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?


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.


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