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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.

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Although you can always apply operators on unspecified functions explicitly like was suggested here by @gingras.ol, most often, physicists will use a symmetry argument to conclude that operators commute. You can do that without calculations but it requires some experience with symmetry arguments.

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 $L_x, L_y, L_z$.

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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.

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  • $\begingroup$ Dear gingras, can we say the the potential is given is central ? Like $V(x^2+y^2)$ or $V(r^2)$ ? $\endgroup$ – user193422 Aug 8 '19 at 5:44
  • $\begingroup$ Yes, if the potential depends only on $r$ it is said to be central. $\endgroup$ – gingras.ol Aug 8 '19 at 15:41

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