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

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

• 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.
– JSdJ
Mar 1 at 14:42

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