Square of the x-coordinate and the z-component of angular momentum
Task number: 4570
Consider the observables \(A=x^2\) and \(B=L_z\).
a) Construct the uncertainty principle for \(\sigma_A \sigma_B\).
b) Evaluate \(\sigma_B\) in the hydrogen state \(\psi_{nlm}\).
c) What can you conclude about \(\left \langle xy \right \rangle\) in this state?
Hint
Recall or look up the general form of the uncertainty relations, the connection between the uncertainty of an observable and its variance in a given state, as well as the procedure for calculating the variance. In addition, revise the rules for computing commutators and for simplifying commutators of composite operators.
Solution a)
To derive the uncertainty relation for the observables \(A\) and \(B\), we must first determine the commutator of their corresponding operators. To evaluate this commutator, we use the formula for the commutator of composite operators, noting that the operator \(\hat x^2\) can be regarded as the product \(\hat x \, \hat x\)
\[ \left [\hat A, \hat B \right ] = \left [\hat x^2, \hat L_z \right ] = \left [\hat x \, \hat x, \hat L_z \right ] = \hat x \left [\hat x, \hat L_z \right ] + \left [\hat x, \hat L_z \right ] \hat x \, . \]For the commutator \(\left [\hat x, \hat L_z \right ]\), we can substitute its value \(-i \hbar y\). A detailed calculation of this commutator can be found in problem Commutators involving components of angular momentum, Solution a) – c). We may also replace the operator \(\hat x\) by \(x\), since \(\hat x = x\).
Upon substitution, we thus obtain
\[ \left [\hat A, \hat B \right ] = x(-i \hbar y) + (-i \hbar y)x = -2i \hbar xy \, . \]Comparing our result with the form of the commutator in the general uncertainty relation (see Solution to hint), we conclude that \(\hat K = -2 \hbar xy\). We can now substitute this into the general form of the uncertainty relation, which, after some algebraic manipulation, leads to
\[ \delta A \, \delta B \geq \frac{1}{2} \left | \left \langle K \right \rangle_{\psi_{\mathrm{nlm}}} \right | = \frac{1}{2} \left | \left \langle -2 \hbar xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | = \frac{1}{2} \left | -2 \hbar \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | = \frac{1}{2} 2 \hbar \left | \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | = \hbar \left | \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | \, . \]Thus, we finally obtain
\[ \delta x^2 \, \delta L_z \geq \hbar \left | \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | \, . \]Solution b)
To calculate the uncertainty \(\sigma_B\), we need to determine the variance \(\left (\sigma_B \right )^2\) of the observable \(B\) in the state described by the wavefunction \(\psi_{\mathrm{nlm}}\). We make use of the fact that the given wavefunction is an eigenfunction of \(\hat L_z\). I.e., \(L_z\) has a sharp value in these states, meaning that the expectation value equals the corresponding eigenvalue. Substituting into the general expression for the variance, we obtain
\[ (\sigma_B)^2 = \left \langle B^2 \right \rangle_{\psi_{\mathrm{nlm}}} - \left \langle B \right \rangle_{\psi_{\mathrm{nlm}}} ^2 = \left \langle L_z^2 \right \rangle_{\psi_{\mathrm{nlm}}} - \left \langle L_z \right \rangle_{\psi_{\mathrm{nlm}}} ^2 = m^2 \hbar ^2 - (m \hbar)^2 = 0 \, . \]Hence, we obtain
\[ \sigma_{L_z} = 0 \, . \]Solution c)
Here, we combine the results from the previous two sections. In part a), we obtained the relation \(\sigma_{x^2} \, \sigma_{L_z} \geq \hbar \left | \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right |\), and in part b) we found that \(\sigma_{L_z} = 0\). Since the uncertainty \(\sigma_{x^2}\) is finite, we can conclude that \(0 \geq \hbar \left | \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | \).
Therefore, for the hydrogen atom states described by \(\psi_{nlm}\), it follows that
\[ \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} = 0 \, . \]Answer
a) The uncertainty relation reads \( \sigma_{x^2} \, \sigma_{L_z} \geq \hbar \left | \left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} \right | \).
b) In the eigenstates of the hydrogen atom, \(\delta L_z = 0\).
c) In the eigenstates of the hydrogen atom, it follows that \(\left \langle xy \right \rangle_{\psi_{\mathrm{nlm}}} = 0\).
