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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 $q_1$:

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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?

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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.

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  • $\begingroup$ Got it. Thank you for the answer. However, this means that if I want to uncompute/reverse a circuit with a reset gate I would need to compute the value of the qubit before resetting and then swap that in when uncomputing, correct? $\endgroup$ – Rehaan Ahmad Jul 10 at 3:11
  • $\begingroup$ @RehaanAhmad You can't use a reset gate to uncompute anything, and if you uncompute something you don't need the reset gate (because the qubit will be in the zero state already). $\endgroup$ – Craig Gidney Jul 10 at 4:23
  • $\begingroup$ Perhaps I misphrased the comment above. Lets say I have a circuit that takes $|v\rangle$ to $U|v\rangle$ where U is a circuit that has some reset gates. Now, I want to make another circuit $U^{\dagger}$, that takes $U|v\rangle$ to $|v\rangle$. In this case, how could I construct the circuit $U^{\dagger}$? I would need to know the value of each of the qubits before they were initially reset in order to do so right? $\endgroup$ – Rehaan Ahmad Jul 10 at 5:16
  • $\begingroup$ Typically you'd implement a $U^\dagger$ that would work for any value, not just one. $\endgroup$ – Craig Gidney Jul 10 at 10:13
  • $\begingroup$ Right, which means that reversing a circuit with the reset gate isn't a straightforward task, correct (as opposed to reversing other more standard gates)? Let's say I have a circuit that is just the reset gate, taking any $|v\rangle$ to $|0\rangle$. There is no easy way to uncompute this circuit, right? (other than having a copy of $|v\rangle$ to begin with and using it) $\endgroup$ – Rehaan Ahmad Jul 10 at 19:19

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