# How does a qubit reset affect the probabilities of the other qubits in Qiskit?

I'm looking at a 2 qubit system, with a reset on the second qubit (qubit=1).

I would have expect that circ.reset(qubit=1) is equal to state.probabilities(qargs=).
But that's not the case.

For example a maximally pure state:

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

i_state = Statevector.from_label('00')
t_state = i_state.evolve(qc)

plot_histogram(data=t_state.probabilities_dict())
plot_histogram(data=t_state.probabilities_dict(qargs=))


results in:
50% $$|00\rangle$$, 50% $$|11\rangle$$ and 50% $$|0\rangle$$, 50% $$|1\rangle$$

qc.reset(qubit=1)


results sometimes in: 100% $$|00\rangle$$ and 100% $$|0\rangle$$
OR sometimes in: 100% $$|01\rangle$$ and 100% $$|1\rangle$$.

What does .reset(qubit=1) actually do?
Does it sample from the probability of qubit=1 and collapse the other, entangled qubit=0 depending on how qubit=1 was sampled?

According to IBM Research Blog:

Internally, these reset instructions are composed of a mid-circuit measurement followed by an x-gate conditioned on the outcome of the measurement.

So, your circuit is equivalent to the following circuit. This should explain to you the result you get:

• If measurement result is $$1$$ the state will collapse to $$|11\rangle$$ and $$X$$ gate will be applied resulting in $$|01\rangle$$
• If measurement result is $$0$$ the state will collapse to $$|00\rangle$$ and $$X$$ gate will not be applied. Final state remains $$|00\rangle$$

In general, the state will collapse to one of the possible states. Then the resetted qubit state will change to become zero if it is not zero already.

• Apr 27, 2022 at 12:35