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.

  • $\begingroup$ I honestly have no idea what to do. I'm a first year mathematical physics graduate student, and I was hoping that someone could offer a link that explains this. $\endgroup$
    – user119264
    Feb 1, 2014 at 4:28
  • $\begingroup$ I'm not saying you need to know how to find the answer, but if you've really done nothing at all, we're not really incentivized to help you. What is it that you don't understand about this question? Is it a term you're not familiar with, or a mathematical procedure you don't know how to apply? Which step confuses you? Some of the advice in our homework policy might be useful in improving the question, regardless of whether it's actually a homework assignment. $\endgroup$
    – David Z
    Feb 1, 2014 at 5:44
  • $\begingroup$ I'm more interested in the mathematical procedure to apply. I wish to know it for the sake of understanding research papers. $\endgroup$
    – user119264
    Feb 1, 2014 at 5:53
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    $\begingroup$ I would like to see the answer to this question. I sketched the graph of $V(x)$ and couldn't see any obvious symmetry. I assume the symmetry group is the group of transformations that leave $V(x)$ unchanged but it isn't obvious to me what these are. $\endgroup$ Feb 1, 2014 at 9:45
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    $\begingroup$ I'll post my comments as an "answer" of sorts, maybe people can comment on it if they find errors. $\endgroup$ Feb 1, 2014 at 20:59

1 Answer 1


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}]

enter image description here enter image description here enter image description here

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 $A_1=0$.

  • $\begingroup$ The question is far away from my knowledge, but does every potential have a associated symmetry group? $\endgroup$
    – jinawee
    Feb 1, 2014 at 22:05
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    $\begingroup$ Not really sure. In Vedensky's group theory course it's shown that the possible degeneracies of quantum states associated with a potential are the dimensionalities of the irreducible representations which are induced by the symmetry group associated with that Hamiltonian (or at least in the absence of accidental degeneracy). In short, degeneracy is usually a consequence of nontrivial groups of symmetries. Whether any potential has a symmetry might depend on how you define "symmetry", and I'm not a physicist (I'm a chemist) so I'm not an expert on this. $\endgroup$ Feb 1, 2014 at 22:15

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