Collection of Solved Problems in Physics

Arbitrary spin state

Task number: 4411

• 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 the relation for calculating the expectation value of a physical observable \(F\) in a state described by the spinor \(\psi\).

• Solution to hint 2

  The expectation value of a physical observable \(F\) in a state described by the spinor \(\psi\) can be determined from

  \[\left\langle F\right\rangle_{\psi}=\frac{\left\langle\psi| \hat{F}\psi\right\rangle}{\left\langle \psi|\psi\right\rangle},\]

  where \(\hat{F}\) is the operator associated with the physical observable \(F\).

• Hint 3

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

• Solution to hint 3

  If the state is described by a normalised spinor \(\psi\), which we decompose 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 \(\lambda_i\) (whose corresponding eigenstate is described by the eigenfunction \(\psi_i\)) is

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

• Hint 4

  The scalar product in the matrix formalism is computed in the same way as the ordinary scalar product

  \[\left\langle\begin{pmatrix} a\\ b \end{pmatrix}\Bigg|\begin{pmatrix} x\\ y \end{pmatrix}\right\rangle=\begin{pmatrix} a & b \end{pmatrix}\mbox{*}\begin{pmatrix} x\\ y \end{pmatrix}=a\mbox{*}x+b\mbox{*}y,\]

  where it is necessary to use the complex conjugate of the first vector.

• Hint 5

  If we wish to decompose an arbitrary state represented by the spinor \(\psi\) into a linear combination of the vectors of an orthonormal basis \(|n_i\rangle\), then the coefficients \(c_i\) of this decomposition are given by the scalar product

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

• Solution

  First, we find the normalised form of the given spinor describing the state of the particle

  \[\psi_n=\frac{\psi}{\sqrt{\left\langle\psi|\psi\right\rangle}}=\frac{\begin{pmatrix} a\\ b \end{pmatrix}}{\sqrt{\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}}}=\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}.\]

  We expand the considered state into a linear combination of the eigenstates of the spin projection operators for a spin‑\(\frac{1}{2}\) particle along the \(x\), \(y\) and \(z\) axes. An overview of these operators, their eigenvalues, and eigenvectors can be found in Solution to hint 1.

  First, let us perform the calculation for the \(z\)‑axis direction

  \[\psi_n=\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}=\frac{a}{\sqrt{a^2+b^2}}\begin{pmatrix} 1\\ 0 \end{pmatrix}+\frac{b}{\sqrt{a^2+b^2}}\begin{pmatrix} 0\\ 1 \end{pmatrix}=\frac{a}{\sqrt{a^2+b^2}}|z+\rangle+\frac{b}{\sqrt{a^2+b^2}}|z−\rangle.\]

  Since the decomposition into the eigenstates of the operator \(\hat{S}_x\) and \(\hat{S}_y\) is not immediately obvious, we calculate the coefficients using the scalar product. If the first vector in the product is complex, its complex conjugate must be taken. Thus

  \[\left\langle x+\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\left\langle\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\frac{a+b}{\sqrt{2}\sqrt{a^2+b^2}},\] \[\left\langle x−\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\left\langle\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ −1 \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\frac{a−b}{\sqrt{2}\sqrt{a^2+b^2}},\] \[\left\langle y+\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\left\langle\frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ i \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\frac{a−ib}{\sqrt{2}\sqrt{a^2+b^2}},\] \[\left\langle y−\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\left\langle\frac{1}{\sqrt{2}}\begin{pmatrix} i\\ 1 \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\frac{−ia+b}{\sqrt{2}\sqrt{a^2+b^2}}.\]

  We can now write

  \[\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}=\frac{a+b}{\sqrt{2}\sqrt{a^2+b^2}}|x+\rangle+\frac{a−b}{\sqrt{2}\sqrt{a^2+b^2}}|x−\rangle,\] \[\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}=\frac{a−ib}{\sqrt{2}\sqrt{a^2+b^2}}|y+\rangle+\frac{−ia+b}{\sqrt{2}\sqrt{a^2+b^2}}|y−\rangle.\]

  Since the spinor describing the particle’s state before measurement is expanded in terms of linear combinations of eigenstates, and all the spinors used are normalised, the probabilities of measuring the eigenvalues of the spin projections along each axis are given by the squared magnitudes of the corresponding coefficients in the linear combination. Thus, along the \(x\) axis

  \[P_{\frac{\hbar}{2}}^{(x)}=\frac{(a+b)^2}{2(a^2+b^2)},\qquad P_{−\frac{\hbar}{2}}^{(x)}=\frac{(a−b)^2}{2(a^2+b^2)},\]

  along the \(y\) axis

  \[P_{\frac{\hbar}{2}}^{(y)}=\frac{a−ib}{\sqrt{2}\sqrt{a^2+b^2}}\frac{a+ib}{\sqrt{2}\sqrt{a^2+b^2}}=\frac{a^2+b^2}{2(a^2+b^2)}=\frac{1}{2},\] \[P_{−\frac{\hbar}{2}}^{(y)}=\frac{a+ib}{\sqrt{2}\sqrt{a^2+b^2}}\frac{a−ib}{\sqrt{2}\sqrt{a^2+b^2}}=\frac{a^2+b^2}{2(a^2+b^2)}=\frac{1}{2},\]

  and along the \(z\) axis

  \[P_{\frac{\hbar}{2}}^{(z)}=\frac{a^2}{a^2+b^2},\qquad P_{−\frac{\hbar}{2}}^{(z)}=\frac{b^2}{a^2+b^2}.\]

  The individual probabilities along each axis always sum to one.

   

  Finally, let us calculate the expectation values of the spin projection along the individual axes using the general relation

  \[\left\langle S_i\right\rangle_{\psi}=\left\langle\psi| \hat{S}_i\psi\right\rangle.\]

  After substituting, we obtain

  \[\left\langle S_x\right\rangle_{\psi}=\left\langle\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1& 0 \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\frac{\hbar}{2(a^2+b^2)}\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} b\\ a \end{pmatrix}=\] \[=\frac{ab\hbar}{a^2+b^2},\] \[\left\langle S_y\right\rangle_{\psi}=\left\langle\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\frac{\hbar}{2}\begin{pmatrix} 0 & −i \\ i& 0 \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}\right\rangle=\frac{\hbar}{2(a^2+b^2)}\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} −ib\\ ia \end{pmatrix}=\] \[=\frac{\hbar}{2(a^2+b^2)}(−iba+iab)=0,\] \[\left\langle S_z\right\rangle_{\psi}=\left\langle\frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix} a\\ b \end{pmatrix}\Bigg|\frac{1}{\sqrt{a^2+b^2}}\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0& −1 \end{pmatrix}\begin{pmatrix} a\\ −b \end{pmatrix}\right\rangle=\frac{\hbar}{2(a^2+b^2)}\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} a\\ −b \end{pmatrix}=\] \[=\frac{\hbar}{2}\frac{a^2−b^2}{a^2+b^2}.\]

• Answer

  If we consider a particle with spin \(\frac{1}{2}\) in a state described by the spinor

  \[\psi=\begin{pmatrix} a\\ b \end{pmatrix},\]

  where \(a\) and \(b\) are real numbers, the probabilities of measuring the projection of spin‑\(\frac{1}{2}\) along the \(x-\), \(y-\), and \(z-\)axes are given by

  \[P_{\frac{\hbar}{2}}^{(x)}=\frac{(a+b)^2}{2(a^2+b^2)},\,\,P_{−\frac{\hbar}{2}}^{(x)}=\frac{(a−b)^2}{2(a^2+b^2)},\] \[P_{\frac{\hbar}{2}}^{(y)}=\frac{1}{2},\,\,P_{−\frac{\hbar}{2}}^{(y)}=\frac{1}{2},\] \[P_{\frac{\hbar}{2}}^{(z)}=\frac{a^2}{a^2+b^2},\,\,P_{−\frac{\hbar}{2}}^{(z)}=\frac{b^2}{a^2+b^2}.\]

  The expectation values of the spin projection along the respective axes are

  \[\left\langle S_x\right\rangle_{\psi}=\frac{ab\hbar}{a^2+b^2},\] \[\left\langle S_y\right\rangle_{\psi}=0,\] \[\left\langle S_z\right\rangle_{\psi}=\frac{\hbar}{2}\frac{a^2−b^2}{a^2+b^2}.\]