Expectation values of the spin components in a general state
Task number: 4414
For the most general normalised spinor \[\chi = \begin{pmatrix} a \\ b \end{pmatrix} \, ,\] where \(a, \, b\) are complex numbers, compute the following:
a) \(\langle S_x \rangle, \, \langle S_x \rangle, \, \langle S_x \rangle,\)
b) \(\langle S^2_x \rangle, \, \langle S^2_y \rangle, \,\) and \(\langle S^2_z \rangle\).
c) Verify that the spin angular momentum squared operator \(S^2\) possesses eigenvalues of the same form as those of the orbital angular momentum squared operator \(L^2\). Determine the expectation value of \(S^2\) in the state \(\chi\).
Hint 1
Recall how the expectation value of the observable \(F\) in a state described by the normalised spinor \(\chi\) can be determined.
Hint 2
Recall the matrix form of the spin-\(\frac{1}{2}\) projections along all three axes \(x, y, z\).
Hint 3
Recall the form of the eigenvalues of the square of the angular momentum operator \(\hat L^2\).
Solution a)
We start from the general expression for the expectation value, into which we substitute and then simplify
\[ \left \langle S_x \right \rangle_{\chi} = \chi^{\dagger} \hat S_x \chi = \begin{pmatrix} a^* & b^* \end{pmatrix} \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \] \[ = \frac{\hbar}{2} \begin{pmatrix} b^* & a^* \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} =\frac{\hbar}{2}({ab}^*+a^*b) \, . \]An analogous procedure is followed for \(\langle S_y \rangle_\chi\) and \(\langle S_z \rangle_\chi\)
\[ \left \langle S_y \right \rangle_{\chi} = \chi^{\dagger} \hat S_y \chi = \begin{pmatrix} a^* & b^* \end{pmatrix} \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \] \[ = \frac{\hbar}{2} \begin{pmatrix} {ib}^* & {-ia}^* \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} =\frac{i\hbar}{2}({ab}^*{-a}^*b) \, , \] \[ \left \langle S_z \right \rangle_{\chi} = \chi^{\dagger} \hat S_z \chi = \begin{pmatrix} a^* & b^* \end{pmatrix} \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \] \[ = \frac{\hbar}{2} \begin{pmatrix} a^* & {-b}^* \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}= \frac{\hbar}{2}({aa}^*-{bb}^*) = \frac{\hbar}{2}(|a|^2 - |b|^2) \, . \]Note: Although complex numbers appear in the expressions above, all three expectation values are real. This can be most easily verified for \(\hat S_z\), where we immediately see that one can measure either the value \(\frac{\hbar}{2}\) with probability \(|a|^2\) or the value \(- \frac{\hbar}{2}\) with probability \(|b|^2\).
Solution b)
First, we calculate the operators corresponding to the squares of the individual spin projections
\[ \hat S_x^2 = \hat S_x \hat S_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \frac{\hbar^2}{4}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{\hbar^2}{4} \hat {\mathbb E} \, , \]where \(\hat {\mathbb E}\) denotes the \(2×2\) identity matrix; this notation is used throughout the task. The calculation is analogous for \(\hat S_y^2\) and \(\hat S_z^2\)
\[ \hat S_y^2 = \hat S_y \hat S_y = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \frac{\hbar^2}{4}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{\hbar^2}{4} \hat {\mathbb E} \, , \] \[ \hat S_z^2 = \hat S_z \hat S_z = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \frac{\hbar^2}{4}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{\hbar^2}{4} \hat {\mathbb E} \, . \]All three operators turn out to be identical and proportional to the identity matrix; therefore, the expectation value of the square of the spin projection along each of the axes \(x, \, y, \, z\) is \(\frac{\hbar^2}{4}\). A detailed calculation can be found in section Comment.
Solution c)
First, we determine the form of the operator \(\hat S^2\). To this end, we use the relation \(\hat S_x^2 + \hat S_y^2 + \hat S_z^2 = \hat S^2\), into which we substitute the previously determined values, yielding
\[ \hat S^2 = \frac{\hbar^2}{4} \hat{\mathbb{E}} + \frac{\hbar^2}{4} \hat{\mathbb{E}} + \frac{\hbar^2}{4} \hat{\mathbb{E}} = \frac{3}{4} \hbar^2 \hat{\mathbb{E}} \, . \]The identity matrix has an eigenvalue of \(1\). It follows that \(\hat S^2\) has a single eigenvalue equal to \(\frac{3}{4} \hbar^2\).
The eigenvalues of the square of the angular momentum operator are of the form \(\hbar^2 l(l+1)\). Substituting \(l = s = \frac{1}{2}\) gives
\[ \hbar^2 s(s+1) = \hbar^2 \frac{1}{2} (\frac{1}{2} + 1) = \hbar^2 \frac{1}{2} \frac{3}{2} = \frac{3}{4} \hbar^2 \, . \]This verifies that the operator for the square of the spin, \(\hat S^2\), has eigenvalues in the same form as the operator for the square of the angular momentum \(\hat L^2\).
Since \(\hat S^2\) has a single eigenvalue equal to \(\frac{3}{4} \hbar^2\), the expectation value of the square of the spin is always \(\left \langle S^2 \right \rangle = \frac{3}{4} \hbar^2\) regardless of the state \(\chi\).
Answer
a) We have determined the expectation values of the spin projections along the axes \(x, \, y, \, z\) in a state described by the normalised spinor \(\chi\) \[\left \langle S_x \right \rangle_{\chi} = \frac{\hbar}{2}({ab}^*+a^*b) \, ,\] \[\left \langle S_y \right \rangle_{\chi} = \frac{i\hbar}{2}({ab}^*-a^*b) \, ,\] \[\left \langle S_z \right \rangle_{\chi} = \frac{\hbar}{2}(|a|^2 - |b|^2) \, .\]
b) We have determined the expectation values of the squares of these projections in the same state \[\left \langle S_x^2 \right \rangle_{\chi} = \left \langle S_y^2 \rangle_{\chi} = \langle S_z^2 \right \rangle_{\chi} = \frac{\hbar^2}{4} \, .\]
c) The eigenvalues of the operator for the square of the spin are of the form \(\hbar ^2 s(s+1)\). Therefore, the operators \(\hat S^2\) and \(\hat L^2\) have eigenvalues in the same form. The expectation value of the square of the spin is always
\[ \left \langle S^2 \right \rangle = \frac{3}{4} \hbar^2 \]regardless of the state \(\chi\).
Comment – detailed calculation of ⟨ Sx2 ⟩χ
Again, we use the general expression for the expectation value, into which we substitute the matrix of the operator and the state, and then simplify.
\[ \left \langle S_x^2 \right \rangle_{\chi} = \chi^{\dagger} \hat S_x^2 \chi = \begin{pmatrix} a^* & b^* \end{pmatrix} \frac{\hbar^2}{4}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \] \[ = \frac{\hbar^2}{4} \begin{pmatrix} a^* & b^* \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \frac{\hbar^2}{4}({aa}^*+{bb}^*) = \frac{\hbar^2}{4}(|a|^2 + |b|^2) \, . \]Noting that we are working with a normalised spinor, i.e., \(|a|^2 + |b|^2 = 1\), we obtain the desired result \[ \left \langle S_x^2 \right \rangle_{\chi} = \frac{\hbar^2}{4} \, . \]
The operators \(\hat S_y^2\) and \(\hat S_z^2\) are identical to \(\hat S_x^2\), therefore \[ \left \langle S_x^2 \right \rangle_{\chi} = \left \langle S_y^2 \right \rangle_{\chi} = \left \langle S_z^2 \right \rangle_{\chi} = \frac{\hbar^2}{4} \, . \]
