A Time Reversal and initial conditions

[hokhani]

TL;DR

How can $\psi^*(x,-t)$ be a solution of the Schrodinger equation with specific initial conditions?

In the solution of Schrodinger equation, $H\psi =i\hbar \frac{\partial \psi}{\partial t}$, the solution $\psi(x,t)$ depends on the initial condition at $t=t_0$. It is mentioned in the literatures that if $H=H^*$ then $\psi^*(x,-t)$ is another solution. However, $\psi^*(x,-t)$ is not consistent with the initial condition. In other words, $\psi^*(x,-t_0) \ne \psi(x,t_0)$! Any help is appreciated.

 

[div_grad]

In other words, $\psi^*(x,-t_0) \ne \psi(x,t_0)$!

Why not? For the simplest case, consider a time-dependent part of the wave function at $t = t_0$ to have a well-known form $\psi(t_0) = \exp\left(-\frac{\mathrm{i}}{\hbar} E t_0\right)$, with $E$ the eigenvalue of the Hamiltonian. Now, write down the corresponding $\psi^*(-t_0)$ and make use of the fact that $E$ is a real number (because it is an eigenvalue of the Hermitian operator). You will immediately see that, in fact, $\psi^*(-t_0) = \psi(t_0)$.

 

[hokhani]

Why not? For the simplest case, consider a time-dependent part of the wave function at $t = t_0$ to have a well-known form $\psi(t_0) = \exp\left(-\frac{\mathrm{i}}{\hbar} E t_0\right)$, with $E$ the eigenvalue of the Hamiltonian. Now, write down the corresponding $\psi^*(-t_0)$ and make use of the fact that $E$ is a real number (because it is an eigenvalue of the Hermitian operator). You will immediately see that, in fact, $\psi^*(-t_0) = \psi(t_0)$.

Thanks, you have considered only the time-dependent part of the simplest case of time-independent Hamiltonian. Consider the free particle case with the initial condition $\psi(x,t_0=0)=e^{ikx}$ and then,
$$\psi(x,t)=e^{- \frac{iEt}{\hbar}} e^{ikx}$$ and so $$\psi^*(x,-t)=e^{- \frac{iEt}{\hbar}} e^{-ikx}.$$ In this case: $\psi^*(x,0)=e^{-ikx} \ne \psi(x,0) =e^{ikx}$.

 

[PeroK]

TL;DR Summary: How can $\psi^*(x,-t)$ be a solution of the Schrodinger equation with specific initial conditions?

In the solution of Schrodinger equation, $H\psi =i\hbar \frac{\partial \psi}{\partial t}$, the solution $\psi(x,t)$ depends on the initial condition at $t=t_0$. It is mentioned in the literatures that if $H=H^*$ then $\psi^*(x,-t)$ is another solution. However, $\psi^*(x,-t)$ is not consistent with the initial condition. In other words, $\psi^*(x,-t_0) \ne \psi(x,t_0)$! Any help is appreciated.

"Initial conditions" means the state of the system at some "initial" time, $t_0$. Unless $t_0 = 0$, then the initial time changes under time reversal. The initial time is now at $-t_0$.

The same applies to parabolic motion under gravity. There is time reversal symmetry, but the time coordinate associated with "initial" conditions must change.

More generally, under any transformation, the coordinates of the initial event change. If an initial event is at the spatial origin and you move the origin of your coordinate system, then the initial event is longer at the origin. That should be clear.

 

[hokhani]

"Initial conditions" means the state of the system at some "initial" time, $t_0$. Unless $t_0 = 0$, then the initial time changes under time reversal. The initial time is now at $-t_0$.

In the example I sent in the post #3 the initial time is $t_0=0$ which gives different initial conditions for the two solutions.

 

[PeroK]

In the example I sent in the post #3 the initial time is $t_0=0$ which gives different initial conditions for the two solutions.

Time reversal in the SDE requires the conjugate wavefunction - as stated.

 

[hokhani]

Time reversal in the SDE requires the conjugate wavefunction - as stated.

Thanks, do you approve that the initial condition for time reversal solution is different from the main solution? If no, can we say that for one initial condition we have two solutions for SDE? If yes, we would have at least double degeneracy for most of time independent Hamiltonians!

 

[PeroK]

Thanks, do you approve that the initial condition for time reversal solution is different from the main solution? If no, can we say that for one initial condition we have two solutions for SDE? If yes, we would have at least double degeneracy for most of time independent Hamiltonians!

You can't ask whether the conjugate wavefunction is a solution and demand that at some initial time it is not the conjugate. The initial condition is the wavefunction at some chosen time. That must also be conjugated.

 

[Demystifier]

TL;DR Summary: How can $\psi^*(x,-t)$ be a solution of the Schrodinger equation with specific initial conditions?

In the solution of Schrodinger equation, $H\psi =i\hbar \frac{\partial \psi}{\partial t}$, the solution $\psi(x,t)$ depends on the initial condition at $t=t_0$. It is mentioned in the literatures that if $H=H^*$ then $\psi^*(x,-t)$ is another solution. However, $\psi^*(x,-t)$ is not consistent with the initial condition. In other words, $\psi^*(x,-t_0) \ne \psi(x,t_0)$! Any help is appreciated.

Nobody said that $\psi^*(x,-t)$ satisfies the same initial conditions as $\psi(x,t)$. It doesn't. $\psi^*(x,-t)$ satisfies the same Schrodinger equation, although with a different initial condition. But since it satisfies the Schrodinger equation, it is a solution. A solution, not the solution.