fzero, I didn't even think of that! A free particle would have zero potential energy, so for L to be conserved it would still have to commute with the Hamiltonian, and that means L and P^2 must commute (2m factor is irrelevant). I thought for sure I was making a mistake somewhere since my mind...
Oh, I found one mistake but it still leads me to the same conclusion. The second equality should be plus, not minus: [A,BC] = [A,B]C+B[A,C]:
\left[ \vec{L} , P^2 \right] = \left[ L^k , P_i P_i \right] = \left[ L^k , P_i \right] P_i + P_i \left[ L^k , P_i \right] = \left( - i \hbar...
I have been told that L and P^2 do not commute, but I don't see why. It seems like the commutator should be zero.
\left[ \vec{L} , P^2 \right] = \left[ L^k , P_i P_i \right]
= \left[ L_k , P_i \right] P_i - P_i \left[ L_k , P_i \right]
= \left( - i \hbar \epsilon_{i}^{km} P_m \right)...
Helmholtz' Theorem starts with the two components in my original post and defines the divergence and curl as:
div[V] = s(r)
and
curl[V] = c(r), where div[c(r)] = 0
But I can't find anything about how we can define a generic vector as two components:
V = -grad[phi] + curl[A], where "phi" is...
I have to show that a generic vector can be decomposed into an irrotational and solenoidal component:
V(r) = -Grad[phi(r)] + Curl[A(r)]
I'm not sure how to start. Do I need to take the curl or div of V and use a vector identity? Any help would be greatly appreciated!