What is the probability $\Pr(\|U-I\|_{\rm op}<\varepsilon)$ of a Haar-random unitary being close to the identity?

Question

If one generates an $n\times n$ Haar random unitary $U$, then clearly $\Pr(U=I)=0$. However, for every $\epsilon>0$, the probability $$\Pr(\|U-I\|_{\rm op}<\varepsilon)$$ should be positive. How can this quantity be computed?

Answer 1

$U-I$ is a normal matrix so $||U-I||_{op}$ is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\phi-1)^2+\sin^2\phi=4\sin^2\frac{\phi}{2}\Rightarrow |\lambda|=2\left|\sin\frac{\phi}{2}\right|,$$ where $e^{i\phi}$ is some eigenvalue of $U$.

Now, the distribution of eigenvalues for a random unitary matrix is known: there are $n$ eigenvalues $e^{i\phi_n}$ each distributed over $-\pi\leq \phi_n\leq \pi$ with probability density $$p(\phi_1,\cdots,\phi_n)=\frac{1}{n!(2\pi)^n}\prod_{1\leq j< k\leq n}\left|e^{i\phi_j}-e^{i\phi_k}\right|^2=\frac{2^n}{n!\pi^n}\prod_{1\leq j< k\leq n}\sin^2\frac{\phi_j-\phi_k}{2}.$$ So now we can express the desired quantity as $$\mathrm{Pr}(||U-I||_{op}\leq\epsilon)=\frac{2^n}{n!\pi^n}\int_{-\pi}^{\pi} d\phi_1\cdots d\phi_n \prod_{1\leq j< k\leq n}\sin^2\frac{\phi_j-\phi_k}{2}\left(2\max\{\left|\sin \frac{\phi_1}{2}\right|,\cdots,\left|\sin \frac{\phi_n}{2}\right|\}\leq \epsilon\right).$$

For the maximum to be less than or equal to $\epsilon/2$, all of the quantities must be less than or equal to $\epsilon/2$. This happens when $|\phi_i|\leq 2\arcsin{\frac{\epsilon}{2}}$, so the result is $$\mathrm{Pr}(||U-I||_{op}\leq\epsilon)=\frac{2^n}{n!\pi^n}\int_{-2\arcsin\frac{\epsilon}{2}}^{2\arcsin\frac{\epsilon}{2}} d\phi_1\cdots d\phi_n \prod_{1\leq j< k\leq n}\sin^2\frac{\phi_j-\phi_k}{2} .$$

Right now I can't evaluate that for arbitrary $n$... the integrals can all be done analytically but they are tedious. For $n=2$ the answer is $$-\frac{4\epsilon^2}{\pi^2}+\frac{\epsilon^4}{\pi^2}+\frac{16}{\pi^2}\arcsin^2\frac{\epsilon}{2}\approx \frac{4\epsilon ^4}{3\pi^2}+O\left(\epsilon ^{6}\right),$$ for $n=3$ the answer is $$\frac{1}{8\pi^3} \left(\left(\epsilon^6-8 \epsilon^4+28 \epsilon^2-48\right) \epsilon^2 \sin ^{-1}\left(\frac{\epsilon}{2}\right)+\left(\epsilon^4-6 \epsilon^2+8\right) \sqrt{4-\epsilon^2} \epsilon^3+64 \sin ^{-1}\left(\frac{\epsilon}{2}\right)^3\right)\approx \frac{4 \epsilon^9}{135 \pi ^3}+O\left(\epsilon^{11}\right),$$ and for $n=$ the answer is (I haven't simplified the trig terms here but they're doable) $$\frac{96 \sin ^3\left(2 \sin ^{-1}\left(\frac{\epsilon}{2}\right)\right) \sin ^{-1}\left(\frac{\epsilon}{2}\right) \left(-5 \epsilon^2+2 \cos \left(6 \sin ^{-1}\left(\frac{\epsilon}{2}\right)\right)+10\right)+4 \epsilon^2 \left(\epsilon^2-4\right) \left(2 \epsilon^8-16 \epsilon^6+53 \epsilon^4-84 \epsilon^2+108\right) \sin ^{-1}\left(\frac{\epsilon}{2}\right)^2+1152 \sin ^{-1}\left(\frac{\epsilon}{2}\right)^4+\sin ^4\left(2 \sin ^{-1}\left(\frac{\epsilon}{2}\right)\right) \left(-172 \cos \left(4 \sin ^{-1}\left(\frac{\epsilon}{2}\right)\right)+\cos \left(8 \sin ^{-1}\left(\frac{\epsilon}{2}\right)\right)-45\right)}{1152 \pi ^4}\approx \frac{\epsilon^{16}}{23625 \pi ^4}+O(\epsilon^18).$$