I wonder if there exists a way to map the square of a 2-body Ising Hamtiltonian (which will make it 4-body) back to a 2-body Hamiltonian that has the same ground state? Let me explain what I mean by an example.

Suppose we have the following 2-body Ising Hamiltonian, $$ H =\sum_{{i,j}\in E} Z_iZ_j\tag{1}. $$ If we square this Hamiltonian we get a 4-body Ising Hamiltonian, $$ H^2 =\sum_{{i,j,k,l}\in E} Z_iZ_jZ_kZ_l\tag{2}. $$ Is it possible to map $H^2$ such that it becomes a 2-body $\sum_{ij} a_{ij} Z_iZ_j$, without changing the grond state, where $a_{ij}$ are some possible constants?

  • $\begingroup$ Can I just check the terminology you're using: your initial Hamiltonian is 2-local in the sense it is comprised of 2-body terms, and these are on neighbouring qubits according to the underlying lattice geometry defined by $G$. The terms in $H^2$ are 4-body terms, but they're not local because a term could arise from two edges which are far apart. So presumably there is no hope in recreating it with a local 2-body Hamiltonian? $\endgroup$
    – DaftWullie
    Commented Jan 19, 2022 at 9:55
  • $\begingroup$ Thanks for your comment. I've updated my description. I mean 2-body and not 2-local, so the qubits doesn't have to be neighboring. $\endgroup$ Commented Jan 19, 2022 at 10:05
  • $\begingroup$ are you familiar with perturbation gadgets? $\endgroup$
    – DaftWullie
    Commented Jan 19, 2022 at 10:19
  • $\begingroup$ No I haven't heard about it. $\endgroup$ Commented Jan 19, 2022 at 10:22

2 Answers 2


There is a technique known as perturbation gadgets (google the term and you'll find plenty of papers) about how to build many-body operations using two-body ones. The key is introducing some ancilla qubits, and using two different scales of strength in the couplings so that you can apply degenerate perturbation theory.

It's been a long time since I read this material, so I'm a little bit hazy on some of the details, but crudely:

Consider, for example, that you want to create a 3-body term $Z_1Z_2Z_3$ between three qubits labelled 1,2 and 3. Introduce three ancillas, labelled A, B and C. If we make a Hamiltonian of the form $$ -\Delta(Z_AZ_B+Z_AZ_C+Z_BZ_C)+(Z_1X_A+Z_2X_B+Z_3X_C), $$ where $\Delta\gg 1$, then the ground state space of the qubits A,B,C is degenerate (neglecting the small terms) between $|000\rangle$ and $|111\rangle$. The only way that the perturbation term can preserve that ground state space is by acting at third order (or higher). The third order product is $Z_1Z_2Z_3X_AX_BX_C$, which basically gives you a term like $(1/\Delta^3)Z_1Z_2Z_3$ as an effective Hamiltonian for the low-energy dynamics.

Change the geometry from a triangle to a square, and you'll get 4-body interactions (although there are probably better methods).

One place that I remember a local version of this coming up was in Kitaev's honeycomb model, where it is proven that in a particular perturbative limit, the two-body interactions on a hexagon yield 4-body terms just as in the toric code.

  • $\begingroup$ Thank you. This was very helpful. $\endgroup$ Commented Jan 19, 2022 at 10:49
  • $\begingroup$ Does that gadget work? Would the Kempe-Kitaev-Regev paper of 2004 not have had coefficients of $-1/8$, $-1/6$ and $1/24$ and various different powers of $\Delta$ to make that work? $\endgroup$ Commented Jan 19, 2022 at 23:03
  • $\begingroup$ @user1271772 I have no idea - what I described was a crude overview and my recollection of the precise details is hazy. But it should be a good starting point for finding out more. $\endgroup$
    – DaftWullie
    Commented Jan 20, 2022 at 9:46
  • $\begingroup$ The problem with perturbation gadgets (here) is that they require non-commuting terms (after all, the perturbation should couple low- and high-energy sector), which seems to contradict the spirit of the question. -- Then, as you point out in the comment below, in the edited version asking for the ground state only, using H or -H is enough. -- Even further, any diagonal Hamiltonian will have classical ground states, which can be obtained in many ways (basically by imposing parallel spins to be parallel and anti-parallel spins to be anti-parallel; the remaining global Z2 symmetry is also ... $\endgroup$ Commented Jan 21, 2022 at 18:05
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    $\begingroup$ @quest I don't think it's a fundamental physical law that interactions must be two-body. Rather, two-body terms are far more common, and tend to be stronger, than any multi-body terms. $\endgroup$
    – DaftWullie
    Commented Mar 24, 2023 at 7:21

The answer by DaftWullie is useful for more general Hamiltonians, but since you are working specifically with the Ising Hamiltonian, then you do not need auxiliary qubits or perturbations which are two of the worst disadvantages of using perturbative gadgets. If you use perturbative gadgets, you might require millions of extra qubits, and coupling strengths between those qubits spanning several orders of magnitude.

Since the Ising Hamiltonian is diagonal (only being comprised of Pauli-$Z$ operatiors), you can simply use classical quadratization techniques, which have been used in the operations research community for decades.

Probably the simplest way (but certainly not the most sophisticated or efficient way) to reduce the Hamiltonian $Z_1Z_2Z_3Z_4$ is to first convert the $Z$ operators (which have eigenvalues +1 and -1) to $B$ operators (which describe precisely the same physics but with the energies of all states shifted up by a constant, or the "zero" of energy moved by up by one unit):

\begin{align} \tag{1} B &\equiv \frac{1}{2}\left(Z + I\right) \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.\tag{2} \end{align}

Then the ground state of a Hamiltonian such as the following (with $\alpha,\beta$ and $\gamma$ being whatever coupling-constants you wish):

$$ H = \alpha B_1B_2B_5 + \beta B_1B_6B_7B_8 - \gamma B_3B_4\tag{3}, $$

can be found by minimizing the following three 2-local Hamiltonians:

\begin{align} H_{0} &= -\gamma B_3B_4\tag{4}\\ H_{10} &= \alpha B_2B_5 - \gamma B_3 B_4 \tag{5}\\ H_{11} &= \alpha B_2B_5 + \beta B_7 B_8- \gamma B_3 B_4. \tag{6} \end{align}

$H_0$ comes from setting $B_1 = 0$, while $H_{10}$ and $H_{11}$ come from setting $\{B_1,B_6\} = \{I,0\}$ and $\{I,I\}$ respectively. If you minimize each of the three Hamiltonians, then check to see which configuration gives the overall minimum energy, you will get precisely the ground state of the original 4-local Hamiltonian by only minimizing three (much simpler) 2-local Hamiltonians.

In the literature this has been called "split-reduction", and notice that the cost of doing this was sub-exponential in the number of splits: in this case we need to minimize three Hamiltonians after doing 3 splits, instead of $2^2=4$.

Alternatively if you used the perturbative gadget in DaftWullie's answer, you would end up with an incredibly more complicated Hamiltonian that not only now has $X$ operators (turning it from stoquastic to non-stoquastic, which makes it significantly harder to minimize whether on quantum annealers or classical simulators), but you'd also need 9 extra qubits: Notice that $Z_1Z_2Z_3$ became a Hamiltonian involving $\{Z_1,Z_2,Z_3,Z_A,Z_B,Z_C\}$ which means 3 extra qubits to quadratize the cubic term, which would be needed twice for the $B_1B_6B_7B_8$ term and again once for the $B_1B_2B_5$ term, leading to 9 extra qubits in total. Increasing the size of the Hilbert space by a factor of $2^9$ is much worse than minimizing 3 far simpler Hamiltonians that are in smaller Hilbert spaces than the original Hamiltonian.

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    $\begingroup$ Just to check: are you only talking about correctly reproducing the ground state? If that were the case, it would be even easier: the ground state of $H^2$ is either the ground state of $H$ or $-H$. Does the method you describe also work if I want to get the time evolution of a state under the action of $H^2$? (In fact, I see you have changed the question to make it reflect only ground state properties. Are we sure that's what the OP wanted? I was answering a different question.) $\endgroup$
    – DaftWullie
    Commented Jan 20, 2022 at 9:53
  • $\begingroup$ @DaftWullie Neither of us answered the original question. OP asked if there's a way to "map H^2 so that it again becomes H", but H^2 and H are not the same Hamiltonian, they are different. OP must be asking how to map a 4-local Hamiltonian into a 2-local one (which is precisely the title) such that the low-lying spectrum is preserved. Almost always this is for AQC, where we're interested in the ground state. About H and -H, I hadn't thought much about the fact that OP talks about literally squaring H. The title got me thinking that they want to map general 4-local to 2-local. $\endgroup$ Commented Jan 20, 2022 at 10:04
  • $\begingroup$ Your question about time-evolution is an interesting one. I wonder if you could point me to a paper where the authors need to quadratize a 4-local Hamiltonian in order to do time-evolution on the 2-local Hamiltonian for the purposes of simulating the time-evolution of a 4-local Hamiltonian? $\endgroup$ Commented Jan 20, 2022 at 10:08
  • $\begingroup$ @DaftWullie There's other non-perturbative gadgets the preserve the full spectrum and would also work perfectly with time-evolution. I gave a simple scheme to preserve the ground state here, but there's more complicated ones to preserve the full spectrum too. $\endgroup$ Commented Jan 21, 2022 at 0:53

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