Can somebody explain what it means when one says a Hamiltonian is SU(2) invariant? I know Heisenberg Hamiltonian is SU(2) invariant but why?

  • 3
    $\begingroup$ It means that if one performs an SU(2) transformation, the Hamiltonian will retain its original form. $\endgroup$
    – Danu
    Oct 14, 2013 at 10:01

1 Answer 1


Heisenberg's Hamiltonian, if I understood what you're referring to, is a Hamiltonian operator for a system whose components are a family of spins. What is relevant here is that $H$ is a sum of terms of the kind $\mathbf{s}_i\cdot\mathbf{s}_j$.

$SU(2)$ is a group of transformations which in this case represents rotations (it is the covering of the $SO(3)$ rotation group and is used for spin, see for example here).

Now, why should we expect the hamiltonian to be invariant with respect to rotations? Because whenever you have a scalar product between two spin vectors, you simply obtain something proportional to $\cos\theta$, where $\theta$ is the angle between the two. This value does not change with a rotation of the reference frame, because involves only the relative position of the two spins. This is the physical explanation of this invariance. Mathematically, one could try to directly perform the transformation on the Hamiltonian.

I'm not sure about this, but I think it should be done as $H' = D H D^\dagger$ where $D$ is the rotation operator $D=\exp(-\frac{i\phi}{\hbar}\mathbf{n}\cdot\mathbf{S})$ (see here).


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