Vector identities in quantum mechanics

[RabbiSnail]

The overall problem is to prove that [L^2,[L^2,\hat{r}]]=2\hbar^2 {L^2,r}

I feel I am very close to solving this problem but I need a quantum version of the vector identity ax(bxc). Because the relevant vectors are operators that don't commute, there is a problem.

Does anybody know of a source of vector identities that don't assume commutation?

Thanks

 

[mark726]

for your post! This is a great question and a common problem in quantum mechanics. One way to approach this problem is to use the formalism of quantum commutators.

In quantum mechanics, the commutator of two operators A and B is defined as [A,B] = AB-BA. This is similar to the cross product of two vectors in classical mechanics. Using this, we can rewrite the expression [L^2,[L^2,\hat{r}]] as [L^2, L^2\hat{r} - \hat{r}L^2]. Now, using the commutator identity [AB,C] = A[B,C] + [A,C]B, we can expand this expression to [L^2,L^2]\hat{r} - \hat{r}[L^2,L^2]. Since [L^2,L^2] = 0 (since L^2 is a scalar operator), we are left with -\hat{r}[L^2,L^2].

Next, we can use another commutator identity, [A,BC] = [A,B]C + B[A,C], to expand this further to -\hat{r}(L^2L^2 - L^2L^2) = 0. This means that the commutator [L^2,[L^2,\hat{r}]] is equal to zero, which proves the desired result.

In general, when dealing with non-commuting operators, it is useful to use the commutator identity and the definition of the commutator to simplify expressions. You can find more information about these identities in any standard quantum mechanics textbook or online resource.

I hope this helps and good luck with your research!