I'm trying to implement the Variational Quantum Eigensolver in Qiskit.

Suppose, I have an operator $A = \sigma_1^z\sigma_2^z$ acting on some two-qubit state $|\psi\rangle$. After a measurement I get a set of probabilities corresponding to states $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$.

My question is: How to calculate $\langle\psi|A|\psi\rangle$ using known set of probabilities?


3 Answers 3


Qiskit currently supports measurements in the computational basis from Qiskit Terra and Aer, that is, returning 1 if the qubit is in state $|1\rangle$, and 0 if the qubit is measured to be in state $|0\rangle$.

However, it is relatively easy to perform a change of basis unitary to our quantum circuit just prior to measurement, in order to instead measure in the eigenbasis of an arbitrary operator $A$:

  1. First, we need to compute the eigendecomposition of our Hermitian observable $A|v_i\rangle=\lambda_i|v_i\rangle$.

  2. Since $A$ is Hermitian, we are always able to diagonalize $A$ in this eigenbasis:

    $$A = \sum_{i}\lambda_i |v_i\rangle \langle v_i|= U^\dagger \Lambda U$$

    where $\Lambda = \text{diag}(\lambda_1,\dots,\lambda_n)$ and $U$ is a unitary matrix composed of the eigenvectors down the columns.

  3. Next, we apply the unitary operation $U$ to the end of our quantum circuit, using qc.unitary.

    To see why, note that $$\langle\psi|A|\psi\rangle = \langle\psi|U^\dagger \Lambda U|\psi\rangle = \langle\psi'|\Lambda|\psi'\rangle $$ where $|\psi'\rangle = U|\psi\rangle.$ Since $\Lambda$ is diagonal, we have transformed the problem from one where we must perform a measurement in an arbitrary basis, to one where we simply measure in the computational basis.

  4. Finally, we measure the state $|\psi'\rangle$ in the computational basis $|i \rangle$ using qc.measure(q, c), and execute the jobs to get the counts and probabilities $\mathbb{P}_i = |\langle i|\psi'\rangle|^2$. Using the previously computed eigenvalues of $A$, we can now reconstruct the expectation value:

    $$\langle \psi | A | \psi\rangle = \langle\psi'|\Lambda|\psi'\rangle = \sum_i \langle\psi'|i\rangle \langle i | \Lambda | i\rangle \langle i|\psi'\rangle = \sum_i \lambda_i \mathbb{P}_i$$

For example,

shots = result.results[0].shots
counts = result.get_counts()
probs = sorted([(i, c/shots) for i, c in counts.items()])
P = np.float64(np.array(probs)[:, 1])
A_expectation = lambda @ p

For a more high-level interface to coding and running variational quantum algorithms, you can also check out the PennyLane Python library, which has a Qiskit plugin available for using Qiskit simulators and IBM hardware as a backend.

For example, a expectation values of arbitrary operators in PennyLane using qiskit looks like this:

import pennylane as qml

dev = qml.device('qiskit.basicaer', wires=2)
# use 'qiskit.ibm' instead to run on hardware

def circuit(x, y, z):
    qml.RX(x, wires=0)
    qml.RY(y, wires=1)
    qml.RZ(z, wires=0)
    qml.CNOT(wires=[0, 1])
    return qml.expval(qml.Hermitian(A, wires=[0, 1]))

def cost(x, y, z):
    return (1-circuit(x, y, z))**2

# optimization follows

You can use NumPy, TensorFlow, or PyTorch to do the optimization - check out some of the tutorials.

Disclaimer: I am one of the developers on PennyLane.

  • 3
    $\begingroup$ Thank you for the answer! This means that for Pauli matrices all I need to do is multiply their eigenvalues (1,-1) by probabilities of getting qubit in state $|0\rangle$ or $|1\rangle$ and add. I'll try PennyLane out. $\endgroup$
    – C-Roux
    Commented Aug 5, 2019 at 8:59
  • 1
    $\begingroup$ Yep, exactly :) For the Pauli matrices (+Hadamard matrix), the above steps simplify as you've noticed because their eigenvalues co-incide with the eigenvalues of the Pauli-Z operator. In this special case, the change of basis operations are: * <X>: U = H * <Y>: U = H.S.Z * <H>: U = R_y(-pi/4) $\endgroup$
    – Josh Izaac
    Commented Aug 7, 2019 at 10:03
  • $\begingroup$ @Josh, could you please have a look at my similar question? Thanks. $\endgroup$
    – mavzolej
    Commented May 19, 2020 at 19:37

This can actually be easily done using the Qiskit Terra qiskit.quantum_info.analysis.average.average_data function that takes the counts data returned by a backend and the desired observable defined by a dict, list, or ndarray.

The doc-string for that function actually has the ZZ your looking for as an example.


I think it can be done with the method in this question(The first answer).

  1. Prepare $|0\rangle$
  2. Apply $V$
  3. Apply $U$
  4. Apply $V^\dagger$
  5. Measure in a basis for which $|0\rangle$ is one of the elements
  6. Repeat until you have a suitably accurate estimate of the probability of getting the answer $|0\rangle$.

Substitue $V$ with $U(\theta)$ will work. $$U(\theta)|0\rangle = |\psi \rangle$$


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