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How to prove that the transpose operation maps an arbitrary qubit to its complex conjugate, $|\psi^*\rangle \rightarrow |\psi\rangle$

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    $\begingroup$ In case you're not aware if it: please note that the transpose is not a valid quantum map, as it is not completely positive. $\endgroup$ – JSdJ Mar 1 at 14:42
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Think about that projector $$ \rho=|\psi\rangle\langle\psi|. $$ Note that this is Hermitian, $\rho^\dagger=\rho$. Take the transpose, $$ {\rho^\dagger}^T=\rho^T $$ but since the hermitian conjugate is the complex conjugate transpose, ${\rho^\dagger}^T=\rho^\star$.

If you want to see what pure state $\rho^\star=|\phi\rangle\langle\phi|$ corresponds to, think of $\rho$ as a matrix and find a non-zero column. That column is proportional to $|\psi\rangle$. The same column of $\rho^\star$ is proportional to $|\phi\rangle$. It should thus be clear that $|\phi\rangle=|\psi^\star\rangle$.

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This question makes more sense in density matrix notation. You can then ask how to prove that $$ (|\psi\rangle\langle \psi|)^T = |\psi^*\rangle\langle \psi^*|.$$ It's not difficult, just write the state in some basis, $|\psi\rangle = \sum_i \alpha_i |i\rangle$, and apply the operations.

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I hope I understoody your question correctly but I think that there is nothing to prove, it is simply a definition.

Lets have a quantum state $|\psi\rangle$ which is described by a column vector. Then quantum state $\langle \psi|$ is defined as $(|\psi\rangle^T)^*$, i.e. coefficients are complex conugate numbers and the column vector is transposed. In the end you, have row vector with complex conjugated coefficients.

Maybe, a thread How does bra-ket notation work? would be helful for getting deeper understanding.

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