Angular momentum and the Hamiltonian
Task number: 4574
Evaluate the commutators \(\left [\hat L_z, \hat r^2 \right ]\) and \(\left [\hat L_z, \hat p^2 \right ]\) (where, of course, \(\hat r^2 = \hat x^2 + \hat y^2 + \hat z^2\) and \(\hat p^2 = \hat p_x^2 + \hat p_y^2 + \hat p_z^2\)).
Show that the Hamiltonian \(\hat H = \frac{\hat p^2}{2m} + \hat V\) commutes with all three components of \(\hat L\), provided that \(\hat V\) depends only on \(r\). (Thus \(\hat H\), \(\hat L^2\), and \(\hat L_z\) are mutually compatible observables.)
Hint
Recall the rules for evaluating commutators and for simplifying commutators of composite operators.
Solution a)
First, we expand \(\hat r^2\) in the commutator as a sum of the squares of the position operators. We then simplify the commutator of composite operators
\[ \left [\hat L_z, \hat r^2 \right ] = \left [\hat L_z, \hat x^2 + \hat y^2 + \hat z^2 \right ] = \left [\hat L_z, \hat x^2 \right ] + \left [\hat L_z, \hat y^2 \right ] + \left [\hat L_z, \hat z^2 \right ] = \] \[ = \hat x \left [\hat L_z, \hat x \right ] + \left [\hat L_z, \hat x \right ] \hat x + \hat y \left [\hat L_z, \hat y \right ] + \left [\hat L_z, \hat y \right ] \hat y + \hat z \left [\hat L_z, \hat z \right ] + \left [\hat L_z, \hat z \right ] \hat z \, . \]The values of all the commutators above are known from Commutators involving components of angular momentum, Solution a) – c). We can therefore substitute them and simplify, which yields the following expression:
\[ \left [\hat L_z, \hat r^2 \right ] = \hat x \left (i \hbar \hat y \right ) + \left (i \hbar \hat y \right ) \hat x + \hat y \left (-i \hbar \hat x \right ) + \left (-i \hbar \hat x \right ) \hat y + \hat z \cdot 0 + 0 \cdot \hat z = \] \[ = i \hbar \left (\hat x \hat y + \hat y \hat x - \hat y \hat x - \hat x \hat y \right ) = 0 \, . \]In the case of the commutator \(\left [\hat L_z, \hat p^2 \right ]\), the procedure is completely analogous, so we present it only briefly
\[ \left [\hat L_z, \hat p^2 \right ] = \hat p_x \left [\hat L_z, \hat p_x \right ] + \left [\hat L_z, \hat p_x \right ] \hat p_x + \hat p_y \left [\hat L_z, \hat p_y \right ] + \left [\hat L_z, \hat p_y \right ] \hat p_y + \hat p_z \left [\hat L_z, \hat p_z \right ] + \left [\hat L_z, \hat p_z \right ] \hat p_z \, . \]The values of all the commutators above are known from Commutators involving components of angular momentum, Solution d) – f). We can therefore substitute them and simplify, yielding the following result
\[ \left [\hat L_z, \hat p^2 \right ] = i \hbar \left (\hat p_x \hat p_y + \hat p_y \hat p_x - \hat p_y \hat p_x - \hat p_x \hat p_y \right ) = 0 \, . \]Since the \(x\) and \(y\) axes are equivalent to the \(z\) axis, there is no physical reason for the commutation relations of the components \(\hat L_x\) and \(\hat L_y\) to differ. Therefore, all components of the angular momentum commute with the operators \(\hat r^2\) and \(\hat p^2\).
A detailed calculation of these commutators for the \(i\)‑th component of the angular momentum operator can be found at the end of this problem in section Angular momentum and the Hamiltonian, Comment – general form of the calculation for part a).
Solution b)
From part a), we know that all components of the angular momentum operator commute with the operators \(\hat r^2\) and \(\hat p^2\).
The kinetic energy operator \(\hat T = \frac{\hat p^2}{2m}\) differs from the squared momentum operator only by the constant factor \(\frac{1}{2m}\), which does not affect commutativity. Therefore, the components of angular momentum operator commute with the kinetic energy operator.
Since we are considering a spherically symmetric potential, the potential energy operator \(\hat V\) depends only on \(r\). Note that \(r=\sqrt{r^2}\). Because the components of the angular momentum operator commute with \(\hat r^2\), they also commute with any function of \(\hat r^2\). Therefore, they commute with the potential energy operator \(V \left (\sqrt{\hat r^2} \right )\).
Since all components of the angular momentum operator commute with both the kinetic and potential energy operators, they also commute with their sum. Consequently, all components of angular momentum operator commute with the Hamiltonian.
Answer
a) All components of the angular momentum operator commute with the operators \(\hat r^2\) and \(\hat p^2\).
b) Since all components of the angular momentum operator commute with \(\hat r^2\) and \(\hat p^2\), they also commute with the Hamiltonian.
Comment – general form of the calculation for part a)
For the \(j\)‑th component of the angular momentum operator, the calculation is very similar to the \(z\)‑component case. Using the Einstein summation convention, we can write the operator \(\hat r^2\) as \(\hat r^2 = \hat x_k \, \hat x_k\). Substituting this into the commutator, we simplify as follows:
\[ \left [\hat L_j, \hat r^2 \right ] = \left [\hat L_j, \hat x_k \, \hat x_k \right ] = \hat x_k \left [\hat L_j, \hat x_k \right ] + \left [\hat L_j, \hat x_k \right ] \hat x_k \, . \]The value of this commutator is known from Commutators involving components of angular momentum, Solution j) – l). We can therefore substitute and simplify as follows:
\[ \left [\hat L_j, \hat r^2 \right ] = \hat x_k i \hbar \varepsilon_{jkl} \hat x_l + i \hbar \varepsilon_{jkl} \hat x_l \hat x_k \, . \]In the second term, we relabel the indices \(k \rightarrow l\) and \(l \rightarrow k\). This allows us to factor and simplify, yielding the result
\[ \left [\hat L_j, \hat r^2 \right ] = \hat x_k i \hbar \varepsilon_{jkl} \hat x_l + i \hbar \varepsilon_{jlk} \hat x_k \hat x_l = i \hbar \hat x_k \hat x_l \left (\varepsilon_{jkl} + \varepsilon_{jlk} \right ) = \] \[ = i \hbar \hat x_k \hat x_l \left (\varepsilon_{jkl} - \varepsilon_{jkl} \right ) = 0 \, . \]Analogously, for the commutator \(\left [\hat L_j, \hat p^2 \right ]\), manipulations lead to commutators of the form \(\left [\hat L_j, \hat p_k \right ] = i \hbar \varepsilon_{jkl} \hat p_l\). The calculation of this commutator can be found in the same problem as above. The evaluation after substitution is also analogous, and is therefore presented only briefly
\[ \left [\hat L_j, \hat p^2 \right ] = \hat p_k \left [\hat L_j, \hat p_k \right ] + \left [\hat L_j, \hat p_k \right ] \hat p_k = \] \[ = i \hbar \hat p_k \hat p_l \left (\varepsilon_{jkl} + \varepsilon_{jlk} \right ) = i \hbar \hat p_k \hat p_l \left (\varepsilon_{jkl} - \varepsilon_{jkl} \right ) = 0 \, . \]
