# Question about squaring a Haar random unitary

Consider an $$n \times n$$ unitary $$U$$, drawn from the Haar measure. I'm trying to find the distribution for $$U^{2}$$. Is it true that $$U^{2}$$ is also Haar random?

Note that for any fixed unitary $$V$$, $$VU$$ and $$UV$$ are both Haar random unitaries, by the translational invariance of the Haar measure. But our case is slightly different where the matrix we are multiplying $$U$$ by also "depends" on $$U$$.

• No. $Tr(U)$ converges in distribution to standard complex gaussian, while $Tr(U^2)$ converges in distribution to standard complex gaussian times $\sqrt{2}$ (this is due to Diaconis-Shahshahani). So $U^2$ cannot be distributed like $U$. Jun 12, 2022 at 21:39

The probability distribution of $$U^p$$ for $$U$$ uniformly distributed with the Haar measure in $$\text{U}(n)$$ has been calculated by Eric Rains in Images of eigenvalue distributions under power maps.
For $$p=2$$ and $$n$$ even the eigenvalue distribution of $$U^2$$ is obtained by taking the union of the eigenvalues of two independent matrices $$U_1$$ and $$U_2$$, uniformly distributed in $$\text{U}(n/2)$$. For $$n$$ odd the two independent matrices are taken from $$\text{U}((n+1)/2)$$ and $$\text{U}((n-1)/2)$$.
So $$U^2$$ is only Haar random in the trivial case $$n=1$$, but not for $$n>1$$. For $$n=2$$, in particular, the two eigenvalues of $$U^2$$ are independent, in contrast to the two eigenvalues of $$U$$.
To see how squaring the $$2\times 2$$ unitary $$U$$ removes the correlations, use that the joint probability distribution of the two eigenvalues $$\lambda_1$$ and $$\lambda_2$$ of $$U$$ is $$P(\lambda_1,\lambda_2)\propto|\lambda_1-\lambda_2|^2=2-2\,{\rm Re}\,\lambda_1\bar{\lambda}_2.$$ Squaring $$U$$ identifies $$\pm\lambda_i$$, so the contributions from the correlator $${\rm Re}\,\lambda_1\bar{\lambda}_2$$ cancel and $$P(\lambda_1^2,\lambda_2^2)$$ becomes independent of $$\lambda_i$$.
• To sanity check, the expected value of $U^{p}(n)$, over the Haar measure, should still be the $n \times n$ identity matrix (it is just a tensor product of $p$ identity matrices of suitable dimensions), is that correct? Jun 13, 2022 at 4:21
• certainly, the expected value of $U^p$ is the identity matrix. Jun 13, 2022 at 6:23