# What is the expectation value of $|\langle \psi|U|\psi \rangle|$ over Haar random states $|\psi\rangle$?

We know the average unitary fidelity, $$\int |\langle \psi|U|\psi \rangle|^2 d\psi$$, has a nice closed-form solution: $$\frac{1+\frac{1}{d}|Tr (U)|^2}{1+d}$$, thanks to Horodecki and Nielsen.

However, I am looking for an analytical expression for the Haar average $$\int |\langle \psi|U|\psi \rangle| d\psi$$, where $$|\psi \rangle\in \mathbb{C}^d$$. Is it computable? I am particularly interested in the case where the unitary $$U$$ is traceless.

Additionally, it would be helpful to have some useful references on this.

Remarks:

1. Without the absolute, one can calculate the integral $$\int \langle \psi|U|\psi \rangle d\psi$$: $$U\int|\psi\rangle\langle\psi|d\psi{=}U.Id/d$$, hence taking trace on both sides, we have the integral reduced to $$Tr(U)/d$$. This is not what I am looking for.
2. One can use the Cauchy-Schwartz inequality on the average gate fidelity to derive an upper bound of the integral in question. However, a closed-form solution would have been helpful.
• Related: this post and this post. Jun 24 at 6:33
• Hello, thanks for sharing the posts. But the connection of the posts to this problem is not quite apparent to me.
– Abir
Jun 24 at 18:31

This partial answer calculates the integral for $$d=2$$. In this case, every traceless unitary $$U$$ is equivalent to the Pauli $$Z$$ up to similarity and global phase, so, by rotational invariance of the Haar measure, we have \begin{align} \int|\langle\psi|U|\psi\rangle|\,d\psi&=\int |\langle \psi|Z|\psi \rangle| \,d\psi\tag1\\ &=\int|\psi_0\overline{\psi_0}-\psi_1\overline\psi_1|\,d\psi\tag2\\ &=\frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\left|\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\right|\sin\theta\,d\phi\,d\theta\tag3\\ &=\frac{1}{2}\int_0^\pi\left|\cos\theta\right|\sin\theta\,d\theta\tag4\\ &=\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta\tag5\\ &=\frac{1}{2}\int_0^{\pi/2}\sin 2\theta\,d\theta\tag6\\ &=\frac{1}{2}.\tag7 \end{align} This agrees with the bound $$\int|\langle\psi|U|\psi\rangle|\,d\psi\leqslant\frac{1}{\sqrt{d+1}}\tag8$$ obtained from Jensen's inequality and Horodecki's formula. It should be possible to generalize the calculation above. For example, for $$d=3$$, we have \begin{align} \int|\langle\psi|U|\psi\rangle|\,d\psi&=\int |\langle \psi|Z|\psi \rangle| \,d\psi\tag9\\ &=\int|\psi_0\overline{\psi_0}+\omega\psi_1\overline\psi_1+\omega^2\psi_2\overline\psi_2|\,d\psi\tag{10} \end{align} where $$Z=\mathrm{diag}(1,\omega,\omega^2)$$ and $$\omega=e^{2\pi i/3}$$. One could use the parametrization of $$\psi$$ from this answer. Note that for larger $$d$$ the spectrum of $$U$$ has continuous degrees of freedom. For example, for $$d=4$$ it can take the form $$\mathrm{diag}(1, -1, z, -z)$$ for any $$z\in\mathbb{C}$$ with $$|z|=1$$.

• thanks. The iterative parameterisation in the linked answer is insightful and would be helpful to calculate for a few lower dimensions.
– Abir
Jun 27 at 12:34
• I'll be curious if your calculations agree with the conjecture that the integral is $1/d$. Please share your results :-) Note that for $d > 3$ the spectrum isn't quite as constrained. For example, for $d=4$, the unitary 𝑈 may be equivalent to any $\mathrm{diag}(a, -a, b, -b)$. Also, if you use the parametrization from my other answer, make sure you get the Jacobian right (I don't think it is the same as the Jacobian of the $n$-sphere). Jun 28 at 2:22
• numerical analysis with 1000 samples suggests it is higher than $1/d$. After curve fitting I am getting something like $0.9065*(d+1.844)^{(-0.5055)}$ :) I think it is $1/\sqrt{d+2}$. I will surely share my answer here, thanks for your input.
– Abir
Jun 28 at 2:58
• I was also looking at this answer (quantumcomputing.stackexchange.com/a/7026/14173). I am not familiar with this approach, do you think one could use a similar parameterisation?
– Abir
Jun 28 at 2:59
• Yes, David Bar Moshe's approach looks like it might work in the general case :-) Jun 28 at 3:06

As a follow-up discussion with Adam, I evaluated the integral till $$d=4$$ for high-dimensional unitary $$Z$$, with $${Z}_{kl}{=}\exp(\frac{i2\pi k}{d})\delta_{kl}$$. For $$d{=}4$$, I haven't considered the other possibility that Adam mentioned, i.e., $$\tilde{U}{=}V_1\sigma_zV_1^\dagger{\oplus}e^{i\alpha}V_2\sigma_zV_2^\dagger$$, $$\sigma_z$$ is Pauli-z.

I referred to this paper to parameterise the normalized, unitary invariant measure $$d\psi$$ on the manifold of $$|\psi\rangle{=}\sum_{j=1}^d \sqrt{r_j}e^{i\theta_j}|j\rangle$$ ($$r_j{\geq}0$$, $$\theta_j{\in}[0,2\pi]$$). The parameterisation is introduced after Section V, Lemma 4. For completeness, the parameterisation for $$d\psi$$ is given by the following Dirac-delta representation:

\begin{align} d{\psi}\equiv\frac{\Gamma(d)}{2\pi^d}\delta\Big(1-\sum_{j=1}^d r_j\Big)\ \prod_{j=1}^d dr_jd\theta_j. \end{align} Here $$\Gamma(d){=}(d-1)!$$ is the Gamma function. First, let us check if the above parameterisation agrees with the $$d{=}2$$ case. In this case, $$|\langle \psi |Z|\psi\rangle|{=}|r_1-r_2|$$, and the integral $$I_{d=2}$$ becomes \begin{align} I_{d=2}&= \frac{1}{4\pi^2}\int_{r_1{=}0}^1\int_{r_2{=}0}^1 |r_1-r_2|\delta\Big(1-r_1-r_2\Big)dr_1dr_2 \int_{0}^{2\pi}\int_{0}^{2\pi} d\theta_1d\theta_2 \\ &=\int_{r_1,r_2{=}0}^1 |r_1-r_2|\delta\Big(1-r_1-r_2\Big)dr_1dr_2\\ &{=}\frac{1}{2}. \end{align} Then I moved on to $$d{=}3$$, in which case, $$|\langle \psi |Z|\psi\rangle|{=}|r_1+\omega r_2 + \omega^2 r_3|$$, with $$\omega{=}\exp(\frac{i2\pi}{3})$$. With little intention to work it out myself, I fed it to Mathematica and found $$I_{d=3}=\frac{1}{3}+\frac{\ln(2+\sqrt{3})}{6\sqrt{3}}{\approx}0.460058$$, not conforming to my guess, $$\frac{1}{\sqrt{d+2}}$$. Similarly, I calculated for $$I_{d=4}{\approx}0.405806$$ with a clumsy analytical expression as shown in the attached screenshot. I cross-verified the evaluation with numerical estimations with $$10^6$$ samples. Here are the list of numerical results for $$d{\in}[2,6]$$, $$[0.4998, 0.4600, 0.4058, 0.3695, 0.3409]$$.

So at this stage, I don't have much intuition about the general analytical expression. However the upper bound of $$\frac{1}{\sqrt{d+1}}$$ is helpful. I will be open to further interesting perspectives on this.