Determining the group associated with a given potential? I'm trying to understand how symmetry groups are related to potentials of the Schrodinger equation. In particular, I wish to know if it is possible to find the symmetry group of this potential 
$$V(x) = A_0 +A_1x +A_2x^2 -\frac{9}{4}x^4$$
where $A_0$,$A_1$,$A_2$ $\in \mathbb{R}$
I've tried to see if it is related to the SO(3) group and the unitary group U(1), but neither seem possible. I asked this question because coming from a pure math background, I am having a really difficult time trying to understand this.
 A: This is admittedly an incomplete answer, as I don't work in this sort of physics, but a couple things can be pointed out. First, what kind of symmetries are you looking for? This is a 1-dimensional example, and it's not periodic, so unless you're looking for something crazy, the easiest thing to look for is an affine symmetry of the form $V(\alpha x+\beta)=V(x)$. Pictures may help:
Plot[x^1 + x^2 + x^4, {x, -1.5, 1.4}]
Plot[-x^1 - x^2 + x^4, {x, -1.5, 1.7}]
Plot[-x^2 + x^4, {x, -1.7, 1.7}]




One might conjecture that the functions either have no affine symmetry or they have reflection symmetry. I won't prove that, but I'll give code (explanation can be found in comments section) that shows this is the case for various inputs of $A_1,A_2$, and $A_4$ (obviously $A_0$ is irrelevant):
rule = x -> (a x + b); rule2 = {A1 -> 6, A2 -> 3, A4 -> 1}; 
Reduce[(((A1 x + A2 x^2 + A4 x^4 == (A1 x + A2 x^2 + A4 x^4 /. rule)) /. rule2) /. x -> 1) && (((A1 x + A2 x^2 + A4 x^4 == (A1 x + A2 x^2 + A4 x^4 /. rule)) /. rule2) /. x -> 2) && (((A1 x + A2 x^2 + A4 x^4 == (A1 x + A2 x^2 + A4 x^4 /. rule)) /. rule2) /. x -> 3), {a, b}]

You will either get only the identity or the identity and parity transform for most values of $A_1,A_2,A_4$. If anyone knows a more rigorous way to show this, or sees that what I wrote is wrong, by all means post away. I think you can prove it by noting that if you assume the symmetry group is finite (which seems reasonable), then you must also have that the set ${x,T(x),T(T(x)),...}$ is finite, where $T(x)=\alpha x+\beta$. Since $T^n(x)=\alpha^n x+\frac{\alpha^n-\alpha}{\alpha-1}\beta=x$ for some value of $n$, you must have $\alpha$ be a root of unity. 
If you temporarily rule out the possibility of complex-valued position, then it follows that the only possible values of $\alpha$ are $\pm 1$.
If $\alpha=1$, then since the potential is not translation-invariant, you must have $\beta=0$, giving the identity transform. 
If $\alpha=-1$, then it's a little more complicated, but it's intuitively obvious that $\beta=0$ is the only possibility, since the potential is visually invariant up to a translation under a reflection if and only if $A_1=0$. So you get parity iff $A_1=0$.
