin{pmatrix} 2 -\sqrt{3} & 1 \\ 1 & 2 + \sqrt{3} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equation by the factor \(\frac{\hbar}{4}\) and write it out component‑wise. This yields a system of two linear equations for the two unknowns

  \[ \left ( 2 -\sqrt{3} \right ) v_1 + v_2 = 0 \, , \] \[ v_1 + \left ( 2 + \sqrt{3} \right ) v_2 = 0 \, . \]

  These equations are linearly dependent. From the first equation we therefore immediately see that one solution of the system is \(v_1 = - 1, \, v_2 = 2 - \sqrt{3}\). In this way we have determined the direction of the eigenvector corresponding to the eigenvalue \(- \frac{\hbar}{2}\). We now normalise it

  \[ \frac{1}{\sqrt{|v_1|^2 + |v_2|^2}} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \frac{1}{\sqrt{1 - 4 \sqrt{3} + 7}} \begin{pmatrix} -1 \\ 2 - \sqrt{3} \end{pmatrix} = \] \[ = \frac{1}{2 \sqrt{2 - \sqrt{3}}} \begin{pmatrix} -1 \\ 2 - \sqrt{3} \end{pmatrix} \, . \]

  We thus obtain the unit eigenvector of the matrix \(P\) corresponding to the eigenvalue \(- \frac{\hbar}{2}\)

  \[ \vec v = \frac{1}{2 \sqrt{2 - \sqrt{3}}} \begin{pmatrix} -1 \\ 2 - \sqrt{3} \end{pmatrix} \, . \]

• Hint 2

  Calculate or look up the matrix form of the operator \(\hat S_\theta\) representing the projection of the spin \(\frac{1}{2}\) along an arbitrary direction in the \(xz\)‑plane, as well as its eigenvectors. The direction is characterised by the angle \(\theta\), which specifies the deviation of this direction from the \(z\)‑axis and takes values from \(-π\) to \(π\). Calculate or look up how the probabilities of measuring the two possible spin‑\(\frac{1}{2}\) projections along this direction are determined.

  Calculate or look up the explicit form of the eigenvectors of the operators \(\hat S_z\) and \(\hat S_x\).

• Solution to hint 2

  The matrix \(\hat S_\theta\) representing the projection of the spin \(\frac{1}{2}\) along an arbitrary direction \(\vec n = (\sin \theta, \, 0, \, \cos \theta)\) in the \(xz\)‑plane is given by

  \[ \hat S_\theta = \hat{\vec S} \cdot \vec n = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & − \cos \theta \end{pmatrix} \, . \]

  A derivation of the explicit form of this matrix can be found in section SG experiment oriented at an arbitrary angle, Solution to hint 1.

  This matrix has unit eigenvectors \(|\theta+\rangle\) and \(|\theta-\rangle\) corresponding to the eigenvalues \(\lambda_{1, 2} = \pm \frac{\hbar}{2}\), given by

  \[ + \frac{\hbar}{2} \, \rightarrow \, |\theta+\rangle = \begin{pmatrix} \cos \frac{\theta}{2} \\ \sin \frac{\theta}{2} \end{pmatrix} \, , \] \[ - \frac{\hbar}{2} \, \rightarrow \, |\theta-\rangle = \begin{pmatrix} - \sin \frac{\theta}{2} \\ \cos \frac{\theta}{2} \end{pmatrix} \, . \]

  Their derivation is presented in section SG experiment oriented at an arbitrary angle, Solution – eigenvalues and eigenvectors.

  The probabilities \(P_+\) and \(P_-\) of measuring the positive and negative spin‑\(\frac{1}{2}\) projections, respectively, along the direction characterised by the angle \(\theta\) are given by

  \[ P_+ = \cos^2 \frac{\theta}{2} \, , \] \[ P_- = \sin^2 \frac{\theta}{2} \, . \]

  Their derivation can be found in section SG experiment oriented at an arbitrary angle, Solution – decomposition of the state entering the measurement.

  The operators \(\hat S_z, \, \hat S_x\), corresponding to the projections of the spin \(\frac{1}{2}\), have unit eigenvectors \(|z+\rangle, \,\) \(|z-\rangle, \,\) \(|x+\rangle, \,\) and \(|x-\rangle\), corresponding to the eigenvalues \(\lambda_{1, 2} = \pm \frac{\hbar}{2}\), given by

  \(+ \frac{\hbar}{2}\)	\(|z+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \, ,\)
  \(- \frac{\hbar}{2}\)	\(|z-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \, ,\)
  \(+ \frac{\hbar}{2}\)	\(|x+\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \, ,\)
  \(- \frac{\hbar}{2}\)	\(|x−\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 1 \end{pmatrix} \, .\)

• Hint 3

  Recall or look up why and how the state of a system prior to a measurement is decomposed into the basis of eigenvectors of the spin projection operator along the axis defined by the orientation of the magnet, and how the probabilities of measuring the individual outcomes are obtained from this decomposition.

• Solution to hint 3

  Owing to the measurement axiom, the state entering the measurement must be decomposed into a linear combination of the eigenstates of the observable being measured. The probabilities of obtaining the individual measurement outcomes are given by the squared modulus of the corresponding coefficients in this linear combination.

• Solution b)

  To determine the direction in the \(xz\)‑plane along which the matrix \(P\) represents a spin‑\(\frac{1}{2}\) projection operator, we compare this matrix with the general spin‑\(\frac{1}{2}\) projection matrix \(\hat S_\theta\) in the \(xz\)‑plane. This matrix is given by (see Hint 2)

  \[ \hat S_\theta = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & − \cos \theta \end{pmatrix} \, . \]

  By comparison we obtain

  \[ P = \hat S_\theta \, , \] \[ \frac{\hbar}{4} \begin{pmatrix} -\sqrt{3} & 1 \\ 1 & \sqrt{3} \end{pmatrix} = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix} \, . \]

  Dividing both sides by \(\frac{\hbar}{2}\) and equating the corresponding matrix elements yields

  \[ \cos \theta = - \frac{\sqrt{3}}{2} \, , \] \[ \sin \theta = \frac{1}{2} \, . \]

  The unique angle \(\theta\) in the interval \(\langle - \pi, \, \pi \rangle\) that satisfies these equations is \(\frac{5}{6} \pi\).

  The matrix \(P\) therefore represents the spin‑\(\frac{1}{2}\) projection operator along a direction in the \(xz\)‑plane deviating from the \(z\)‑axis by an angle of \(\frac{5}{6}\pi\).

  For verification, we compare the eigenvectors \(\vec u, \, \vec v\) with the eigenvectors of the general spin‑\(\frac{1}{2}\) projection matrix \(\hat S_\theta\) in the \(xz\)‑plane. These vectors are given by (see Hint 2)

  \[ + \frac{\hbar}{2} \, \rightarrow \, |\theta+\rangle = \begin{pmatrix} \cos \frac{\theta}{2} \\ \sin \frac{\theta}{2} \end{pmatrix} \, , \] \[ - \frac{\hbar}{2} \, \rightarrow \, |\theta-\rangle = \begin{pmatrix} - \sin \frac{\theta}{2} \\ \cos \frac{\theta}{2} \end{pmatrix} \, . \]

  Substituting \(\theta = \frac{5}{6} \pi\) yields

  \[ + \frac{\hbar}{2} \, \rightarrow \, \left |\frac{5}{6} \pi+ \right \rangle = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} = \vec u \, , \] \[ - \frac{\hbar}{2} \, \rightarrow \, \left |\frac{5}{6} \pi- \right \rangle = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} - 1 \\ 2 - \sqrt{3} \end{pmatrix} = \vec v \, . \]

  We observe that the eigenvectors coincide.

  Note: For this verification it would have been sufficient to compute any nonzero multiples of \(\vec u, \, \vec v\), as any scalar multiple of an eigenvector is also an eigenvector corresponding to the same eigenvalue.

• Solution c)

  The measurement axiom states that the description of a state must be decomposed into a linear combination of the eigenstates of the spin projection along the direction of measurement. We therefore express the state represented by the vector \(\vec u\) in the basis of eigenvectors of \(\hat S_z\) as

  \[ \vec u = c_1 |z+\rangle + c_2 |z-\rangle \, , \]

  where \(c_1, \, c_2\) are as yet unknown complex coefficients.

  Substituting the vectors \(\vec u, \, |z+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \, |z-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) allows us to determine the unknown coefficients

  \[ \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} 0 \\ 1 \end{pmatrix} \, . \]

  We immediately see that

  \[ c_1 = \frac{\sqrt{2 - \sqrt{3}}}{2} \, , \] \[ c_2 = \frac{1}{2\sqrt{2 - \sqrt{3}}} \, . \]

  The probabilities of measuring the spin‑\(\frac{1}{2}\) projections along the chosen direction are given by the squared magnitude of the corresponding coefficients, i.e.,

  \[ P_+ = |c_1|^2 \doteq 6{,}7 ~ \% \, , \] \[ P_- = |c_2|^2 \doteq 93{,}3 ~ \% \, . \]

  We know that the matrix \(P\) describes the spin‑\(\frac{1}{2}\) projection operator along a direction in the \(xz\)‑plane deviating from the \(z\)‑axis by an angle of \(\frac{5}{6} \pi\). The probabilities of measuring the individual spin projections can therefore also be determined (see Hint 2) as

  \[ P_+ = \cos^2 \frac{\theta}{2} = \cos^2 \frac{5}{12} \pi \doteq 6{,}7 ~ \% \, , \] \[ P_- = \sin^2 \frac{\theta}{2} = \sin^2 \frac{5}{12} \pi \doteq 93{,}3 ~ \% \, . \]

  We see that this result agrees with the previous calculation.

  For the eigenvector \(\vec v\), the procedure is analogous. However, we can exploit the fact that this vector differs from \(\vec u\) only in the ordering of components and a sign. The corresponding coefficients \(d_1, \, d_2\) in the decomposition \(\vec v = d_1 |z+\rangle + d_2 |z-\rangle\) are thus

  \[ d_1 = - c_2 = \frac{- 1}{2\sqrt{2 - \sqrt{3}}} \, , \] \[ d_2 = c_1 = \frac{\sqrt{2 - \sqrt{3}}}{2} \, . \]

  The probabilities of measuring the spin projections are then

  \[ P_+ = |d_1|^2 \doteq 93{,}3 ~ \% \, , \] \[ P_- = |d_2|^2 \doteq 6{,}7 ~ \% \, . \]

  The values of the probabilities are interchanged for this second direction, which can be understood as a measurement along an axis oriented in the opposite direction.

  Geometrically, these values can be interpreted such that the positive direction is “very close” to \(|z-\rangle\) and the negative direction is “very close” to \(|z+\rangle\).

• Solution d)

  In this part, we proceed analogously to the previous section, with the difference that the decomposition now takes the form

  \[ \vec u = f_1 |x+\rangle + f_2 |x-\rangle \, , \]

  where \(f_1, \, f_2\) are as yet unknown complex coefficients.

  We substitute the vectors \(\vec u, \, |x+\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \, |x-\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ 1 \end{pmatrix}\). Since this decomposition is not immediately “visible” as in the previous section, the coefficients \(f_1, \, f_2\) are determined using the scalar product

  \[ f_1 = \left \langle x+ | \, \vec u \right \rangle \, , \] \[ f_2 = \left \langle x- | \, \vec u \right \rangle \, . \]

  Let us calculate the coefficient \(f_1\)

  \[ f_1 = \left \langle x+ | \, \vec u \right \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} = \] \[ = \frac{2-\sqrt{3}+1}{2\sqrt2\sqrt{2-\sqrt{3}}} = \frac{3-\sqrt{3}}{2\sqrt{4-2\sqrt{3}}} \, . \]

  Now, using the identity \(4-2\sqrt{3} = (\sqrt{3}-1)^2\), we simplify

  \[ f_1 = \frac{3-\sqrt{3}}{2\sqrt{(\sqrt{3}-1)^2}} = \frac{\sqrt{3}(\sqrt{3}-1)}{2(\sqrt{3}-1)} = \frac{\sqrt{3}}{2} \, . \]

  The calculation of the coefficient \(f_2\) is very similar, so we provide it here briefly

  \[ f_2 = \left \langle x- | \, \vec u \right \rangle = \frac{-2 + \sqrt{3} + 1}{2\sqrt{2}\sqrt{2 - \sqrt{3}}} = \frac{\sqrt{3}-1}{2\sqrt{4-2\sqrt{3}}} = \frac{1}{2} \, . \]

  We now determine the probabilities of measuring the two spin‑\(\frac{1}{2}\) projections along the given direction. The probabilities of measuring each eigenvalue are equal to the squared modulus of the corresponding coefficient, i.e.,

  \[ P_+ = |f_1|^2 = \frac{3}{4} = 75 ~ \% \, , \] \[ P_- = |f_2|^2 = \frac{1}{4} = 25 ~ \% \, . \]

  We know that the matrix \(P\) describes the spin‑\(\frac{1}{2}\) projection operator along a direction in the \(xz\)‑plane deviating from the \(z\)‑axis by an angle of \(\frac{5}{6} \pi\). Equivalently, this direction is deviated from the \(x\)‑axis by an angle of \(\alpha = \frac{\pi}{3}\). The probabilities of measuring the individual spin projections can therefore also be determined as

  \[ P_+ = \cos^2 \frac{\alpha}{2} = \cos^2 \frac{\pi}{6} = \frac{3}{4} = 75 ~ \% \, , \] \[ P_- = \sin^2 \frac{\alpha}{2} = \sin^2 \frac{\pi}{6} = \frac{1}{4} = 25 ~ \% \, , \]

  which agrees with the general relation in Hint 2.

  For the eigenvector \(\vec v\), the procedure is analogous. However, we can use the fact that a negative projection along the given axis is equivalent to a positive projection along the oppositely oriented axis.

  Hence, the probabilities of measuring the individual spin projections are reversed compared to the previous case, i.e.,

  \[ P_+ = \frac{1}{4} = 25 ~ \% \, , \] \[ P_- = \frac{3}{4} = 75 ~ \% \, . \]

• Answer

  a) The eigenvalues \(\lambda_{1, 2}\) of the matrix \(P\) and their corresponding eigenvectors \(\vec u, \, \vec v\) are

  \[ \lambda_1 = + \frac{\hbar}{2} \,\, \rightarrow \,\, \vec u = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} 2 - \sqrt{3} \\ 1 \end{pmatrix} \, , \] \[ \lambda_2 = - \frac{\hbar}{2} \,\, \rightarrow \,\, \vec v = \frac{1}{2\sqrt{2 - \sqrt{3}}} \begin{pmatrix} - 1 \\ 2 - \sqrt{3} \end{pmatrix} \, . \]

  b) The matrix \(P\) represents the spin‑\(\frac{1}{2}\) projection operator along a direction in the \(xz\)‑plane deviated from the \(z\)‑axis by an angle of \(\frac{5}{6} \mathrm{\pi}\).

  c) The eigenvectors \(\vec u, \, \vec v\) of the matrix \(P\) expressed in the basis of eigenvectors of \(\hat S_z\) are

  \[ \vec u = \frac{\sqrt{2 - \sqrt{3}}}{2} |z+\rangle + \frac{1}{2\sqrt{2 - \sqrt{3}}} |z-\rangle \, , \] \[ \vec v = - \frac{1}{2\sqrt{2 - \sqrt{3}}} |z+\rangle + \frac{\sqrt{2 - \sqrt{3}}}{2} |z-\rangle \, . \]

  The probabilities of measuring the individual spin projections along the \(z\)‑axis in these states are

  • for the eigenvector \(\vec u\)
    • \(P_+ = \cos^2 \frac{5}{12} \pi \doteq 6{,}7 ~ \% \),
    • \(P_- = \sin^2 \frac{5}{12} \pi \doteq 93{,}3 ~ \%\),
  • for the eigenvector \(\vec v\)
    • \(P_+ = \sin^2 \frac{5}{12} \pi \doteq 93{,}3 ~ \% \),
    • \(P_- = \cos^2 \frac{5}{12} \pi \doteq 6{,}7 ~ \%\).

  d) The eigenvectors \(\vec u, \, \vec v\) of the matrix \(P\) expressed in the basis of eigenvectors of \(\hat S_x\) are

  \[ \vec u = \frac{\sqrt{3}}{2} |x+\rangle + \frac{1}{2} |x-\rangle \, , \] \[ \vec v = - \frac{1}{2} |x+\rangle + \frac{\sqrt{3}}{2} |x-\rangle \, . \]

  The probabilities of measuring the individual spin projections along the \(x\)‑axis in these states are

  • for the eigenvector \(\vec u\)
    • \(P_+ = \cos^2 \frac{\pi}{6} = \frac{3}{4} = 75 ~ \% \),
    • \(P_- = \sin^2 \frac{\pi}{6} = \frac{1}{4} = 25 ~ \%\),
  • for the eigenvector \(\vec v\)
    • \(P_+ = \sin^2 \frac{\pi}{6} = \frac{1}{4} = 25 ~ \% \),
    • \(P_- = \cos^2 \frac{\pi}{6} = \frac{3}{4} = 75 ~ \%\).

  Note: The angle \(\frac{\pi}{6}\) corresponds to the angle between the direction associated with the states having a sharp spin value and the direction along which the measurement is performed.