Vector product of angular momentum
Task number: 4573
Evaluate the vector product of the angular momentum operator with itself.
Compare the result with the classical evaluation of the vector product of the angular momentum vector with itself and explain any differences.
Hint – how to proceed
In this problem, it is recommended to determine the individual components of the vector product step by step.
It is also possible to calculate the \(i\)‑th component using the general expression with the Levi‑Civita symbol, but this approach is considerably more complicated.
Solution – calculation using operators
We determine the components of the vector product step by step. Starting with the first component, we write
\[ \left ( \hat{\vec L} \times \hat{\vec L} \right )_1 = \varepsilon_{1jk} \hat L_j \hat L_k = \hat L_2 \hat L_3 - \hat L_3 \hat L_2 \, . \]We now notice that the expression on the right‑hand side is the commutator of the second and third components of the angular momentum operator. The result of this commutator is known from Commutators involving components of angular momentum, Answer. Substituting it, we obtain
\[ \left ( \hat{\vec L} \times \hat{\vec L} \right )_1 = \left [\hat L_2, \hat L_3 \right ] = i \hbar \hat L_1 \, . \]Similarly, the remaining two components of the vector product are determined as
\[ \left ( \hat{\vec L} \times \hat{\vec L} \right )_2 = \varepsilon_{2jk} \hat L_j \hat L_k = \hat L_3 \hat L_1 - \hat L_1 \hat L_3 = \left [\hat L_3, \hat L_1 \right ] = i \hbar \hat L_2 \, , \] \[ \left ( \hat{\vec L} \times \hat{\vec L} \right )_3 = \varepsilon_{3jk} \hat L_j \hat L_k = \hat L_1 \hat L_2 - \hat L_2 \hat L_1 = \left [\hat L_1, \hat L_2 \right ] = i \hbar \hat L_3 \, . \]Altogether we obtain
\[ \hat{\vec L} \times \hat{\vec L} = i \hbar \left (\hat L_1, \hat L_2, \hat L_3 \right ) = i \hbar \hat{\vec L} \, . \]Solution – classical calculation
Let us now examine the classical case, i.e., \(\vec L \times \vec L\).
In classical calculation, the vector product of a vector with itself is zero, i.e., \(\vec L \times \vec L = \vec 0\).
The difference between the quantum and classical results arises from the non‑commutativity of the angular momentum operator components.
Answer
We have determined
\[ \hat{\vec L} \times \hat{\vec L} = i \hbar \hat{\vec L} \, , \] \[ \vec L \times \vec L = \vec 0 \, . \]The difference in these results is due to the non‑commutativity of the angular momentum operator components, an effect that does not appear in the classical calculation.
