Physics Asked by Jack Ceroni on December 7, 2020
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 $|E_nrangle$. Let $H’$ be the time-reversed Hamiltonian, with eigenstates $|E_n’rangle$, where $|E_n’rangle = Theta |E_nrangle$ 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:
$$H’ |E_n’rangle = -Theta H Theta |E_n’rangle = Theta H |E_nrangle = Theta E_n |E_nrangle = E_n |E_n’rangle$$
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:
$$e^{-i gamma H’} = e^{i gamma Theta H Theta} = displaystylesum_{n = 0}^{infty} frac{(i gamma Theta H Theta)^n}{n!}$$
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:
$$displaystylesum_{n = 0}^{infty} frac{(i gamma Theta H Theta)^n}{n!} = displaystylesum_{n = 0}^{infty} (-i Y K) frac{(-i gamma H )^n}{n!} (i Y K) = Y K Big( displaystylesum_{n = 0}^{infty} frac{(-i gamma H )^n}{n!} Big) Y K = Y K e^{-i gamma H} Y K$$
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.
Answered by Jack Ceroni on December 7, 2020
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