I have the simplified Ising model. The Hamiltonian is given by $$ \mathcal{H} = -\mathrm{J}\sum_{<ij,i' j'>} \sigma_{ij} \sigma_{i'j'}. $$ Where the sum over $<ij,i'j'>$ means just the sum over nearest neighbors and $\mathrm{J}$ is a constant.

I am trying to recreate the algorithm described in the paper 'Geometric cluster Monte Carlo simulation'. To do this I need a 'symmetry transformation' that is self-inverse and a global symmetry of $\mathcal{H}$. The paper says these can be 'relfections, inversions, translations, ..'. I wrote a Python implementation of this algorithm and it didn't work. I am worried I am misunderstanding what this transformation has to be.

Suppose I have $L\times L$ spins (with periodic boundary) on a square lattice. Spin $\sigma_{ab}$ is placed in the position $(a,b)$ in the lattice. Let $T$ be the following transformation $$ T:\{0,1,..,L-1\}^2\rightarrow\{0,1,..,L-1\}^2, \\T(a,b)=(L-1-a, L-1-b). $$

So $T$ is a point inversion through the center of the lattice.

To me self-inverse means $$f(f(x))=x.$$ This is satisfied by $T$, since $$ T(T(a,b))=T(L-1-a,L-1-b)=(L-1-(L-1-a),L-1-(L-1-b))=(a,b). $$ Global symmetry of $\mathcal{H}$, I thought is $\mathcal{H}$(original latice) $= \mathcal{H}$(transformed lattice). Meaning I should get the same value for the Hamiltonian if I compute it with the original spin configuration as when I compute it with the $T$ transformed spin configuration. $T$ satisfies this condition since the spin $\sigma_{ab}$ affects $\mathcal{H}$ only in four terms, the terms with its neighbors. The neighbors of $\sigma_{ab}$ are the same after the transformation, they just swap positions with opposing neighbors of $\sigma_{ab}.$ So the contribution of $\sigma_{ab}$ to $\mathcal{H}$ before and after the transformation is the same.

Is this correct? Is this what is meant by 'self-inverse and global symmetry of the Hamiltonian'?

EDIT: I contacted the author and he was kind enough to look at my code. I simply made an error in the code. $T$ from above is indeed an example to what was meant by such transforms.


1 Answer 1


The only meaning of self-inverse (also called involutions) I can find or think of is the one you described (as a ref., see https://math.stackexchange.com/questions/1960078/what-are-the-domains-of-self-inverse-functions)

You're also right about the global symmetry of the Hamiltonian, it's perfect.

There is just one thing bothering me... Translation is not self-inverse (although still being a global symmetry). Maybe he is talking about a "weak" self-inverse condition, being:

"There exist n $\in \mathbb{N}$ such that $f(f(...f(x)...) = f^{(n)}(x) = x$." $\hspace{1cm}$('normal' self-inverse case being n=2)

In general, we need to apply the translation L times so that we get back to where we started.


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