# Anticoncentration for two independent random quantum circuits in parallel

Consider two Haar random $$n$$ qubit unitaries, $$U_1$$ and $$U_2$$. Consider the quantum state

$$|\psi\rangle = (U_1 \otimes U_2) |0^{2n}\rangle.$$

Let $$p_x = |\langle x| \psi \rangle|^{2}$$, for $$x \in \{0, 1\}^{2n}$$. Does the probability distribution $$p$$ anti-concentrate (anticoncentration is as defined in equation $$1$$ here)?

I'm going to rewrite the $$p_x$$ from your question as $$p_{xy}$$. So, we have $$p_{xy}=|\langle x|U_1|0^n\rangle|^2\ |\langle y|U_2|0^n\rangle|^2$$ Note that this is two independent probabilities $$p_x$$ and $$p_y$$.
Now, the probability that $$p_{xy}>\alpha'$$, which we write as $$\mathbb{P}(p_{xy}>\alpha')$$ certainly satisfies $$\mathbb{P}(p_{xy}>\alpha')>\mathbb{P}(p_{x}>\sqrt{\alpha'})^2$$ because if both the independent probabilities $$p_x$$ and $$p_y$$ are bigger than $$\sqrt{\alpha'}$$, the probability of the produce being larger than $$\alpha$$ (which includes some events not counted in the independent case) is certainly larger.
Now, from the paper you cite (theorem 5, $$\epsilon=0$$), we know $$\mathcal{P}\left((p_x>\frac{\alpha}{N}\right))\geq\frac{(1-\alpha)^2}{2}$$ So, equate $$\sqrt{\alpha'}=\frac{\alpha}{N}$$ and follow through to find the constant $$\beta'$$ such that $$\mathbb{P}(p_{xy}>\alpha')>\beta',$$ the definition of being anti-concentrated.