# Eigenvalues of an time-ordered exponential operator

Let's consider a simple 1-qubit time-dependent Hamiltonian: $$H(t) = \delta(t) \sigma_x + \sigma_z \ ,$$

where $$\delta(t)$$ is some time-continuous (real-valued) function. Evolving $$H(t)$$ continuously in time from $$t = 0$$ to $$t = T$$, gives a unitary operation at final time $$T$$: $$U(T) = \mathcal{T}\exp \left( -i \int_0^T H(t) dt \right) \ .$$

Can we upper bound a sum of eigenvalues of $$U(T)$$, i.e. $$\text{Tr} \ [U(T)]$$, in terms of $$T$$ and $$\lvert \delta(t) \rvert$$ ?

I'm not sure what you mean by |δ(t)| , but your integral, before time-ordering, collapses to $$\int_\epsilon ^T \!\! dt ~ \sigma_z + \int^\epsilon _0 \!\! dt ~\delta(t) ~\sigma_x = (T-\epsilon) \sigma_z +\tfrac{1}{2} \sigma_x ,$$ where I have adopted the half-Gaussian convention B3 in here, and ε is to be taken to 0 after time ordering. It then trivially follows that $$U(T)= e^{-iT\sigma_z} e^{-i\sigma_x/2}= (\cos T-i\sigma_z \sin T) (\cos 1/2 -i \sigma_x \sin 1/2 ),$$ whose trace is obviously $$2\cos T ~ \cos 1/2$$.
• I doubt that $\delta$ was supposed to be the Dirac delta function here. Only op can say for sure, though. Sep 2, 2023 at 13:56
• Hi @CosmasZachos, thank you for your comment. As the first comment suggested, $\delta(t)$ I meant by any continuous function. I apologize for the confusion. (I will change my question) Sep 3, 2023 at 3:27
I doubt that any useful bound can be found here, for the following reason. Note first that for any unitary operator $$U$$ in $$n$$ dimensions, we have $$|\operatorname{tr}(U)| \leq n . \tag 1$$ This is because each eigenvalue has the absolute value one. Second, consider $$\delta(t) = 0$$. Then, $$\operatorname{tr}(U(T)) = 2\cos T$$. The bound (1) is therefore sharp already. Third, $$\operatorname{tr}(U(T))$$ will be oscillating between 0 and 2 for any constant $$\delta$$. Increasing $$\delta$$ thus does not generally decrease the upper bound.
Of course, this doesn't exclude the possibility that a better bound may exist. It only seems unlikely to me too get one without actually computing $$U(T)$$, which will only be possible for special choices of $$\delta$$.