Direction of time-evolution in time reversed wave functions

[hokhani]

TL;DR

Is time evolution identical for a wave function and its time reversed?

Consider the time reversal operator as $\Theta$ so that $\psi_r(t)=\Theta \psi(t)$ where $\psi_r(t)$ is time reversed of $\psi(t)$. What is the direction of time evolution for $\psi_r(t)$? In other words, if we have $\psi(t)=exp(-i\frac{Ht}{\hbar}) \psi(0)$, do we also have $\psi_r(t)=exp(-i\frac{Ht}{\hbar}) \psi_r(0)$ or not?

 

[Ben vdP]

Just a thought. Shouldn't you be transforming to the minus of the time parameter?

Have you checked the following page, it is to elaborate to repeat here:

https://en.wikipedia.org/wiki/T-symmetry

 

[hokhani]

Just a thought. Shouldn't you be transforming to the minus of the time parameter?

Have you checked the following page, it is to elaborate to repeat here:

https://en.wikipedia.org/wiki/T-symmetry

Thank you for attention to this matter. The function $\psi_r(t)$ implies time reversed wave function at $-t$.
Also, I didn't see anything about the time evolution of the time reversed wave function in your link.

 

[PeterDonis]

the time reversal operator

Isn't what you appear to think it is.

I would suggest reading section 13.3 of Ballentine.

 

[Ben vdP]

Thank you for attention to this matter. The function $\psi_r(t)$ implies time reversed wave function at $-t$.
Also, I didn't see anything about the time evolution of the time reversed wave function in your link.

I don't want to wrestle with Latex for complex conjugation of the wave function and replacing time with minus time and such.
At some point I may have to.

You could also consult the following link especially page 11:

https://bingweb.binghamton.edu/~suzuki/QM_Graduate/Time_reversal_I_operator.pdf

 

[hokhani]

Isn't what you appear to think it is.

I would suggest reading section 13.3 of Ballentine.

Thanks, it was a very good elementary source. But I didn't find there the answer to my question. let me explain clearly:
In the classical example of a particle falling in the ground gravity, consider the initial point of free fall at $t=0$ with $v_0=0$ and the position $y=0$. The final point at $t$ with $v(t)$. In the time reversed configuration, we do $t\rightarrow -t$. So, the backward movie shows the motion from $-t$ to $0$ and hence the direction of evolution seems to be from $-t$ to $0$.
However, in the text by Robert G. Littlejohn (bohr.physics.berkeley.edu/classes/221/9697/timerev.pdf), Eq. (83), the time evolution for the reversed time configuration (with wave function $\psi_r(t)$ implying the time reversed situation at time $-t$) is:
$$\psi_r(t)=exp(- \frac{iHt}{\hbar}) \psi_r (0)$$ where implies evolution from $0$ to $-t$. So, this time evolution is in the opposite direction of the backward movie. In fact, I think there is a discrepancy between the correspondence of time reversal and the backward movie, despite use of this correspondence in a lot of texts.
I would be grateful for any help.

 

[PeterDonis]

the backward movie shows the motion from $-t$ to $0$

No, it doesn't. The backward movie has time running backwards, from larger to smaller values, instead of from smaller to larger values. That means from $0$ to $- t$.

 

[hokhani]

No, it doesn't. The backward movie has time running backwards, from larger to smaller values, instead of from smaller to larger values. That means from $0$ to $- t$.

Thanks, but suppose you take a movie from time $0$ to $t$. If you run it backwards, in fact you are playing the frames back from $t$ to $0$. In other words, according to time reversal, $t\rightarrow -t$ you are playing from $-t$ to $0$ with the transformations $x(-t)\rightarrow x(t)$ and $v(-t)\rightarrow -v(t)$.
Furthermore, in a visual representation, take for example the free fall of a particle in the gravity. If time reversed frames is from $0$ to $- t$, the particle must fall towards the ground while its velocity is always increasing in the other direction!
Could you please point out where I might have gone wrong?

 

[PeterDonis]

suppose you take a movie from time $0$ to $t$. If you run it backwards, in fact you are playing the frames back from $t$ to $0$.

Yes, but that's not what you're doing when you do "time reversal" $t \to - t$.

according to time reversal, $t\rightarrow -t$ you are playing from $-t$ to $0$ with the transformations $x(-t)\rightarrow x(t)$ and $v(-t)\rightarrow -v(t)$.

Yes. But that's not what you described earlier. The result of the transformation you describe here is not a movie that goes from $t$ to $0$ instead of $0$ to $t$. It's a movie that goes from $0$ to $-t$ instead of $0$ to $t$. The two movies actually look the same visually as far as the sequence of events is concerned; the only difference is the time labels.

 

[hokhani]

The two movies actually look the same visually as far as the sequence of events is concerned; the only difference is the time labels.

Please see the figure in the link: https://i.postimg.cc/F15gwHgr/TR.png
The motion of a falling particle is shown in direct time and reversed time. In the reversed time frame, if the time evolution is from $0$ to $-t$ we would have motion of a particle falling towards the ground but with the velocity increasing in the other direction which is not visually the same as the real event. Could you please help me identify where I might be incorrect?

 

[PeterDonis]

In the reversed time frame, if the time evolution is from $0$ to $-t$ we would have motion of a particle falling towards the ground but with the velocity increasing in the other direction

I'm not sure what you mean by this, but let me try to rephrase it. For simplicity, I'll assume that at $t = 0$ in the original solution, we have $v = 0$ and $x = h$, where $h$ is the height above the ground at which the fall starts.

Then in the time reversed solution, at $t = 0$ we still have $\bar{v} = 0$ and $\bar{x} = h$ at $t = 0$ (where I've put bars over the transformed functions in the time reversed solution). But now we have $\bar{x}$ decreasing and $\bar{v}$ increasing as $t$ decreases (where in the original solution, we had $x$ decreasing and $v$ increasing as $t$ increases).

which is not visually the same as the real event.

What "real event"? What do you think should be "visually the same", but isn't?

I think you're confused because, as I've already commented, "time reversal" is not what you appear to think it is.

 

[PeterDonis]

Thread closed for moderation.

 

[berkeman]

Based on the OP not responding to feedback, the thread will remain closed.