Collection of Solved Problems in Physics

SG experiment oriented at an arbitrary angle

Task number: 4409

• Hint 1

  Recall or look up the matrix form of the operator \(\hat S_\theta\) representing the spin‑\(\frac{1}{2}\) projection along an arbitrary direction in the \(xz\) plane. 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 \(π\).

• Solution to hint 1

  To determine this matrix, we first construct the operator for the spin‑\(\frac{1}{2}\) projection along the given direction \(\vec n\), which in the \(xz\) plane can be expressed in terms of the angle \(\theta\). The unit vector in the direction \(\vec n\) has the form

  \[ \vec n = \left ( \sin \theta, \, 0, \, \cos \theta \right ) \, . \]

  To obtain the spin‑\(\frac{1}{2}\) projection along this direction, we use the scalar product

  \[ \hat S_\theta = \hat{\vec S} \cdot \vec n = \hat S_x \sin \theta + \hat S_z \cos \theta \, . \]

  We now substitute the operators for the two spin projections and add them, which yields

  \[ \hat S_\theta = \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \sin \theta + \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & −1 \end{pmatrix} \cos \theta = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & − \cos \theta \end{pmatrix} \, . \]

• Hint 2

  Recall or look up how the eigenvalues and eigenvectors of matrices are determined.

• Solution to hint 2

  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 order.

  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.

• Hint 3

  Recall or look up how to decompose the state entering the measurement, and how the probabilities of obtaining the individual measurement outcomes can be determined 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

  Electrons with a spin projection along the \(z\)‑axis equal to \(+ \frac{\hbar}{2}\) enter the Stern–Gerlach apparatus. The eigenvector corresponding to this eigenvalue is \(\left |z+ \right \rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix}\).

  A beam of electrons in this state then passes through a magnet oriented at an angle \(\theta\). According to the measurement axiom, the state entering the measurement must first be expressed as a linear combination of the eigenstates of the measured observable.

• Solution – eigenvalues and eigenvectors

  To do this, we need to determine the eigenvalues and corresponding eigenvectors of the matrix \(\hat S_\theta\) representing the spin‑\(\frac{1}{2}\) projection along the direction characterised by the angle \(\theta\). This matrix has the form (see Hint 1)

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

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

  \[ \det \left ( \hat S_\theta - \lambda \hat{\mathbb{E}} \right ) = \det \left ( \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & − \cos \theta \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) = \] \[ = \det \begin{pmatrix} \frac{\hbar}{2} \cos \theta - \lambda & \frac{\hbar}{2} \sin \theta \\ \frac{\hbar}{2} \sin \theta & − \frac{\hbar}{2} \cos \theta - \lambda \end{pmatrix} = \lambda^2 - \frac{\hbar^2}{4} \cos^2 \theta - \frac{\hbar^2}{4} \sin^2 \theta = 0 \, . \]

  We now use the well‑known identity \(\sin^2 x + \cos^2 x = 1\) and the difference of squares formula to obtain

  \[ \lambda^2 - \frac{\hbar^2}{4} \cos^2 \theta - \frac{\hbar^2}{4} \sin^2 \theta = \lambda^2 - \frac{\hbar^2}{4} \left ( \cos^2 \theta + \sin^2 \theta \right ) = \] \[ = \lambda^2 - \frac{\hbar^2}{4} = \left (\lambda - \frac{\hbar}{2} \right ) \left (\lambda + \frac{\hbar}{2} \right ) = 0 \, . \]

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

  We now focus on the eigenvector corresponding to the positive spin projection \(+ \frac{\hbar}{2}\). The eigenvector \(\vec u\) associated with this eigenvalue \(\lambda\) is determined by substituting into the equation \(\left ( \hat S_\theta - \lambda \hat{\mathbb{E}} \right ) \cdot \vec u = \vec o\) and performing the following simplification

  \[ \left ( \hat S_\theta - \lambda \hat{\mathbb{E}} \right ) \cdot \vec u = \left ( \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & − \cos \theta \end{pmatrix} - \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \] \[ = \frac{\hbar}{2} \begin{pmatrix} \cos \theta - 1 & \sin \theta \\ \sin \theta & \cos \theta - 1 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equation by the factor \(\frac{\hbar}{2}\) and expand it into its component equations. This yields a system of two linear equations with two unknowns

  \[ \left ( \cos \theta - 1 \right ) u_1 + \sin \theta \, u_2 = 0 \, , \] \[ \sin \theta \, u_1 - \left ( \cos \theta + 1 \right ) u_2 = 0 \, . \]

  These equations are linearly dependent. From the second equation, we immediately see that one solution of the system is \(u_1 = \cos \theta + 1, \, u_2 = \sin \theta\). This determines the direction of the eigenvector corresponding to the eigenvalue \(+ \frac{\hbar}{2}\). We now normalise it

  \[ \frac{1}{\sqrt{|u_1|^2 + |u_2|^2}} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \frac{1}{\sqrt{\cos^2 \theta + 2 \cos \theta + 1 + \sin^2 \theta}} \begin{pmatrix} \cos \theta + 1 \\ \sin \theta \end{pmatrix} = \] \[ = \frac{1}{\sqrt{2 + 2 \cos \theta}} \begin{pmatrix} \cos \theta + 1 \\ \sin \theta \end{pmatrix} \, . \]

  Using the double‑angle trigonometric identities, the above expression can be rewritten as

  \[ \frac{1}{\sqrt{2 + 2 \cos \theta}} \begin{pmatrix} \cos \theta + 1 \\ \sin \theta \end{pmatrix} = \] \[ = \frac{1}{\sqrt{2 \left (1 + \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} \right )}} \begin{pmatrix} \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} = \] \[ = \frac{1}{\sqrt{4 \cos^2 \frac{\theta}{2}}} \begin{pmatrix} 2 \cos^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} \, . \]

  Taking the square root and dividing through, we obtain the normalised eigenvector in the form

  \[ \frac{1}{\sqrt{4 \cos^2 \frac{\theta}{2}}} \begin{pmatrix} 2 \cos^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} = \frac{1}{2 \cos \frac{\theta}{2}} \begin{pmatrix} 2 \cos^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} = \begin{pmatrix} \cos \frac{\theta}{2} \\ \sin \frac{\theta}{2} \end{pmatrix} \, . \]

  Since this eigenvector corresponds to the eigenvalue \(+ \frac{\hbar}{2}\), i.e., the positive spin projection along the direction characterised by the angle \(\theta\), we will denote this vector henceforth as

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

  For the eigenvalue \(- \frac{\hbar}{2}\), the procedure is analogous. The calculation of the corresponding eigenvector can be found at the end of this task in the section Comment – calculation of the second eigenvector. Here, we determine it using the fact that eigenvectors must be orthogonal. The normalised eigenvector corresponding to the eigenvalue \(- \frac{\hbar}{2}\), i.e., the negative spin projection along the direction characterised by the angle \(\theta\), is thus

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

• Solution – decomposition of the state entering the measurement

  We now express the state \(| z+ \rangle\) entering the measurement as a linear combination of the eigenstates of the measured observable

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

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

  Substituting the corresponding eigenvectors, we obtain

  \[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} = c_1 \begin{pmatrix} \cos \frac{\theta}{2} \\ \sin \frac{\theta}{2} \end{pmatrix} + c_2 \begin{pmatrix} - \sin \frac{\theta}{2} \\ \cos \frac{\theta}{2} \end{pmatrix} \, . \]

  Expanding this equality component‑wise gives a system of two linear equations with the two unknowns

  \[ 1 = c_1 \cos \frac{\theta}{2} - c_2 \sin \frac{\theta}{2} \, , \] \[ 0 = c_1 \sin \frac{\theta}{2} + c_2 \cos \frac{\theta}{2} \, . \]

  Solving the system yields \(c_1 = \cos \frac{\theta}{2}, \, c_2 = - \sin \frac{\theta}{2} \, \).

  The probabilities of obtaining the two spin projections along the given direction are then given by the squared modulus of the corresponding coefficients

  \[ P_+ = |c_1|^2 = \cos^2 \frac{\theta}{2} \, , \] \[ P_- = |c_2|^2 = \sin^2 \frac{\theta}{2} \, . \]

• Answer

  The probabilities of measuring the spin‑\(\frac{1}{2}\) projections along the direction tilted by an angle \(\theta\) from the \(z\)‑axis \((\varphi = 0)\) in the state \(|z+\rangle\) are

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

• Comment – calculation of the second eigenvector

  In this section, we focus on the eigenvector corresponding to the negative spin projection \(- \frac{\hbar}{2}\). The eigenvector \(\vec v\) associated with this eigenvalue \(\lambda\) is determined by substituting into the equation \(\left ( \hat S_\theta - \lambda \hat{\mathbb{E}} \right ) \cdot \vec v = \vec o\) and performing the following simplification

  \[ \left ( \hat S_\theta - \lambda \hat{\mathbb{E}} \right ) \cdot \vec v = \left ( \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & − \cos \theta \end{pmatrix} + \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \] \[ = \frac{\hbar}{2} \begin{pmatrix} \cos \theta + 1 & \sin \theta \\ \sin \theta & - \cos \theta + 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \, . \]

  We divide this equality by the factor \(\frac{\hbar}{2}\) and expand it into its component equations. This yields a system of two linear equations with two unknowns

  \[ \left ( \cos \theta + 1 \right ) v_1 + \sin \theta \, v_2 = 0 \, , \] \[ \sin \theta \, v_1 + \left ( 1 - \cos \theta \right ) v_2 = 0 \, . \]

  These equations are linearly dependent. From the second equation, we immediately see that one solution of the system is \(v_1 = \cos \theta - 1, v_2 = \sin \theta\). This determines the direction of the eigenvector corresponding to the eigenvalue \(- \frac{\hbar}{2}\). We now normalise it

  \[ \frac{1}{\sqrt{|v_1|^2 + |v_2|^2}} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \frac{1}{\sqrt{\cos^2 \theta - 2 \cos \theta + 1 + \sin^2 \theta}} \begin{pmatrix} \cos \theta - 1 \\ \sin \theta \end{pmatrix} = \] \[ = \frac{1}{\sqrt{2 - 2 \cos \theta}} \begin{pmatrix} \cos \theta - 1 \\ \sin \theta \end{pmatrix} \, . \]

  Using the double‑angle trigonometric identities, the above expression can be rewritten as

  \[ \frac{1}{\sqrt{2 - 2 \cos \theta}} \begin{pmatrix} \cos \theta - 1 \\ \sin \theta \end{pmatrix} = \] \[ = \frac{1}{\sqrt{2 \left (1 - \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} \right )}} \begin{pmatrix} \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} - \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} = \] \[ = \frac{1}{\sqrt{4 \sin^2 \frac{\theta}{2}}} \begin{pmatrix} -2 \sin^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} \, . \]

  Taking the square root and dividing through, we obtain the normalised eigenvector in the form

  \[ \frac{1}{\sqrt{4 \sin^2 \frac{\theta}{2}}} \begin{pmatrix} -2 \sin^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} = \frac{1}{2 \sin \frac{\theta}{2}} \begin{pmatrix} -2 \sin^2 \frac{\theta}{2} \\ 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \end{pmatrix} = \begin{pmatrix} - \sin \frac{\theta}{2} \\ \cos \frac{\theta}{2} \end{pmatrix} \, . \]

  Since this eigenvector corresponds to the eigenvalue \(- \frac{\hbar}{2}\), i.e., the negative spin projection along the direction characterised by the angle \(\theta\), we will denote this vector henceforth as

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

• Comment – calculation of the decomposition coefficients using the scalar product

  The sought coefficients \(c_1, \, c_2\) of the decomposition

  \[ |z+\rangle = c_1 |\theta +\rangle + c_2 |\theta -\rangle \]

  can be determined using the scalar product

  \[ c_1 = \langle \theta + | \, z+ \rangle = \begin{pmatrix} \cos \frac{\theta}{2} & \sin \frac{\theta}{2} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \cos \frac{\theta}{2} \, , \] \[ c_2 = \langle \theta - | \, z+ \rangle = \begin{pmatrix} - \sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = - \sin \frac{\theta}{2} \, . \]

  Both methods—solving a system of two linear equations for two unknowns or using the scalar product—are correct and yield the same results.

• Link

  In the problem Spinový stav daný úhlem θ (Czech version only), the reverse situation is considered, i.e., the spin projection along the \(z\)‑axis is measured in the state \(|\theta +\rangle\) (it is not stated there, however, that this is an eigenstate of \(\hat S_{\theta}\)). Comparing the results, we see that the measurement probabilities are the same. This is because the denotations of the directions do not matter. Only the relative angle between the direction corresponding to the eigenstate with positive spin projection and the orientation of the Stern–Gerlach apparatus is relevant.