# Can any linear but non-unitary “time-evolution operator” be normalized to a unitary one?

A comment to this answer to another question states

I would imagine that for any linear non-unitary time-evolution operator, I can find a unitary one that will yield the same expectation values for every [physical state], which makes non-unitary time-evolution with manual normalization equal to unitary time evolution with standard normalization.

Is this correct?

No. For any particular initial state $$|\psi_0\rangle$$, we can manually normalize the hypothetical non-unitary but linear time-evolution operator $$\hat{O}(t)$$ in such a way that the manually normalized operator $$\hat{O}_n(t) \equiv N_{\psi_0}(t)\, \hat{O}(t)$$ produces a time-evolved trajectory $$|\psi(t)\rangle = \hat{O}_n(t) |\psi_0\rangle$$ with constant norm. But the key point is that the manual normalization function $$N_{\psi_0}(t)$$ necessarily depends on the particular initial state $$|\psi_0\rangle$$; there is not in general any manually normalized version of $$\hat{O}(t)$$ that preserves the norm along the trajectories for all initial states, like a unitary time-evolution operator does. The unitarity of the time-evolution operator is therefore a much stronger requirement than mere linearity, and you can't manually normalize an arbitrary linear time-evolution operator to a unitary one. (But note that the physical interpretation of unitarity is somewhat obscure in the projective-space formalism, where physical states don't have norms.)

As a simple example, consider the hypothetical linear but non-unitary time-evolution operator

$$\hat{O}(t) = \left( \begin{array}{cc} 1 & i \omega t \\ 0 & 1 \end{array}\right).$$

This operator trajectory is a one-parameter Lie group, i.e. it satisfies the composition property $$\hat{O}(t_2) \hat{O}(t_1) = \hat{O}(t_2 + t_1)$$. It preserves the norm of the initial state $$(1, 0)$$, so the manual normalization function for that initial state is the trivial $$N(t) \equiv 1$$. But the operator scales the norm of the initial state $$(0, 1)$$ over time as $$\sqrt{1 + (\omega t)^2}$$, so the manual normalization function for that initial state is $$N(t) = 1/\sqrt{1 + (\omega t)^2}$$. You can't normalize $$\hat{O}(t)$$ to simultaneously preserve the norm of both initial states. (Relatedly, the generator of this Lie group $$i \frac{d\hat{O}}{dt}|_{t=0} = \left( \begin{array}{cc} 0 & -1 \\ 0 & 0 \end{array}\right)$$ is not Hermitian.)

Given a linear (but possibly non-unitary) time-evolution operator $$\hat O(t)$$, "manual normalization" would mean to consider the time evolution $$|\psi(t)\rangle = \frac{\sqrt{\langle \psi_0 | \psi_0 \rangle}}{\sqrt{\langle \psi_0 | \hat O(t)^\dagger \hat O(t) |\psi_0\rangle}} \, \hat O(t) |\psi_0\rangle .$$ It is clear that this map $$|\psi_0\rangle \mapsto |\psi(t)\rangle$$ is non-linear in general (except if $$\hat O(t)^\dagger \hat O(t)$$ is a multiple of the identity). In other words, we can only "repair" normalization at the cost of linearity.

• Great answer, but it should be $$|\psi(t)\rangle = \sqrt{\frac{\langle \psi_0 | \psi_0 \rangle}{\langle \psi_0 | \hat O(t)^\dagger \hat O(t) |\psi_0\rangle}} \hat O(t) |\psi_0\rangle.$$ Unitary time evolution means constant norm, not unit norm (and you forgot the square root). – tparker Oct 4 at 11:31
• @tparker Thanks, I did forget the square root! Also, I assumed that the initial state is normalized to simplify the expression, but should have mentioned that. – Noiralef Oct 5 at 12:56
• Respectfully, I think that the assumption that $|\psi_0\rangle$ is normalized messes up your whole answer. If you only give the time-evolution map for normalized $|\psi_0\rangle$, then there's no way of evaluating whether or not that map is linear - which is the whole point of your answer - because a linear combination of normalized kets is not in general normalized. (Indeed, your formula superficially appears to be nonlinear even if $O^\dagger O = 1$.) I think that you need to give the general map in order for your answer to really make sense. – tparker Oct 5 at 14:49
• @tparker I have added the general map, since it is definitely not wrong. I want to add one more comment though. Let $H_\sim$ be the space of physical (i.e., normalized) states. It does indeed make sense to discuss whether a map $\hat X: H_\sim \to H$ is linear: it requires that $\hat X(\alpha \phi_1 + \beta \phi_2) = \alpha \hat X(\phi_1) + \beta \hat X(\phi_2)$ for $|\alpha|^2 + |\beta|^2 = 1$. Violation of this condition is imo the crucial point here. If it is satisfied, the map can be trivially extended to a linear map $H \to H$ by adding the normalization factor. – Noiralef Oct 6 at 4:54
• That's true mathematically, but requiring every state to pass through a normalized stage before linear combinations can be taken is very cumbersome if we're imagining an alternative theory to quantum mechanics in which time evolution is nonunitary. There might be some order-of-operations ambiguities about whether you normalize before or after taking linear combinations, I haven't thought about it much. It just seems to me that if time evolution is nonunitary then we need to be a little more careful about equating "physical" and "normalized". – tparker Oct 6 at 12:59