Collection of Solved Problems in Physics

Spin projections – minimum uncertainty

Task number: 4406

• Hint 1

  Recall or look up how the uncertainty of an observable is related to the variance of the same observable in a given state. Furthermore, recall how the variance is determined.

• Solution to hint 1

  The variance \(\left (\sigma_F \right )^2\) of the observable \(F\) in a state described by the spinor \(\chi\) is the square of its uncertainty \(\sigma_F\). It can be determined using the relation

  \[ \left ( \sigma_F \right )^2 = \left \langle F^2 \right \rangle_{\chi} - \left \langle F \right \rangle_{\chi} ^2 \, . \]

• Hint 2

  Recall the matrix form of the spin-\(\frac{1}{2}\) projections along all three axes \(x, \, y, \, z\) and the relation for calculating the expectation value of a physical observable in a state described by the normalised spinor \(\chi\).

• Solution to hint 2

  The matrices representing the spin-\(\frac{1}{2}\) projections along the individual axes are given by \[(\hat S_x, \hat S_y, \hat S_z) = \frac{\hbar}{2} \left(\begin{pmatrix} 0 &1 \\ 1& 0 \end{pmatrix},\begin{pmatrix} 0 &−i \\ i& 0 \end{pmatrix},\begin{pmatrix} 1 &0 \\ 0& −1 \end{pmatrix}\right) \, .\]

  The expectation value of an observable \( F\) in a state described by the normalised spinor \(\chi\) can be determined from \[\left \langle F \right \rangle_{\chi} = \left \langle \chi \, \Big | \, \hat F \chi \right \rangle = \chi^{\dagger} \hat F \chi \, ,\] where \(\hat F\) is the operator associated with the observable \(F\), and the dagger denotes the Hermitian adjoint, i.e., the transpose of the matrix followed by complex conjugation.

• Hint 3

  Recall the algebraic, trigonometric, and exponential forms of a complex number. Furthermore, recall how a complex number can be converted from one form to another.

• Solution to hint 3

  The algebraic form of a non‑zero complex number \(z\) is defined as \(z = c + id\), where \(c, \, d\) are real numbers and \(i\) is the imaginary unit.

  The trigonometric form of a non‑zero complex number \(z\) is defined as \(z = |z| (\cos\varphi + i\sin\varphi)\), where \(|z|\) is the modulus of the complex number \(z\) and \(\varphi\) is its argument. By comparing the real and imaginary parts of these two forms, we obtain the conversion relations between the algebraic and trigonometric forms

  \[ c = |z| \cos \varphi \, , \] \[ d = |z| \sin \varphi \, . \]

  Squaring and summing these relations gives

  \[ c^2 + d^2 = |z|^2 \, , \]

  from which the modulus of the complex number can be determined as

  \[ |z| = \sqrt{c^2 + d^2} \, . \]

  Dividing the relations for the algebraic‑to‑trigonometric conversion yields

  \[ \mathrm{tg}\, \varphi = \frac{d}{c}, \]

  from which the argument \(\varphi\) of the complex number \(z\) can be determined.

  By defining \(e^{i \varphi} = \cos\varphi + i\sin\varphi\), a non‑zero complex number \(z\) can also be expressed in the exponential form \(z = |z| e^{i\varphi}\), where \(|z|\) is the modulus and \(\varphi\) is the argument of \(z\).

  All these forms are equivalent, i.e.,

  \[ z = c + id = |z| (\cos\varphi + i\sin\varphi) = |z| e^{i\varphi}. \]

• Solution

  A detailed calculation of the expectation values of the spin projections along the individual axes, as well as the expectation values of the squares of the spin projections along each axis for a spin‑\(\frac{1}{2}\) particle in a general state described by the normalised spinor \[\chi = \begin{pmatrix} a \\ b \end{pmatrix} \, ,\] can be found in task Expectation values of the spin components in a general state, Solution a).

  Here we summarise only the resulting values

  \[\left \langle S_x \right \rangle_{\chi} = \frac{\hbar}{2}({ab}^*+a^*b) \, ,\] \[\left \langle S_y \right \rangle_{\chi} = \frac{i\hbar}{2}({ab}^*-a^*b) \, ,\] \[\left \langle S_z \right \rangle_{\chi} = \frac{\hbar}{2}(|a|^2 - |b|^2) \, ,\] \[\left \langle S_x^2 \right \rangle_{\chi} = \left \langle S_y^2 \right \rangle_{\chi} = \left \langle S_z^2 \right \rangle_{\chi} = \frac{\hbar^2}{4} \, .\]

  Since multiplying complex numbers is most straightforward in exponential form, we convert the numbers \(a, \, b\) and their complex conjugates \(a^*, \, b^*\) into this form \(a = |a|e^{i \varphi_a}\), \(b = |b|e^{i \varphi_b}\), \(a^* = |a|e^{-i \varphi_a}\), \(b^* = |b|e^{-i \varphi_b}\) and express the necessary products as

  \[ {ab}^* = |a|e^{i\varphi_a}|b|e^{-i\varphi_b} = |a||b|e^{i(\varphi_a-\varphi_b)} \, , \] \[ a^*b = |a|e^{-i\varphi_a}|b|e^{i\varphi_b} = |a||b|e^{i(\varphi_b-\varphi_a)} \, . \]

  If now we denote the difference of the arguments \(\varphi_a-\varphi_b\) by \(\varphi\), these products can be written as

  \[ {ab}^* = |a||b|e^{i\varphi} \, , \] \[ a^*b = |a||b|e^{-i\varphi} \, . \]

  We now substitute into the expressions for the expectation values of the spin projections and simplify

  \[ \left \langle S_x \right \rangle_{\chi} = \frac{\hbar}{2}({ab}^*+a^*b) = \frac{\hbar}{2}(|a||b|e^{i\varphi} + |a||b|e^{-i\varphi}) = \] \[ = \frac{\hbar}{2} |a||b| (\cos\varphi + i\sin\varphi + \cos\varphi - i\sin\varphi) = \hbar |a||b| \cos\varphi \, , \] \[ \left \langle S_y \right \rangle_{\chi} = \frac{i\hbar}{2}({ab}^*-a^*b) = \frac{i\hbar}{2}(|a||b|e^{i\varphi} - |a||b|e^{-i\varphi}) = \] \[ = \frac{i\hbar}{2} |a||b| (\cos\varphi + i\sin\varphi - \cos\varphi + i\sin\varphi) = -\hbar |a||b| \sin\varphi \, . \]

  Next, we determine the variances \((\sigma_{S_x})^2\) and \((\sigma_{S_y})^2\) by substituting into the general relation and symplifying

  \[ \left ( \sigma_{S_x} \right )^2 = \left \langle S_x^2 \right \rangle_{\chi} - \left \langle S_x \right \rangle_{\chi} ^2 = \frac{\hbar^2}{4} - (\hbar |a||b| \cos\varphi)^2 = \frac{\hbar^2}{4} - \hbar^2 |a|^2|b|^2 \cos^2\varphi \, , \] \[ \left ( \sigma_{S_y} \right )^2 = \left \langle S_y^2 \right \rangle_{\chi} - \left \langle S_y \right \rangle_{\chi} ^2 = \frac{\hbar^2}{4} - (-\hbar |a||b| \sin\varphi)^2 = \frac{\hbar^2}{4} - \hbar^2 |a|^2|b|^2 \sin^2\varphi \, . \]

  From here, we could take the square root and substitute into the uncertainty relation for further simplification. However, it is more convenient to first square the uncertainty relation \(\sigma_{S_x} \, \sigma_{S_y} \geq \frac{\hbar}{2} \left | \left \langle S_z \right \rangle_{\chi} \right |\) and then substitute and simplify. Since we are looking for the conditions under which equality holds, we replace the inequality sign with an equality sign in the following

  \[ \left ( \sigma_{S_x} \right )^2 \left ( \sigma_{S_y} \right )^2 = \frac{\hbar^2}{4} \left \langle S_z \right \rangle_{\chi}^2 \, , \] \[ \left(\frac{\hbar^2}{4} - \hbar^2 |a|^2|b|^2 \cos^2\varphi\right) \left(\frac{\hbar^2}{4} - \hbar^2 |a|^2|b|^2 \sin^2\varphi\right) = \frac{\hbar^2}{4} \left (\frac{\hbar}{2}(|a|^2 - |b|^2)\right)^2 \, , \] \[ \frac{\hbar^2}{4} \frac{\hbar^2}{4} (1-4|a|^2|b|^2 \cos^2\varphi)(1-4|a|^2|b|^2 \sin^2\varphi) = \frac{\hbar^2}{4} \frac{\hbar^2}{4} (|a|^4 - 2|a|^2|b|^2 + |b|^4) \, . \]

  We now divide the equation by \(\frac{\hbar^4}{16}\) and expand the brackets on the left‑hand side

  \[ 1 - 4|a|^2|b|^2(\cos^2\varphi + \sin^2\varphi) + 16|a|^4|b|^4\cos^2\varphi\sin^2\varphi = |a|^4 - 2|a|^2|b|^2 + |b|^4 \, . \]

  Next, we simplify using the relation \(\cos^2\varphi + \sin^2\varphi = 1\)

  \[ 1 + 16|a|^4|b|^4\cos^2\varphi\sin^2\varphi = |a|^4 + 2|a|^2|b|^2 + |b|^4 \, , \] \[ 16|a|^4|b|^4\cos^2\varphi\sin^2\varphi = \left (|a|^2 + |b|^2 \right)^2 - 1 \, . \]

  We now use the fact that we are working with a normalised spinor, i.e., \(|a|^2 + |b|^2 = 1\). Substituting and simlifying, we obtain the final relation from which the conditions for equality can be determined

  \[ 16|a|^4|b|^4\cos^2\varphi\sin^2\varphi = 0 \, , \] \[ |a|^2|b|^2\cos\varphi\sin\varphi = 0 \, . \]

  This equality holds

  • in the special case, where \(a = 0 \lor b = 0\). In this case, the state corresponds to a sharp value of the spin projection \(\hat S_z\).
  • for \(a, b \neq 0 \, \land \, \sin \varphi = 0\), i.e., \(\varphi = 0 \, \lor \, \varphi = \pi\).
    • If \(\varphi = 0\), the numbers \(a, b\) have the same phase. Thus, for \(a \in \mathbb{R}^+\), we have \(b \in \mathbb{R}^+\).
    • If \(\varphi = \pi\), the numbers \(a, b\) have opposite phases. Thus, for \(a \in \mathbb{R}^+\), we have \(b \in \mathbb{R}^-\).
  • for \(a, b \neq 0 \, \land \, \cos \varphi = 0\), i.e., \(\varphi = \pm \frac{\pi}{2}\).
    • If \(\varphi = \pm \frac{\pi}{2}\), the numbers \(a, b\) are mutually phase‑shifted by \(\frac{\pi}{2}\).
      Thus, for \(a \in \mathbb{R}^+\), \(b\) is a purely imaginary number.

• Answer

  The minimum product of the uncertainties of the spin projections \(\hat S_x\) and \(\hat S_y\) occurs in the following cases

  • the numbers \(a, b\) have the same phase, i.e., for \(a \in \mathbb{R}^+_0\), also \(b \in \mathbb{R}^+_0\),
  • the numbers \(a, b\) have opposite phases, i.e., for \(a \in \mathbb{R}^+_0\), \(b \in \mathbb{R}^-_0\),
  • the numbers \(a, b\) are phase‑shifted by \(\frac{\pi}{2}\), i.e., for \(a \in \mathbb{R}^+_0\), \(b\) is a purely imaginary number.

• Comment

  Since this exercise deals with the phase difference between the numbers \(a, b\), it is perfectly legitimate to assume \(a \in \mathbb{R}_0^+\) from the start of the calculation.

  This choice can be made because the description of the state is not unique. We have the freedom to multiply the spinor by a constant, so there always exists a spinor representing the given state for which \(a \in \mathbb{R}_0^+\) holds.

  In this case, the calculation is only slightly simplified, and in the end, we would obtain the phase of the number \(b\). Both approaches are correct and are of approximately the same level of difficulty.