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I'm trying to understand the effect of "measurement" as a gate. But my question is more elementary than that. Consider the following simple quantum circuit:

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  1. $(X,Y)$ begins in the state $1 | 10 \rangle$ .

  2. A 2 Qubit Hadamard Gate $H_2$ is applied to both Qubits resulting in the state $$\frac{1}{2} |00 \rangle + \frac{1}{2} | 01 \rangle - \frac{1}{2}| 10\rangle - \frac{1}{2} | 11 \rangle $$

  3. Now a single qubit Hadamard gate is applied $X$. Here's where I get lost. I would like to believe there is a qubit description of $X$ alone to be able to apply this gate to.

I know that from a measurement standpoint if $(X,Y)$ is in the state $a_0 |00\rangle + a_1 | 01\rangle + a_2 |10 \rangle + a_3 |11 \rangle$ that the state of $X$ can be described as $\sqrt{|a_0|^2 + |a_1|^2} |0\rangle + \sqrt{|a_2|^2 + |a_3|^2}|1\rangle$. But this is NOT the quantum state of $X$. It's just a probabilistic description of what we expect to observe with a particular probability.

In some cases such as a bell state $ a_0 |00 \rangle + a_1 | 11 \rangle $ it's easy to reason that the quantum state of $X$ is $a_0 | 0 \rangle + a_1 | 1 \rangle$. But a state that has all 4 components non-zero like the Hadamard configuration above does NOT give me any obvious way to take such a projection. Is it even meaningful to ask "what is the qubit description of $X$?". It's also worth mentioning that if $(X,Y) = O_1 \otimes O_2 $ then there might be a suitable tensor decomposition such that $O_1$ describes $X$.

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    $\begingroup$ The search strings you want to Google are "partial trace" or "reduced density matrix" $\endgroup$ Commented May 3 at 15:13
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    $\begingroup$ In the example you give, the qubits are separable throughout, so there is always a good description of just qubit X (say). Onn the other hand, you say "In some cases such as a bell state, it's easy...". It might seem easy, but whatever reasoning you have is false. There is not a pure state description of X in that case because the system is entangled. $\endgroup$
    – DaftWullie
    Commented May 3 at 15:50
  • $\begingroup$ @SidharthGhoshal if you apply $H$ to the first qubit of a Bell state you do not only observe $|0\rangle$, but rather you get $|0\rangle$ or $|1\rangle$ with 50/50 probability. In fact, the same is true if you apply any local unitary operation to either qubit before the measurement $\endgroup$
    – glS
    Commented May 6 at 21:45
  • $\begingroup$ I tried to simulate this with cirq and it looks like you are right. My mental models are totally incorrect about these things. $\endgroup$ Commented May 6 at 23:07

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