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Basically the title. From what I know, starting from the $|{+0}\rangle$ state where the reduced density matrix of the first qubit $|+\rangle$ is $\frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, we can create the Bell state $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, in which the reduced density matrix becomes having zero off-diagonal terms $\frac{1}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. How is this described mathematically? Is it related to open quantum system?

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  • $\begingroup$ I think the $|+\rangle$ state is $\frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$, which is why its density matrix is like so? Am I correct? $\endgroup$ May 16 at 11:05
  • $\begingroup$ yes, you are correct $\endgroup$
    – qubitzer
    May 16 at 14:02

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The action $\mathcal{E}$ of CNOT on just the control qubit is the composition of two functions: the action \begin{align} \mathcal{C}(\rho)=\text{CNOT}\,\rho\,\text{CNOT}\tag1 \end{align} of CNOT on both qubits followed by the partial trace \begin{align} \mathcal{P}(\rho)=\mathrm{tr}_2(\rho)\tag2 \end{align} to discard the target qubit. We can use $\text{CNOT}|{+0}\rangle=\frac{1}{\sqrt2}\sum_{i\in\{0,1\}}|ii\rangle$ to compute it as follows \begin{align} \mathcal{E}(|{+0}\rangle\langle{+0}|)&=\mathcal{P}(\mathcal{C}(|{+0}\rangle\langle{+0}|))\tag3\\ &=\mathrm{tr}_2(\text{CNOT}|{+0}\rangle\langle{+0}|\text{CNOT})\tag4\\ &=\frac12\sum_{i,j\in\{0,1\}}\mathrm{tr}_2(|ii\rangle\langle jj|)\tag5\\ &=\frac12\sum_{i,j\in\{0,1\}}|i\rangle\langle j|\cdot\langle i|j\rangle\tag6\\ &=\frac12\sum_{i\in\{0,1\}}|i\rangle\langle i|\tag7\\ &=\frac12\begin{bmatrix}1&0\\0&1\end{bmatrix}\tag8 \end{align} which is indeed the maximally mixed state.

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  • $\begingroup$ Well I understand this approach, but this doesn't describe what happen to ONLY the control qubit. Is there some sort of operator that act ONLY on the density matrix of the control qubit? Obviously it would be non-unitary (or even non-Hermitian?), but I wonder if there's a picture that only takes the density matrix of the control qubit as variable. $\endgroup$ May 16 at 13:27
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In the ZX calculus, the CNOT gate factors into a Z type node for the control linked to an X type node for the target. The Z type node (the "control part of the operation") has three ports: $i$ the control-input line, $o$ the control-output line, and $c$ the to-other-qubit coupler line.

enter image description here

The tensor of the Z type node is $|0_i0_o0_c\rangle + |1_i1_o1_c\rangle$.

The stabilizer flow generators of the Z type node are $+Z_iZ_o$, $+Z_iZ_c$, and $+X_iX_oX_c$.

These are two very useful mathematical representations of the just-the-control-of-a-CNOT operation, that fit into larger frameworks for simulating quantum processes.

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This depends on the question if you are allowing to use the information about the initial state of qubit 2.

I am pointing this out because the information about qubit 2 is no longer contained in the reduced density matrix that you want to start with $\rho_1 = |+\rangle \langle +|$

If you know the initial state of qubit 2 (and are willing to use it), the mathematical operations to retrieve the final state of qubit 1 are (as you already know, I guess):

  • include qubit 2 in your mathematical description again (simple tensor product)
  • apply CNOT
  • trace out qubit 2

If you do not know the original state of qubit 2 it is impossible to describe what CNOT does to qubit 1.

You can see this by the following example:

  • Let's say qubit 1 is (like in your case) in the state $\rho_1 = |+\rangle \langle +|$
  • Let's further say, qubit 2 is also in the state $\rho_2 = |+\rangle \langle +|$
  • Applying CNOT on them has no effect
  • Finally, tracing out qubit 2 would leave you with $\rho_1 = |+\rangle \langle +|$ and not with the maximally mixed state

Thus, your final result depends on the initial state of qubit 2

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  • $\begingroup$ I think perhaps it's not the action of the CNOT, but of the CNOT and the target qubit. What I'm looking for is some sort of Hamiltonian picture that described how a state lose its purity. $\endgroup$ May 17 at 8:17
  • $\begingroup$ Okay, in this case I refer to Craig Gidney's answer. ZX-calculus is a powerful tool for many things. For example, you can diagramatically visualize enganglement with it. And entanglement between the two qubits is the reason why qubit 1 loses its purity. $\endgroup$
    – qubitzer
    May 17 at 22:10

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