1
$\begingroup$

For context, I am reading this paper.

Basically, the paper makes reference to "evolving with respect to the time-reversed Hamiltonian". I'm slightly unclear as to what this actually means. Here is my logic:


Let $H$ be some Hamiltonian with eigenstates $|E_n\rangle$. Let $H'$ be the time-reversed Hamiltonian, with eigenstates $|E_n'\rangle$, where $|E_n'\rangle \ = \ \Theta |E_n\rangle$ with $\Theta$ being the time-reversal operator:

$$\Theta \ = \ i Y K$$

where $K$ is the complex conjugation operator, and $Y$ is the Pauli-Y. It follows that $H'$ can be written as:

$$H' \ = \ -\Theta H' \Theta$$

Since, given some $|E_n'\rangle$, we will have:

$$H' |E_n'\rangle \ = \ -\Theta H \Theta |E_n'\rangle \ = \ \Theta H |E_n\rangle \ = \ \Theta E_n |E_n\rangle \ = \ E_n |E_n'\rangle$$

So the time-reversed states are eigenstates of this Hamiltonian. In addition, energy doesn't change under time-reversal, and as you can see, we retain the same energy $E_n$ for a given eigenstate under time-reversal, thus we have uniquely defined the time-reversed Hamiltonian.

I then went on to define time-evolution under this Hamiltonian:

$$e^{-i \gamma H'} \ = \ e^{i \gamma \Theta H \Theta} \ = \ \displaystyle\sum_{n \ = \ 0}^{\infty} \frac{(i \gamma \Theta H \Theta)^n}{n!}$$

We know that $-\Theta^2 \ = \ \mathbb{I}$, as time-reversing a quantum state twice is equivalent to doing nothing. Therefore, all of the middle terms in the product expansion of the numerator of each term will cancel, and we will have:

$$\displaystyle\sum_{n \ = \ 0}^{\infty} \frac{(i \gamma \Theta H \Theta)^n}{n!} \ = \ \displaystyle\sum_{n \ = \ 0}^{\infty} (-i Y K) \frac{(-i \gamma H )^n}{n!} (i Y K) \ = \ Y K \Big( \displaystyle\sum_{n \ = \ 0}^{\infty} \frac{(-i \gamma H )^n}{n!} \Big) Y K \ = \ Y K e^{-i \gamma H} Y K$$


This clearly makes no sense: time-evolution must be unitary and $K$ is anti-unitary.

I highly doubt that my thinking is correct, as I watched a talk given by one of the authors of the paper, and he said that the time-reversed Hamiltonian is "usually" the same as the original Hamiltonian. Where am I going wrong?

$\endgroup$
0
$\begingroup$

I think I may have found the answer in these lecture notes, which I will summarize in case anyone else has the same question:

We start with the definition of time-reversal, which says that $\Theta |\Psi(t)\rangle \ = \ |\Psi(-t)\rangle$. We have:

$$|\Psi(t)\rangle \ = \ e^{-iHt/\hbar}|\Psi(0)\rangle$$

We will also have:

$$\Theta |\Psi(-t)\rangle \ = \ |\Psi(t)\rangle \ = \ e^{-iHt/\hbar} |\Psi(0)\rangle \ = \ e^{-iHt/\hbar} \Theta |\Psi(0)\rangle$$

If we do a change of variables $t \rightarrow -t$ in the first equation, we have:

$$|\Psi(-t)\rangle \ = \ e^{iHt/\hbar} |\Psi(0)\rangle$$

Substituting into the second equation, we have:

$$\Theta |\Psi(-t)\rangle \ = \ \Theta e^{iHt/\hbar} |\Psi(0)\rangle \ = \ e^{-iHt/\hbar} \Theta |\Psi(0)\rangle$$

Which leaves us with:

$$\Theta e^{iHt/\hbar} \ = \ e^{-iHt/\hbar} \Theta \ \Rightarrow \ e^{iHt/\hbar} \ = \ - \Theta e^{-iHt/\hbar} \Theta \ = \ Y K e^{-iHt/\hbar} Y K$$

Thus, the operator we end-off with is in fact unitary, given by:

$$U \ = \ e^{iHt/\hbar}$$

Which makes sense, as this is the original time-evolution operator, but with the sign changed, signifying evolution backward in time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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