Collection of Solved Problems in Physics

Spin-1/2 projection along the direction (1, 0, 1)

Task number: 4412

• Hint 1

  Recall or look up the matrix forms of the spin‑\(\frac{1}{2}\) projection operators along the \(x\), \(y\) and \(z\) axes, and derive or look up their eigenvalues and the corresponding eigenvectors.

• Solution to hint 1

  The spin‑\(\frac{1}{2}\) projection operators along the \(x\), \(y\), and \(z\) axes are given by

  \[\hat{\vec{S}}=\frac{\hbar}{2}\left(\begin{pmatrix} 0 &1 \\ 1& 0 \end{pmatrix},\begin{pmatrix} 0 &−i \\ i& 0 \end{pmatrix},\begin{pmatrix} 1 &0 \\ 0& −1 \end{pmatrix}\right).\]

  The eigenvalues of each matrix are \(\lambda_{1{,}2}=\pm\frac{\hbar}{2}\), and the corresponding eigenvectors are

  \[|x+\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \end{pmatrix},\, |x−\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ −1 \end{pmatrix},\] \[|y+\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ i \end{pmatrix},\, |y−\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} i \\ 1 \end{pmatrix},\] \[|z+\rangle=\begin{pmatrix} 1\\ 0 \end{pmatrix},\, |z−\rangle=\begin{pmatrix} 0 \\ 1 \end{pmatrix}.\]

• Hint 2

  Recall or look up how to calculate the probability of measuring an eigenvalue of the operator \(\hat{F}\) in the state described by the wavefunction \(\psi\).

• Solution to hint 2

  If the state is described by a normalised wavefunction \(\psi\), which we expand as a linear combination of mutually orthogonal and normalised eigenfunctions \(\psi_i\) of the operator \(\hat{F}\), i.e.,

  \[\psi=\sum_{i}c_i\psi_i \, ,\]

  then the probability \(P_{\lambda_i}\) of measuring the eigenvalue \(\lambda_i\) (with the corresponding eigenstate described by the eigenfunction \(\psi_i\)) is given by

  \[P_{\lambda_i}=|c_i|^2 \, .\]

• Hint 3

  If we wish to express an arbitrary state described by the wavefunction \(\psi\) as a linear combination of the vectors of an orthonormal basis \(|n_i\rangle\), then the coefficients \(c_i\) in this expansion are given by the scalar product

  \[c_i=\langle n_i|\psi\rangle.\]

• Solution

  The initial state of the electron is an eigenstate corresponding to the negative spin‑\(\frac{1}{2}\) projection along the \(x\)‑axis, i.e.,

  \[|x−\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ −1 \end{pmatrix}.\]

  First, we normalise the vector specifying the direction \((1,\,0,\,1)\)

  \[\vec{n}=\frac{1}{\sqrt2}(1,\,0,\,1).\]

  Next, we determine the spin‑\(\frac{1}{2}\) projection operator along the direction \(\vec{n}\)

  \[\hat{S_\vec{n}}=\hat{\vec{S}}·\vec{n}=\frac{\hbar}{2}\left(\begin{pmatrix} 0 &1 \\ 1& 0 \end{pmatrix},\begin{pmatrix} 0 &−i \\ i& 0 \end{pmatrix},\begin{pmatrix} 1 &0 \\ 0& −1 \end{pmatrix}\right)·\frac{1}{\sqrt2}(1{,}0,1)=\frac{\hbar}{2\sqrt{2}}\begin{pmatrix} 1 &1 \\ 1& −1 \end{pmatrix}.\]

  Let us find the eigenvalues of the matrix \(\hat{S_\vec{n}}\)

  \[\begin{vmatrix} \frac{\hbar}{2\sqrt{2}}−\lambda &\frac{\hbar}{2\sqrt{2}} \\ \frac{\hbar}{2\sqrt{2}}& −\frac{\hbar}{2\sqrt{2}}−\lambda \end{vmatrix}=\] \[−\left(\frac{\hbar^2}{8}−\lambda^2+\frac{\hbar^2}{8}\right)=−\left(\frac{\hbar^2}{4}−\lambda^2\right)=0,\]

  from which it follows that

  \[\lambda^2=\frac{\hbar^2}{4},\] \[\lambda=\pm\frac{\hbar}{2}.\]

  This result is as expected, since the spin‑\(\frac{1}{2}\) projection along any direction can take only the values \(\pm\frac{\hbar}{2}\).

  Let us now find the eigenvector corresponding to the eigenvalue \(\lambda=\frac{\hbar}{2}\), i.e., the spinor describing the state with a positive spin‑\(\frac{1}{2}\) projection along the direction \(\vec{n}\)

  \[\frac{\hbar}{2\sqrt{2}}\begin{pmatrix} 1−\sqrt{2} &1 \\ 1& −1−\sqrt{2} \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}=\vec{o},\] \[(1−\sqrt{2})a+b=0,\]

  choosing \(a=1\), the eigenvector is

  \[|\vec{n}+\rangle=\begin{pmatrix} 1\\ \sqrt{2}−1 \end{pmatrix}.\]

  Let us normalise this vector

  \[\langle\vec{n}+|\vec{n}+\rangle=1+2-2\sqrt{2}+1=4−2\sqrt{2},\]

  thus the normalised vector corresponding to the positive spin‑\(\frac{1}{2}\) projection along \(\vec{n}\) is

  \[|\vec{n}+\rangle=\frac{1}{\sqrt{2(2−\sqrt{2})}}\begin{pmatrix} 1\\ \sqrt{2}−1 \end{pmatrix}.\]

  We can now calculate the coefficient of the linear combination for decomposing the original state into the eigenstates of the matrix \(\hat{S_\vec{n}}\) using the scalar product

  \[\langle \vec{n} +|x−\rangle = \frac{1}{\sqrt{2(2−\sqrt{2})}}\begin{pmatrix} 1 & \sqrt{2}−1 \end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ −1 \end{pmatrix}=\frac{1}{2\sqrt{2−\sqrt{2}}}(1−\sqrt{2}+1)=\] \[=\frac{2-\sqrt{2}}{2\sqrt{2−\sqrt{2}}}=\frac{\sqrt{2−\sqrt{2}}}{2},\]

  so the probability of measuring a positive spin‑\(\frac{1}{2}\) projection along the direction \(\vec{n}\) is

  \[P_+=\left(\frac{\sqrt{2−\sqrt{2}}}{2}\right)^2=\frac{2−\sqrt{2}}{4}\doteq0{,}15.\]

• Answer

  Considering an electron whose spin projection along the \(x\)‑axis is negative, the probability of measuring a positive value for the spin projection along the direction \((1,\,0,\,1)\) is

  \[P_+=\frac{2−\sqrt{2}}{4}\doteq0{,}15.\]