Observables, Hamiltonians & Real Applications

Turn 0/1 counts into ⟨Z⟩, measure X and Y by rotating first, and add Pauli terms into the energy that chemistry and optimisation minimise.

Advanced14 min read8 sections · 2 questions+45

By the end you will be able to

01Compute ⟨Z⟩ = P(0) − P(1) from a histogram.
02Measure ⟨X⟩ and ⟨Y⟩ with a basis change before the measurement.
03Write a Hamiltonian as a weighted sum of Pauli strings and estimate its energy term by term.

A measurement gives bits, but chemistry and optimisation ask for numbers such as an energy or a cost. The bridge is an observable: give each outcome a value and average over the shots. For Z the rule is simple: outcome 0 counts as +1 and outcome 1 counts as −1.

⟨ Z ⟩ = ⟨ ψ ∣ Z ∣ ψ ⟩ = (+ 1) P (0) + (- 1) P (1) = P (0) - P (1) \approx \frac{n _{0} - n _{1}}{N}

⟨Z⟩ is +1 for |0⟩, −1 for |1⟩ and 0 for |+⟩. With N shots, n₀ zeros and n₁ ones, the estimate is (n₀ − n₁)/N.

01Measuring X and Y

Observable	Gates before the measurement	Why
Z	none	the hardware measures Z
X	h	H swaps the X and Z axes
Y	sdg, h	S† turns the Y axis into the X axis, then H as above

Hands-on

⟨Z⟩ and ⟨X⟩ of one state

ry(pi/3) tilts the qubit 60° from |0⟩. Simulate: P(0) ≈ 0.75 and P(1) ≈ 0.25, so ⟨Z⟩ ≈ 0.5. Now add h q[0]; before the measurement and simulate again: P(0) ≈ 0.933, so ⟨X⟩ = 2·0.933 − 1 ≈ 0.87, the exact value being √3/2.

OpenQASM 3

OPENQASM 3.0;
include "stdgates.inc";
qubit[1] q;
bit[1] c;
ry(pi/3) q[0];
c[0] = measure q[0];

Open in the IDE Open in the Playground

02Several qubits: Pauli strings

A Pauli string such as Z₀Z₁ multiplies the ±1 values of its qubits: outcomes with an even number of 1s count +1, odd ones −1. For the Bell state (|00⟩ + |11⟩)/√2 every shot is 00 or 11, so ⟨Z₀Z₁⟩ = +1, while ⟨Z₀⟩ alone is 0. Measuring X₀X₁ means an h on both qubits first; the Bell state gives +1 there too.

Hands-on

Correlations as expectation values

Simulate: only 00 and 11, so ⟨Z₀Z₁⟩ = +1. Add h q[0]; h q[1]; before the measurement: still only 00 and 11, so ⟨X₀X₁⟩ = +1 as well. Two independent qubits in |+⟩|+⟩ would give ⟨Z₀Z₁⟩ = 0.

OpenQASM 3

OPENQASM 3.0;
include "stdgates.inc";
qubit[2] q;
bit[2] c;
h q[0];
cx q[0], q[1];
c = measure q;

Open in the IDE Open in the Playground

03Hamiltonians: energy as a sum of Pauli terms

H = i \sum c_{i} P_{i} \Rightarrow ⟨ H ⟩ = i \sum c_{i} ⟨ P_{i} ⟩

Expectation values are linear: estimate each Pauli string (strings that share a measurement basis can share the same shots) and add them with their weights. This is the job of the Estimator primitive.

Chemistry: once electrons are mapped to qubits, a molecule's energy becomes a sum of Pauli strings. The smallest textbook case, H₂ in a minimal basis, can be reduced to two qubits with terms such as Z₀, Z₁, Z₀Z₁, X₀X₁ and Y₀Y₁. VQE searches for the θ that makes ⟨H⟩ as low as possible; that minimum approximates the ground-state energy.
Optimisation: in MaxCut an edge (i, j) contributes (1 − ZᵢZⱼ)/2, which is 1 when the two nodes are on different sides and 0 otherwise. Summing over the edges gives the cost Hamiltonian that QAOA works with. Portfolio selection and scheduling problems are written the same way, as weighted Z and ZZ terms.

Knowledge check

2 questions · one try each counts

Question 1

1000 shots give 900 zeros and 100 ones. What is the estimate of ⟨Z⟩?

Question 2

H = 2Z − X and the state is |+⟩. What is ⟨H⟩?

Observable

Gates before the measurement

Why

none

the hardware measures Z

H swaps the X and Z axes

sdg, h

S† turns the Y axis into the X axis, then H as above

02Several qubits: Pauli strings

03Hamiltonians: energy as a sum of Pauli terms

H = i \sum c_{i} P_{i} \Rightarrow ⟨ H ⟩ = i \sum c_{i} ⟨ P_{i} ⟩

Chemistry: once electrons are mapped to qubits, a molecule's energy becomes a sum of Pauli strings. The smallest textbook case, H₂ in a minimal basis, can be reduced to two qubits with terms such as Z₀, Z₁, Z₀Z₁, X₀X₁ and Y₀Y₁. VQE searches for the θ that makes ⟨H⟩ as low as possible; that minimum approximates the ground-state energy.
Optimisation: in MaxCut an edge (i, j) contributes (1 − ZᵢZⱼ)/2, which is 1 when the two nodes are on different sides and 0 otherwise. Summing over the edges gives the cost Hamiltonian that QAOA works with. Portfolio selection and scheduling problems are written the same way, as weighted Z and ZZ terms.