At what depth and for what architecture are random quantum circuits $1$-designs?

I was confused about something related to quantum $$1$$ designs.

Let us recap two facts we know about random circuit ensembles that form a $$1$$ design.

1. $$1$$ design, for a quantum circuit over $$n$$ qubits, means that the density matrix of the ensemble is equal to the maximally mixed state over $$n$$ qubits.
2. If for depth $$d$$, a random circuit ensemble is a $$1$$ design, then it is also a $$1$$ design at depth $$d+1$$ and beyond.

Now, consider the following random circuit ensemble, taken from Figure 1 (a) here.

As mentioned in the paper, each violet box is drawn independently from the Haar measure on $$\mathbb{U}(4)$$.

Now, consider a circuit constructed like this, whose depth is exactly $$1$$. That is, we start with the all zeros state, apply a random gate to each of the $$n/2$$ pairs of qubits, and then measure in the standard basis.

It is not hard to work out that the ensemble is a $$1$$ design.

However, I am not sure how to prove that the ensemble remains a $$1$$ design (or even an approximate $$1$$ design) when we increase the depth to $$2$$ and beyond. The calculations for the depth $$1$$ case do not immediately generalize.

What is going on here?

Note that it is known that such architectures become $$2$$ designs (and hence, $$1$$ designs too) after linear depth. However, I am interested in the behaviour in constant or $$\log n$$ depths.

Additionally, the fact that the depth $$1$$ case is a $$1$$ design seemed to rely heavily on the type of gate set chosen. What if we change the gate set to another universal gate set --- like all single qubit gates and the CNOT gate?

Does it still remain a $$1$$ design (or an approximate $$1$$ design) or is the gate set in question very special?

• Presumably, if you've already got something that's $I$, $UI U^\dagger=I$ for every choice of $U$? Oct 21, 2021 at 7:36
• Yes, that makes sense. But when you’re writing out the expression for a 1 design, how would you write down the expression for the two different layers, for a depth 2 circuit? Also, do you think this property (that even a depth 1 circuit is a 1 design) is special only to a particular choice of the gate set? As stated in the question, for an example of an alternate gate set, how about all qubit gates and the CNOT gate? Oct 21, 2021 at 7:49
• For the brickwork circuits you're considering, it is useful to define a moment operator for each of two layers, i.e. for even and odd. The total moment operator after $2k$ layers then factorises and becomes $(M_\mathrm{even} M_\mathrm{odd})^{k}$. Otherwise, @DaftWullie 's argument is what you need. This property is not special to this choice of gate set, just think about Pauli/Weyl operators: they completely factorise and give a 1-design. Oct 22, 2021 at 8:01

To study unitary $$t$$-designs, we define the moment operator with respect to a probability measure $$\nu$$ as $$M_t(\nu) := \int_{U(d)} U^{\otimes t} (\cdot) (U^{\otimes t})^\dagger d\nu(U) \simeq \int_{U(d)} (U \otimes \bar U)^{\otimes t} d\nu(U).$$ Often, designs are defined via subsets $$D\subset U(d)$$ endowed with some "canonical" measure. For instance, if $$D$$ is finite one usually takes the normalised counting measure which turns the above integral into a finite sum that averages over the elements in $$D$$.
Then, $$\nu$$ is a unitary $$t$$-design if and only if its moment operator agrees with the one of the Haar measure on the unitary group: $$M_t(\nu) = M_t(\mu_\mathrm{Haar}).$$
If we consider random circuits, where we draw the first gate w.r.t to a measure $$\nu_1$$, the second gate w.r.t. $$\nu_2$$ and so on, the total distribution is given by a convolution of measures $$\nu = \nu_1*\nu_2*\dots*\nu_k.$$ The respective moment operator factorises $$M_t(\nu) = \int (U \otimes \bar U)^{\otimes t} d\nu(U) = (U_1\cdots U_k \otimes \bar U_1\cdots \bar U_k)^{\otimes t} d\nu_1(U_1)\cdots d\nu_k(U_k) \\ = M_t(\nu_1)\cdots M_t(\nu_k)$$
The brickwork circuits you're considering consist of two distinct layers, the even and odd ones, and we can write the total measure as $$\nu = (\nu_\mathrm{even}*\nu_\mathrm{odd})^{*k}$$ for $$2k$$ layers.
Now for your second question about alternate gate sets. I am assuming that you're asking specifically about 1-designs. Note that the Haar-random moment operator can be written as $$M_1(\mu_\mathrm{Haar},U(d^n)) = \int_{U(d^n)} U(\cdot) U^\dagger d\mu_\mathrm{Haar}(U) = \mathrm{tr}(\cdot ) \frac{I}{d^n} = M_1(\mu_\mathrm{Haar}, U(d))^{\otimes n},$$ i.e. it factorises. Hence, if we want to construct a unitary 1-design on $$n$$ qudits, it is sufficient to simply pick a 1-design on every qudit individually.