This feels like it could be a undergrad/grad-school quantum mechanics course level problem, or potentially something pretty interesting. I'd be happy with either answer, but I don't know which one is true.

Consider spin-1/2 quantum spins, and let us denote the Pauli operators as $X_i$, $Y_i$, and $Z_i$ etc. for the $i$-th spin. We have $N$ spins in total, and let's say we can group them into $n$ groups of $k$ spins, i.e. $kn=N$.

If we redefine $\tilde{X}_i := \sum_{j\in i\mathrm{-th~group}}^k X_i$ etc. by summing up those $k$ spins, and then defining a Hamiltonian only using those "coarse grained" terms, we can obtain a Hamiltonian that looks as if the elementary quantum degree of freedom with spin-$k/2$.

For example, something like this: \begin{equation} \tilde{H}=\sum_{i,j\leq n} w_{ij} \left(\tilde{X}_i \tilde{X}_j + \tilde{Y}_i \tilde{Y}_j\right) = \sum_{i,j\leq n} w_{ij} \left(\left(\sum^kX_i\right)\left(\sum^kX_j\right) + \left(\sum^kY_i\right)\left(\sum^kY_j\right)\right). \end{equation} The right hand side is only there to show the definition, and the sum is taken over the coarse-graining "group".

In this case, the coarse-grained operators $\tilde{X}_i$ etc. all satisfy the spin-$k/2$ angular momentum commutation relations; e.g. \begin{equation} [\tilde{X}_j,\tilde{Y}_j]=2i\tilde{Z}_j . \end{equation} Therefore, algebraically the Hamiltonian $\tilde H$ is the same as the Hamiltonian with $n$ spin-$k/2$s as elementary degrees of freedom, which we can call $H_{J=k/2}$ for now.

Now, the ground state $|\mathrm{GS}\rangle$ of Hamiltonian $\tilde H$ always seem to have the "coarse-grained spin" to be maximal, i.e. if you define $\tilde S_i^2 := \tilde X_i^2 + \tilde Y_i^2 +\tilde Z_i^2$, then $\langle\mathrm{GS}|\tilde S_i^2|\mathrm{GS}\rangle = k(k+2)$ for all $i=1,2,\ldots,n$. For me, this seems very natural because this simply matches up with the ground state of $H_{J=k/2}$, but I can't find a rigorous proof for this. Another thing is that for general eigenstates of $\tilde H$, this is not always true, so it seems like you would have to use the fact that you are only interested in the ground state somehow, which makes it harder to prove I think.

Does anyone know a good reference for this? It feels like maybe it's something to do with group representations of spins etc, but I'm not sure. I'd also very much welcome a proof if someone can provide that.

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    $\begingroup$ For clarity, please let me add one constraint for $\tilde{H}$ that it can only be linear for any coarse-grained spin term. So we allow terms like $\tilde{X}_i\tilde{Y}_j\tilde{X}_k$, but don't allow terms like $\tilde{X}_i^2$, because that will trivially reduce the coarse grained spin size. $\endgroup$ Jan 11 at 3:53

1 Answer 1


I don't know any relevant references. And I'm not sure that the statement about the ground state is always true. But I want to offer some thoughts about when and why this statement might be true.

Quite often it can be shown that the ground state of a quantum spin model is non-degenerate in some invariant subspace of the Hamiltonian and the corresponding wave function is positive in some basis. Such properties follow from the Perron-Frobenius theorem when the non-diagonal elements of a Hamiltonian matrix are non-positive and the matrix is irreducible. Examples are the quantum Ising model and the (anisotropic) Heisenberg model. A Hamiltonian composed of coarse-grained operators obviously commutes with any permutations of spins within blocks. At the same time, in nontrivial cases the Hamiltonian does not commute with $\tilde{Z}_i$. Thus, if spin permutations do not change the invariant subspace, and you were able to show that the ground state is nondegenerate in the subspace and its wave function is positive, then the ground state wave function must be symmetric with respect to any spin permutations within the blocks. Due to the positivity and uniqueness of the ground state wave function, it simply cannot change sign during intrablock spin permutations and must be symmetric. It is known that symmetric states of systems of spins $1/2$ are states with maximum $\tilde{S}^2_i$. This is because $\tilde{S}^2_i$ is expressed as a linear combination of spin permutation operators. So in this case the statement under discussion is true.

A note about invariant subspaces. Often the Hamiltonian has some symmetry, for example, it commutes with $\sum_i \tilde{Z}_i$. In this case, the complete Hamiltonian matrix is not irreducible, and the ground state may be degenerate. But the Hamilton matrix limited to an invariant subspace may be irreducible, and the ground state in this subspace may be non-degenerate.

  • $\begingroup$ Thank you for your reply! I have a question though: What you say about the Perron-Frobenius theorem and example of the Heisenberg confuses me. For simplicity, if we consider two qubits with ferromagnetically interacting Heisenberg terms, the GS of that is three-fold degenerate. What am I missing here? $\endgroup$ Feb 5 at 18:36
  • $\begingroup$ @Jun_Gitef17 You do not take into account the commutation of the Hamiltonian with $S^z$. I tried to explain this point in my answer. In your example, the Hamilton matrix is not irreducible. Therefore, the ground state may be degenerate. And it is degenerate, since the Heisenberg model is symmetric with respect to all rotations, and not just rotations around the Z axis. But if we consider the blocks of the Hamiltonian matrix corresponding to a fixed $S^z$, then each of these blocks is irreducible. And you can check that the ground state corresponding to a fixed value of $S^z$ is not degenerate. $\endgroup$
    – Gec
    Feb 5 at 19:04
  • $\begingroup$ @Jun_Gitef17 A system of $N$ spins $1/2$ has $N+1$ different possible values of $S^z$. Consequently, the ground state of any Heisenberg model with $N$ spins $1/2$ can be at most $N+1$-fold degenerate. And for ferromagnetic Heisenberg models of spins $1/2$, the degeneracy of the ground state is indeed $N+1$. $\endgroup$
    – Gec
    Feb 5 at 19:10
  • $\begingroup$ Thank you for your detailed answer! Another question is that for the PF theorem condition, would it be OK to say that the non-positive condition is exactly the same as conditions for Stoquastic (= sign-problem free) Hamiltonians? $\endgroup$ Feb 5 at 19:38
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    $\begingroup$ Thanks. I was curious because I also work on quantum Monte Carlo, and was familiar with the concept. Interesting thing is that many times, although stoquasticity plays a crucial role in how a quantum system could be simulated classically/probabilistically, the physics almost doesn't seem to care. If my intuition is right, it seems that your argument for stoquastic cases should have a natural extension to general cases, although the PF theorem doesn't straight forwardly admit that. $\endgroup$ Feb 5 at 20:00

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