# Which one of them is the time-reversed wave-function, $\psi^{\ast }\left( x,t\right)$ or $\psi^{\ast}\left( x,-t\right)$?

If the wave function $$\psi\left( x,t\right)$$ is a solution of the spinless time-independent Schr$$\ddot{\mathrm{o}}$$dinger equation, $$i\hbar\frac{\partial}{\partial t}\psi\left( x,t\right) =\left[ -\frac {\hbar^{2}}{2m}\nabla^{2}+V\left( \mathbf{r}\right) \right] \psi\left( x,t\right)$$ then, $$\psi^{\ast}\left( x,-t\right)$$ is also the solution $$i\hbar\frac{\partial}{\partial t}\psi^{\ast}\left( x,-t\right) =\left[ -\frac{\hbar^{2}}{2m}\nabla^{2}+V\left( \mathbf{r}\right) \right] \psi^{\ast}\left( x,-t\right)$$ and can be defined as the time reversed wave function of $$\psi\left( x,t\right)$$

$$\psi_{r}\left( x,t\right) =\psi^{\ast}\left( x,-t\right)$$

However, in many discussions about the time-reversed operation, the time reversed wave function $$\psi_{r}\left( x,t\right)$$ is obtained by applying the time reversal operator $$K$$, which is the complex conjugate of the wave function,

$$\psi_{r}\left( x,t\right) =K\psi\left( x,t\right) =\psi^{\ast}\left( x,t\right)$$

So my question is, which one is the time reversed wave function $$\psi^{\ast }\left( x,t\right)$$ or $$\psi^{\ast}\left( x,-t\right) ?$$

The general expression for the time-reversal operator $$T=UK$$ (Eq. (4.4.14) in Modern Quantum Mechanics by J. J. Sakurai), where $$U$$ is a unitary operator and $$K$$ is the complex conjugation operator. For spinless case, one can choose $$U=1$$, so $$T=K$$.

• $\psi^*(x,t)$ was not always guaranteed to be a solution, so you might as well ignore that.(my guess is they meant $t$ going backwards, thus if you had an zero off set, it's equivalent to $-t$ where $t$ going forward.) But could you give a reference of your equations and claims? – ShoutOutAndCalculate May 31 '19 at 22:39
• Wavefunctions don't have to be expressed as functions of $x$ and $t$, so any general definition of a time-reversal operator that assumes $x$ as an input would be kind of odd. E.g., you could be working in the momentum basis. Note that taking a function $f(t)$ and sending it to $f(-t)$ is a unitary operation. So maybe the answer to your question is that both of the things you suggest could be considered the time-reversed wavefunction, depending on what you choose for $U$. Not posting this as an answer because I don't understand this well, not sure I'm right. – user4552 Oct 27 '19 at 20:22

## 1 Answer

As per your reference, it seems that you have mistaken anti-unitary operators for the time reversal operator. The time reversal operator is a kind of anti-unitary operator. The general expression for an anti-unitary operator is, as you had mentioned, on page 269 equation 4.4.14 of J.J Sakurai's book: $$\theta = U K$$ Where $$\theta$$ is an anti-unitary operator, U is a unitary operator and K is the complex conjugation operator. You can't simply take U as the identity, as even though this is an anti-unitary operator, it is not necessarily the time reversal operator.

• For particles with spins, one cannot simply take U as identity. For spinless particles, why can one not choose U as the identity? If one cannot choose U as the identity, then what would be the expression for the time-reversal operator for spinless particles? – Hanks Jun 2 '19 at 16:15
• This doesn't really seem to address the question. – user4552 Oct 27 '19 at 20:23