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 $\lvert \delta(t) \rvert$ ?


2 Answers 2


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 $2\cos T ~ \cos 1/2$.

  • 1
    $\begingroup$ I doubt that $\delta$ was supposed to be the Dirac delta function here. Only op can say for sure, though. $\endgroup$
    – Noiralef
    Sep 2, 2023 at 13:56
  • $\begingroup$ @Noiralef Indeed, the lack of context is deafening. There are few other functions sanely denoted that way, even though your doubt would resolve the absolute value mystery... $\endgroup$ Sep 2, 2023 at 14:11
  • $\begingroup$ 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) $\endgroup$
    – Mohan
    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 $\delta$.


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