Exercise 2

  1. Refer to p.141 in the textbook. There we learned that \[ ( \Delta \mathbf{A})^2 = \braket{\Psi | \bar{\mathbf{A}}^2 | \Psi} = \braket{\bar{\mathbf{A}}^2} \] The same holds true for \(\Delta \mathbf{B}\): \[ ( \Delta \mathbf{B})^2 = \braket{\Psi | \bar{\mathbf{B}}^2 | \Psi} = \braket{\bar{\mathbf{B}}^2} \]

  2. Now let’s prove \([\mathbf{\bar{A}}, \mathbf{\bar{B}}] = [\mathbf{A}, \mathbf{B}]\): \[ \begin{flalign*} [\mathbf{\bar{A}}, \mathbf{\bar{B}}] &= [\mathbf{A} - \braket{\mathbf{A}}, \mathbf{B} - \braket{\mathbf{B}}] \\ &= (\mathbf{A} - \braket{\mathbf{A}})(\mathbf{B} - \braket{\mathbf{B}}) - ((\mathbf{B} - \braket{\mathbf{B}}) (\mathbf{A} - \braket{\mathbf{A}})) \\ &= \mathbf{A} \mathbf{B} + \braket{\mathbf{A}} \braket{\mathbf{B}} - \mathbf{A} \braket{\mathbf{B}} - \braket{\mathbf{A}}\mathbf{B} - (\mathbf{B}\mathbf{A} - \mathbf{B} \braket{\mathbf{A}} - \braket{\mathbf{B}} \mathbf{A} + \braket{\mathbf{B}} \braket{\mathbf{A}}) \\ &= \mathbf{A} \mathbf{B} + \braket{\mathbf{A}} \braket{\mathbf{B}} - \mathbf{A} \braket{\mathbf{B}} - \braket{\mathbf{A}}\mathbf{B} - \mathbf{B}\mathbf{A} + \mathbf{B} \braket{\mathbf{A}} + \braket{\mathbf{B}} \mathbf{A} - \braket{\mathbf{B}} \braket{\mathbf{A}} \end{flalign*} \] Since \(\braket{\mathbf{A}}\) and \(\braket{\mathbf{B}}\) are just numbers, the sequence of multiplication doesn’t matter. Hence it follows \[ [\mathbf{\bar{A}}, \mathbf{\bar{B}}] = [\mathbf{A}, \mathbf{B}] \]

  3. Refer to Eq. 5.12 in the textbook and plug in \(\braket{\mathbf{\bar{A}}}\) and \(\braket{\mathbf{\bar{B}}}\): \[ \begin{flalign*} \sqrt{\braket{\mathbf{\bar{A}^2}} \braket{\mathbf{\bar{B}^2}}} \ge \frac{1}{2} | \braket{\Psi | [\mathbf{\bar{A}}, \mathbf{\bar{B}} ] | \Psi } \end{flalign*} \] With 1) and 2) it follows \[ \Delta \mathbf{A} \, \Delta \mathbf{B} \ge \frac{1}{2} | \braket{\Psi | [\mathbf{A}, \mathbf{B} ] | \Psi } \]