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I'm making a quantum circuit with qubits and quantum gates. While I'm doing it, I have some problem with it. My calculation process is below. enter image description here

As you can see, start qubit is $|0 \rangle$ and after 'X' gate, the result will be $|1 \rangle$. And I checked if the result of quantum circuit and density matrix form of $|1 \rangle$ are different. As you can see, they are different.

Here are my questions.

  1. Why it occurs?
  2. If they are different, is my calculation wrong?
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When you evolve a pure state under the action of a gate $U$, it evolves from $$ |\psi\rangle\rightarrow U|\psi\rangle. $$ However, a density matrix such as $\rho=|\psi\rangle\langle\psi|$ evolves differently. You must calculate $$ \rho\rightarrow U\rho U^\dagger. $$ So, in this case, you must calculate $$ \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)=\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right). $$

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  • $\begingroup$ Thank you for reply. Why we have to multiply conjugate transpose matrix? I wonder the reason. Could you tell me about it? $\endgroup$
    – 김동민
    Oct 30 '20 at 8:45
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    $\begingroup$ Simply because the bra of $U|\psi\rangle$ is $(U|\psi\rangle)^\dagger = \langle \psi |U^\dagger$. Therefore the evolution $|\psi\rangle\mapsto U|\psi\rangle$ is in the density matrix picture $|\psi\rangle\langle \psi |\mapsto U|\psi\rangle\langle \psi |U^\dagger$ $\endgroup$ Oct 30 '20 at 9:21
  • $\begingroup$ Could you be more in detail? I can't understand now. $\endgroup$
    – 김동민
    Oct 30 '20 at 9:46
  • $\begingroup$ You know that to convert a state $|\psi\rangle$ into a density matrix, you rewrite it as $\rho=|\psi\rangle\langle\psi|$. How do you rewrite the state $U|\psi\rangle$ as a density matrix? $\endgroup$
    – DaftWullie
    Oct 30 '20 at 10:42
  • $\begingroup$ Okay! I got this. Thank you $\endgroup$
    – 김동민
    Oct 30 '20 at 11:09

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