Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question converning multiplets when describing atoms.

Let $H=\sum\limits_{k=1}^{N} (p_{k}^2 - \frac{Z}{|x_{k}|} + \sum\limits_{i<k}^{1..N} \frac{1}{|x_{i} - x_{k}|}$ be a (self-adjoint) operator on the Hilbertspace of totally antisymmetrized wave functions of N particles H^(N)_{a}.

In my notes it is written that SO(3) x SU(2) is a symmetry group of this hamiltonian. The group SO(3) x SU(2) consists of pairs (R,A)$\in SO(3) x SU(2)$ and has a unitary representation U on H^(N)_{a} given by:

$(U(R,A)\phi)(x_{1} s_{1} ,...., x_{N} s_{n}) = As_{1}s'_{N} ... A s_{N} s'_{N}\phi(R^{-1}x_{1} s_{1} s'_{1} ... R^{-1} x_{N} s'_{N}) $

Here comes my problem. I don't really understand this definition. It's not given in my notes what the s' are and I don't understand what $As_{1}s'_{N}$ means...A vague guess might be that you multiply the vector $s_{1}$ in $C^{2}$ is multiplied by the matrix A in SU(2) and then scalar mulitplied by $s'_{N}$.

Butit's not clear to me..Can anyone help me to understand this definition better?

In addition to that I thought that if a group G is a symmetry group of a Hamiltonian H then it means that [H,U]=0 for ANY representation U of the group G. But in the above example one only gives one particular representation.

I'd be really happy if someone could help me. Thanks in advance.

share|cite|improve this question

Hopefully the $SO(3)$ part of the symmetry is clear: it's just a rotation that acts on the position part of the wavefunction.

The $SU(2)$ part comes from the fact that these are (supposedly) spin 1/2 particles. Since each of these is described by two spin states we act on them by using a special unitary matrix acting on a two-dimensional complex space ${\mathbb C}^2$. But this is precisely a matrix from $SU(2)$ (in other words, this is a tautological representation where the matrix is represented by itself). Now, when acting on $N$ particles we take a tensor product of these representations. Therefore we get a representation ${\mathbb C}^2 \otimes \cdots \otimes {\mathbb C}^2$. I think what you've written must contain some typo but generally it has a correct structure of containing the matrix $A$ $N$ times and each time it acts on a pair of two spin-vectors.

As for the second question: it's certainly not true that the Hamiltonian commutes with any representation. This is because not every Hilbert space carries every representation of the group. It only contains some of them. The general form of the theorem (due to Wigner) is the following: it the system is invariant under a symmetry then there exists an unitary of anti-unitary operator that represents it on the Hilbert space. In other words, it only asserts the existence of some representation. Obviously then, since most of the representations aren't even there they surely can't commute with Hamiltonian.

Perhaps you were confused by some standard example like Coulomb potential problem. In this case every representation of $SO(3)$ (indexed by a non-negative integer) is realized on the Hilbert space. But it's easy to construct a Hilbert space where this fails, e.g. the ${\mathbb C}^2$ above only carries spin 0 and spin 1/2 representations of $SU(2)$ and no other (since reps with higher spin are more than two-dimensional).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.