# Time reversal and basis independence

It is generally assumed that to time reverse a state, one just takes the complex conjugate of the wave function.

This is apparently not basis-independent.

For example, if we take $|\psi_0 \rangle$ as a basis vector in some basis, then the time reversal operator $K$ acts on it as

$K|\psi_0 \rangle = | \psi_0 \rangle$.

Now let us take $|\phi_0 \rangle = e^{-i \theta } |\psi_0 \rangle$, $\theta \in \mathbb{R}$, then

$K|\psi_0 \rangle = K (e^{i \theta } | \phi_0 \rangle ) = e^{-i \theta } | \phi_0 \rangle \neq |\psi_0 \rangle$.

Therefore, the simple complex conjugate recipe is not basis-independent. How can this be reconciled with the idea that it represents time reversal?

• Did you mean to ask how it can be that if $|\psi\rangle$ and $\alpha|\psi\rangle$ represent the same state, but if $K|\psi\rangle = |\psi\rangle$ and $\alpha$ is non-real then $K(\alpha|\psi\rangle) \ne \alpha|\psi\rangle$? Jun 19, 2014 at 15:53
• I mean, in the basis of $|\psi_0 \rangle$, $|\psi_0 \rangle$ is a real vector, and therefore, according to the complex conjugation recipe, i get $K|\psi_0\rangle = |\psi_0 \rangle$. However, in the basis of $|\phi_0 \rangle$, $|\psi_0\rangle$ is not real, and therefore, according to the recipe, $K |\psi_0 \rangle \neq |\psi_0 \rangle$. In different basis, you get different results! But the basic idea of physics is that any result should be coordinate-independent. Jun 19, 2014 at 16:06
• Change of basis means $U |\psi_0\rangle$ and $U K U^{-1}$ In this new basis $T \neq K$. Jun 20, 2014 at 3:32