# Commutation with unspecified potential function

Instead of a potential given like $$V(r) = k r^2$$ or $$V(r) = y^2$$ , if the potential is given like in the form a function but not clearly specified, can we tell that if that commutes with the hamiltonian or not?

For example: We have given a potential $$V(x^2 +y^2)$$ and if we want to know if this commutes with any operator, for example angular momentum $$L_z$$, can we say that?

The reason I'm saying is because I think I could deal with the commutation if it is specified like,

$$V(r)= 1 / r^2$$ or $$V(r)= r^2 +5\times1 / r^2$$

I need to learn how I can deal with the unspecified function.

For example, the operator $$L_z$$ is an operator that generates rotations around the $$z$$ axis. This means that any potential that is invariant under such rotations of space will commute with $$L_z$$. Even if you do not know the form of $$V(x^2+y^2)$$ you still know that it is symmetric around the $$z$$ axis because it depends only in the distance from $$z$$. Thus, you can immediately conclude that it commutes with rotations around $$z$$ and therefore $$L_z$$.
Similarly, when you have a potential $$V(r)$$ that depends only on the distance $$r$$ from the origin, it is symmetric with rotation around any axis so it commutes with any operator that consists of any combination of generators of rotations $$L_x, L_y, L_z$$.
If you can write your function as a power serie, then you can say something. For example, expand $$V(x^2+y^2)$$ around $$x^2+y^2=0$$ to obtain $$V(x^2+y^2) = \sum_{n=0}^{\infty} \frac{1}{n!} \left[\frac{d^n}{d(x^2+y^2)^n} V(x^2+y^2)\right]\Big \lvert_{x^2+y^2=0} (x^2+y^2)^n.$$ Then, you can apply your commutator on the $$(x^2+y^2)^n$$. For example, $$[\hat{L}_z, V(x^2+y^2)] = \sum_{n=0}^{\infty} \frac{1}{n!} \left[\frac{d^n}{d(x^2+y^2)^n} V(x^2+y^2)\right]\Big \lvert_{x^2+y^2=0} [\hat{L}_z, (x^2+y^2)^n].$$ Then you deal with a commutator with specified ingredients instead of a general function.
• Dear gingras, can we say the the potential is given is central ? Like $V(x^2+y^2)$ or $V(r^2)$ ? – user193422 Aug 8 '19 at 5:44
• Yes, if the potential depends only on $r$ it is said to be central. – gingras.ol Aug 8 '19 at 15:41