How to prove that the transpose operation maps an arbitrary qubit to its complex conjugate?

Question

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

Answer 1

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 $|\phi\rangle=|\psi^\star\rangle$.

Answer 2

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.

Answer 3

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.