5

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 $T=K$.

Hanks
  • 51
  • $\psi^*(x,t)$ 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. –  Oct 27 '19 at 20:22
  • 5
    Does this answer your question? Schrödinger's equation - Time reversal – Rishi Jul 27 '21 at 07:30

2 Answers2

0

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.

Sparsh Mishra
  • 657
  • 2
    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 02 '19 at 16:15
  • This doesn't really seem to address the question. –  Oct 27 '19 at 20:23
0

In the quantum mechanics the operators don't act on the functions of $(t,x)$. They act on the functions of $(x)$. So you can't define the time reversal operator as changing the direction of time. You simply define it as some anti-linear transformation that in general case may include some linear operator $\hat{T}$, \begin{equation} \mathcal{T}\psi(x)=\hat{T}\psi^\star(x) \end{equation}

However this anti-linear transformation implies the reversal of the direction of time. How? The time dependence of the wavefunction in the Schrodinger picture is obtained with help of the evolution operator, \begin{equation} \psi_t(x)=\exp\Big(-i\hat{H}t\Big)\psi_0(x) \end{equation} If you act with $\mathcal{T}$ then thanks to its anti-linearity, \begin{equation} \mathcal{T}\psi_t(x)=\exp\Big(+i\hat{H}^Tt\Big)\mathcal{T}\psi_0(x) \end{equation} where $\hat{H}^T=\mathcal{T}\hat{H}\mathcal{T}$ - the time-reversed Hamiltonian. So if $\hat{H}=\hat{H}^T$ you may write $\mathcal{T}(\psi_t(x))=(\mathcal{T}\psi)_{-t}(x)$.

Similarly the evolution of the operators in the Heisenberg picture, \begin{equation} \mathcal{T}\hat{O}_t\mathcal{T}=\mathcal{T}\exp\Big(+i\hat{H}t\Big)\hat{O}\exp\Big(-i\hat{H}t\Big)\mathcal{T}=\exp\Big(-i\hat{H}^Tt\Big)\mathcal{T}\hat{O}\mathcal{T}\exp\Big(+i\hat{H}^Tt\Big) \end{equation}

I.e. both this objects evolve in the reversed time direction with the time-reversed Hamiltonian $\hat{H}^T$. If you consider the operator $\mathcal{T}\hat{O}\mathcal{T}=\hat{O}$ and $\hat{H}=\hat{H}^T$ then $\mathcal{T}\hat{O}_t\mathcal{T}=\hat{O}_{-t}$.

OON
  • 8,324