Unitary Time Evolution and Reversibility

Question

I have a system as follows: $$\frac{d\mathbf{x}}{dt} = -iA \mathbf{x},$$

which describes the time evolution of variables $\mathbf{x}$ according to the matrix $A$. Now, $A$ is not Hermitian, but has real eigenvalues. Furthermore, this system satisfies a conservation law and is time-reversible. However, its time evolution is not unitary since $A$ is not Hermitian. How is this possible? In my mind, I always equated unitary time evolution to time-reversibility, but it seems like it's not the case here.

To add to my confusion, I can find a variable transformation $\mathbf{x}=C \mathbf{y}$, and the new system becomes: $$\frac{d\mathbf{y}}{dt} = -iC^{-1}AC \mathbf{y},$$

where the transformed matrix $C^{-1}AC$ is Hermitian. What is going on here? It seems like unitary time evolution is a subset of reversible time evolution. Is that correct?

Thanks in advance for your help!

Answer 1

It's not necessary for a general time-reversible system to have the form of a unitary evolution. This is basically the answer to your question, but to make it more clear I'll give a trivial example:

$$ \frac{dx}{dt} = 1 $$

i.e. $x(t) = t$. This system is clearly time-reversible, with the reverse dynamics given by ${dx}/{dt} = -1$. However the evolution operator is not unitary. In fact, it's not even linear (it's affine).

One might object that the reverse dynamics aren't given by the same equation as the forward dynamics. But one can simply construct the combined forward+reverse system:

$$ \frac{d}{dt}(x, y) = (-1, 1) $$

(With solution $(x, y)(t) = (t, -t)$). And now the forward and reverse dynamics have the same equation, you just need to swap $x$ and $y$.

And this brings up an important point: When speaking of time-reversibility, one has to make precise the variable transformation that's required. Example: Newtonian mechanics is time-reversible, with the same equations for the forward and backward direction, if you negate all the momenta while keeping the positions the same. It may not be time-reversible under other variable transformations. Unitary evolution is a special case where the time-reversing transformation is conjugation. The reason we place special importance on unitary evolution is because of the role of unitary evolution in quantum theory.

Another objection might be that we first need to linearize the system before talking about unitary evolution. We can make the system linear by appending the constant:

$$ \frac{d}{dt}(x, z) = (z, 0) $$

i.e.:

$$ \frac{d}{dt}(x, z)^\top = \begin{bmatrix}0&1\\0&0\\\end{bmatrix}(x, z)^\top $$

And now if we start at any $z$, we will get the same dynamics as before e.g. $z=1$ just gives us our previous system. And now the time-reversal transformation just becomes:

$$ (x^-, z^-) = (x^+, -z^+) $$

You can think of x as the 'position' and z as the 'momentum'.

Under certain assumptions, it's possible to represent any time-invertible system as a unitary system. In fact, you can even represent non-time invertible systems as unitary systems, through the addition of ancilla variables, but I suspect this isn't what you're asking.