Question regarding measurement induced phase transitions

Question

It is often claimed in the literature of random quantum hybrid circuits that a measurement induced phase transition is invisible in trajectory averages of linear observables and can only be seen in trajectory averages of non-linear observables (see here, here and here). Romain Vasseur in his BSS lectures (see here), makes the following argument to justify this claim:

Let's consider a brick-wall circuit in which we have the following pattern of evolution: first we apply a Haar random unitary upon even links, then a few measurements are randomly made followed by Haar random unitaries applied to the odd links. The trajectory average i.e. average over different measurement outcomes can be written as a quantum channel: $$\bar{\rho}_t = \sum_{\{m\}} |\psi_{m} \rangle \langle \psi_{m} | = N_t(\rho)=\sum_{\{m\}} K_m \rho K_m^{\dagger}$$ where the Kraus operators are of the form: $$K_m = \prod_{\tau=1}^t \hat{\Pi}^{m_T^{'}} U_\tau^{\text {odd }} \hat{\Pi}^{m_T} U_\tau^{\text {even }}$$ here,

• $\hat{\Pi}^{m_T^{'}}$: the projector onto measurement outcome $m_T^{'}$
• $U_\tau^{\text {odd }}$: Haar-random unitary acting on odd links
• $U_\tau^{\text {even }}$: Haar-random unitary acting on even links

The central claim is that linear quantities such as $\langle O \rangle$, $$\langle O \rangle = \sum_{\{m\}} p_m \frac{\left\langle\psi_m|O| \psi_m\right\rangle}{\left\langle\psi_n \mid \psi_n\right\rangle}=\text{Tr}{\left(\bar{\rho}*t O\right)}$$ cannot "see" measurement induced phase transitions as at long times, the trajectory averaged density matrix becomes an infinite temperature state: $$\rho_t \rightarrow \rho_{\infty}=\frac{1}{d^L} \mathbb{I}$$ This can be shown by taking the following limit: $$\lim_{t\rightarrow \infty} \bar{\rho}_t = \lim_{t\rightarrow \infty} \sum_{\{m\}} K_m \rho K_m^{\dagger}$$

My questions are the following:

1. How does one compute this limit?
2. How does one generalize this argument beyond brick-wall circuits? Is there rather a more intuitive way to understand this?

Answer 1

By definition of the Kraus operators $K_m$, trajectory average is given by the channel defined by the equation, $$ \bar{\rho}_t = \sum_{\{m\}} \prod_{\tau = 1}^{t} \left(\hat{\Pi}^{m_T^{'}} U_\tau^{\text{odd }} \hat{\Pi}^{m_T} U_\tau^{\text{even }}\right) \rho \left( U_\tau^{\dagger \text{ even }} \hat{\Pi}^{m_T} U_\tau^{\dagger \text{ odd }} \hat{\Pi}^{m_T^{'}} \right).$$

Now to evaluate the limit for $t \to \infty$, we must consider the Haar averages as the specifics of the random unitaries essentially wash out at this limit, leaving only the averaged effect. Given a Haar-random unitary $U$, the expectation value of the linear operator $U \mathcal{O} U^\dagger$ with respect to the Haar-measure is given by (see here), $$ \mathbb{E}_U (U \mathcal{O} U^\dagger) \equiv \int_{\mathbf U(d)} d\mu(U) U \mathcal{O} U^\dagger = \mathrm{Tr}(\mathcal{O})\frac{\mathbb{I}}{d}. $$

Using the above expression we compute Haar averages over even and odd links to get, $$ \lim_{t \to \infty} \bar{\rho}_t = \lim_{t \to \infty} \sum_{\{m\}} \prod_{\tau = 1}^{t} \frac{\mathbb{I}}{d} \hat{\Pi}^{m_T^{'}} = \lim_{t \to \infty} \frac{\mathbb{I}}{d^t} = \frac{\mathbb{I}}{d^L} $$

As you can see from the above expression, with increasing $t$, the state gets increasingly mixed. At $t \to \infty$, we can physically intuit that this must converge to the maximally mixed state.

I believe this answer should give some intuition to the result, as well as possible extensions therof, as it has been derived mostly using physical principles.