Chapter 2: The Diagonal Z₂ Approach

In this chapter, you’ll learn to detect diagonal Z₂ symmetries and understand the machinery behind sector fixing.

In This Chapter

What you’ll learn: How to algorithmically detect diagonal Z₂ qubits, what sectors are, and how fixing a sector modifies the Hamiltonian
Why this matters: Before you can taper, you must know which qubits are safe to remove and what choices you have
Try this next: Jump to Chapter 3 — FockMap Implementation to code it up.

Detection: Which Qubits Are Z₂ Symmetric?

Given a Hamiltonian $\hat{H} = \sum_\alpha c_\alpha \sigma_\alpha$, we say qubit $j$ is diagonal Z₂ symmetric if:

\[\forall \alpha: \sigma_\alpha[j] \in \{\mathrm{I}, \mathrm{Z}\}\]

In words: on every single Pauli term $\sigma_\alpha$, the Pauli operator at position $j$ is either the identity $\mathrm{I}$ or $\mathrm{Z}$, never $\mathrm{X}$ or $\mathrm{Y}$.

Algorithmic Check

For each qubit index $j = 0, 1, \ldots, n-1$:

Inspect every term in the Hamiltonian
Extract the Pauli operator at position $j$
If all operators are I or Z, then $j$ is diagonal Z₂ symmetric
If any operator is X or Y, then $j$ is not diagonal Z₂ symmetric

Example: In the Hamiltonian $\hat{H} = 0.8 \,\mathrm{ZIZI} - 0.4 \,\mathrm{ZZII} + 0.3 \,\mathrm{IIZZ}$:

Qubit	Position 0	Position 1	Position 2	Position 3	Diagonal?
Term 1: ZIZI	Z	I	Z	I	—
Term 2: ZZII	Z	Z	I	I	—
Term 3: IIZZ	I	I	Z	Z	—
Result	Z, Z, I → ✓	I, Z, I → ✓	Z, I, Z → ✓	I, I, Z → ✓	All 4 diagonal

All qubits are candidates for removal.

Sectors: Choosing Eigenvalues

For a diagonal Z₂ symmetric qubit $j$, we can fix its Z eigenvalue to be either $+1$ (spin-up) or $-1$ (spin-down).

A sector is a choice of eigenvalue for each qubit we plan to remove.

Mathematically, a sector is a list of pairs:

\[\text{sector} = \{(j_1, \lambda_1), (j_2, \lambda_2), \ldots\}\]

where $j_i$ is a qubit index and $\lambda_i \in {+1, -1}$ is the chosen eigenvalue of $Z_{j_i}$.

Example sectors for the 4-qubit Hamiltonian:

sector = [(0, +1), (1, +1)] — “Fix qubits 0 and 1 to spin-up; leave 2 and 3”
sector = [(1, -1), (3, +1)] — “Fix qubit 1 to spin-down, qubit 3 to spin-up; leave 0 and 2”
sector = [(0, +1), (1, +1), (2, +1), (3, +1)] — “Fix all four to spin-up; reduce to 0 qubits (scalar)”

Modification: How Fixing a Sector Changes Terms

When you fix a sector, each Pauli term is modified as follows:

For each $(j, \lambda)$ in the sector:
- If term $\sigma_\alpha$ has I at position $j$: no change
- If term $\sigma_\alpha$ has Z at position $j$: multiply the coefficient by $\lambda$
Remove qubit $j$ from the tensor product (delete its position from every Pauli string)

Worked Example

Start with: $\hat{H} = 0.8 \,\mathrm{ZIZI} - 0.4 \,\mathrm{ZZII} + 0.3 \,\mathrm{IIZZ}$

Fix sector [(1, +1), (3, -1)] — qubits 1 and 3 to eigenvalues +1 and −1 respectively.

Process term by term:

Original	Qubit 1	Factor	Qubit 3	Factor	New Coeff	Remaining
$0.8 \mathrm{ZIZI}$	I	×1	I	×1	$0.8$	$\mathrm{ZZ}$
$-0.4 \mathrm{ZZII}$	Z	×(+1)	I	×1	$-0.4$	$\mathrm{ZI}$
$0.3 \mathrm{IIZZ}$	I	×1	Z	×(−1)	$-0.3$	$\mathrm{IZ}$

Result: $\hat{H}' = 0.8 \,\mathrm{ZZ} - 0.4 \,\mathrm{ZI} - 0.3 \,\mathrm{IZ}$

Multiple Sectors: Different Physics in Each

For a system with $k$ removable qubits, there are $2^k$ possible sectors. Each sector gives a valid, eigenvalue-preserving tapered Hamiltonian, but they may represent different quantum numbers or molecular states.

To recover the ground state of the full problem:

Compute eigenvalues in all $2^k$ sectors
Find the global minimum across sectors

Sign Conventions and Validation

To avoid bugs:

Always validate before tapering — check that all qubits in your sector list are actually diagonal Z₂ symmetric
Check sector eigenvalues — must be exactly $+1$ or $-1$, not other values
Trace coefficients carefully — when a Z eigenvalue is $-1$, that factor multiplies the coefficient. This is not a sign error; it is the physics.

Previous: Chapter 1 — Why Tapering?

Next: Chapter 3 — FockMap Implementation