Question About How Qiskit Reset Gate Affects Other Entangled Qubits

Question

I am trying to understand how the reset gate in Qiskit affects qubits its entangled with. Consider the following circuit with qubits $q_0$ and $q_1$:

Where circuit240 takes $|0\rangle$ to $a|0\rangle + b|1\rangle$ and circuit 244 takes $|0\rangle$ to $c|0\rangle + d|1\rangle$. Right before the reset gate on qubit $q_1$ the state of this circuit is $$\frac{1}{\sqrt{2}}(c|1\rangle + d|0\rangle)|0\rangle + \frac{1}{\sqrt{2}}(a|1\rangle + b|0\rangle)|1\rangle \tag{1}$$

I'm not quite sure how to mathematically represent what this quantum reset gate does to the quantum entangled state. For example, I tried a few tests with this circuit where I played with the values $a$ and $c$ and tested how the quantum reset gate affects the measurement of the qubit $q_0$. It seems that whether or not the quantum reset gate is added, it doesn't affect the measurements of the other entangled qubit. Does this generally hold?

Furthermore, when I take the qubit expression above and just reset qubit $q_1$to $|0\rangle$ I get the following:

$$\frac{1}{\sqrt{2}}(c|1\rangle + d|0\rangle)|0\rangle + \frac{1}{\sqrt{2}}(a|1\rangle + b|0\rangle)|0\rangle \tag{2}$$

$$= \frac{1}{\sqrt{2}}((a + c)|1\rangle + (b + d)|0\rangle)|0\rangle \tag{3}$$

But, mathematically, the probability of measuring $q_0$ as $|0\rangle$ in $(3)$ is not the same as the probability in qubit expression $(1)$(even though the tests show that removing the reset gate did not change the probability of measuring a $|0\rangle$ in the qubit $q_0$. What is the correct way to represent what the qubit reset gate does to an entangled qubit?

Answer 1

A reset gate is equivalent to a swap gate between the target qubit and a new ancilla qubit in the $|0\rangle$ state. So you can replace your question with "how does swapping a qubit Q with an fresh ancilla qubit affect the qubits Q is entangled with?" or "how does discarding Q affect the qubits Q is entangled with". And the answer is that, for all intents and purposes, it doesn't affect them at all.

There is no test you can do on the qubits entangled with Q that can determined whether or not Q was discarded, or whether or not Q was swapped for a fresh ancilla qubit. Therefore there is no test you can do on the qubits entangled with Q (that don't involve measuring Q's value) whose outcome depends on whether or not Q was reset.

Answer 2

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

As I understand it, the reset will collapse the qubit to either $|0\rangle$ or $|1\rangle$ because of the measurement. Due to the superposition this will also collapse the qubit entangled with the reset qubit

source:
https://www.ibm.com/blogs/research/2021/02/quantum-mid-circuit-measurement/