5
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is certainly not basis independent, it holds in the position representation (for time-independent Hamiltonians). In a different basis the time-reversal operator takes a different form (like any operator, although this operator is anti-linear rather than linear).

$\endgroup$
5
  • $\begingroup$ Of course an operator takes a different form in a different basis. But, its action should not change. Its action should not vary from basis to basis. $\endgroup$ Jun 19, 2014 at 15:11
  • 1
    $\begingroup$ I am not sure I understand what you mean... In a different basis the time reversal operator does not act as complex conjugation anymore. $\endgroup$
    – doetoe
    Jun 19, 2014 at 15:38
  • $\begingroup$ 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 $K(\alpha|\psi\rangle) \ne \alpha|\psi\rangle$? $\endgroup$
    – doetoe
    Jun 19, 2014 at 15:53
  • $\begingroup$ 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. $\endgroup$ Jun 19, 2014 at 16:06
  • 1
    $\begingroup$ Change of basis means $U |\psi_0\rangle$ and $U K U^{-1} $ In this new basis $T \neq K$. $\endgroup$
    – Bubble
    Jun 20, 2014 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.