I would like to find density matrix after the measurement. The toy code:

import qiskit
from qiskit.quantum_info import DensityMatrix

circuit = qiskit.QuantumCircuit(1, 1)
circuit.measure(0, 0)

results in the following error on the second print:

QiskitError: 'Cannot apply instruction with classical bits: measure'

How can I make it work? As I understand, being well-defined (and taking classical probabilities into account) after the measurement is exactly the advantage of the density matrix over the state vector.

I do know how to use save_density_matrix, and it indeed works. However, it only produces empirical results. E.g. when the true entry is 1/2, it might instead print 0.496.

It's more or less my first day working with qiskit and hearing about density matrix, so sorry if the question doesn't make sense. Ideally, I want to have the following information: a list of possible state vectors in the current state, with corresponding probabilities. As I understand, the density matrix is the closest alternative.


1 Answer 1


Accessing the Statevector $| \psi \rangle$ or the DensityMatrix $\rho = | \psi \rangle \langle \psi |$ of the quantum state prepared by a given circuit actually makes sense only before measurement, since measuring a qubit (in the usual computational basis) will make its state collapse into either $| 0 \rangle$ or $| 1 \rangle$. More in general, if you apply a measurement operation on each qubit of an $n$-qubits system you make its state collapse into $| \psi \rangle = | bs \rangle$, where $bs$ is a bit string in $\left \{0, 1\right \}^n$. This is why, in Qiskit, trying to access the state of the system after a call to circuit.measure is not even possible and you got that error.

However, what you might want to do is to access the density matrix of a subsystem of your circuit by using the so-called partial trace operation. Given the state $\rho$ in the Hilbert space $H_A \otimes H_B$, its partial trace with respect to the system $B$, denoted by $\rho^A$, is called the reduced state of $\rho$ on system $A$ and defined by $\rho^A = \mathrm{Tr}_B \rho$. In general, this density matrix represents a mixed state (can't be written as a statevector) and it corresponds to the state of subsystem $B$ you would get after a measure on subsystem $A$ (consistent with outcomes statistics).

In Qiskit, to compute the density matrix $\rho_A$, you can use the partial_trace(state, qargs) function, where state is your original $\rho$ and qargs is a list of qubits indices in the subsytem $B$ to trace out. Here is a simple example:

from qiskit import QuantumCircuit
from qiskit.quantum_info import DensityMatrix, partial_trace

qc = QuantumCircuit(3)
qc.cx(0, 1)
qc.cx(0, 2)

rho = DensityMatrix(qc)
rho_a = partial_trace(state=rho, qargs=[1, 2])
  • $\begingroup$ "DensityMatrix ρ=|ψ⟩⟨ψ| of the quantum state prepared by a given circuit actually makes sense only before measurement": you are talking about pure states. Density matrix can represent mixed states, which, according to the link, is defined after measurement. $\endgroup$
    – Dmitry
    Feb 24, 2023 at 16:47
  • $\begingroup$ Yes exactly! But measurement taken on a different subsystem than the one whose density matrix you want to get: this is exactly what the partial trace operation performs. However, in your toy code you are trying to access the density matrix on the same qubit you have just measured $\endgroup$ Feb 24, 2023 at 18:29
  • $\begingroup$ I see, sorry, I'm pretty new, so I didn't understand your answer. I think I got a general idea now, but I don't really understand what exactly going on (e.g. why we need cnot here). Can I ask you please to elaborate on what happens in the code? Also, will it work if I want to find the density matrix after performing a second measurement? $\endgroup$
    – Dmitry
    Feb 24, 2023 at 19:50
  • $\begingroup$ The code here is just an example to prepare an entangled 3-qubits state. I really suggest you to take a look to the Qiskit Textbook and you will find all your answers with the corresponding code. $\endgroup$ Feb 24, 2023 at 21:31
  • $\begingroup$ By the way, performing a second measurement on a qubit doesn't make sense because its state would have already collapsed to a classical 0 or 1. $\endgroup$ Feb 24, 2023 at 21:35

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