5
$\begingroup$

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.
$\endgroup$
2
  • $\begingroup$ Related: this post and this post. $\endgroup$
    – narip
    Commented Jun 24, 2023 at 6:33
  • $\begingroup$ Hello, thanks for sharing the posts. But the connection of the posts to this problem is not quite apparent to me. $\endgroup$
    – Abir
    Commented Jun 24, 2023 at 18:31

2 Answers 2

5
$\begingroup$

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 $|z|=1$.

$\endgroup$
5
  • $\begingroup$ thanks. The iterative parameterisation in the linked answer is insightful and would be helpful to calculate for a few lower dimensions. $\endgroup$
    – Abir
    Commented Jun 27, 2023 at 12:34
  • 1
    $\begingroup$ 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). $\endgroup$ Commented Jun 28, 2023 at 2:22
  • $\begingroup$ 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. $\endgroup$
    – Abir
    Commented Jun 28, 2023 at 2:58
  • $\begingroup$ 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? $\endgroup$
    – Abir
    Commented Jun 28, 2023 at 2:59
  • $\begingroup$ Yes, David Bar Moshe's approach looks like it might work in the general case :-) $\endgroup$ Commented Jun 28, 2023 at 3:06
1
$\begingroup$

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.

Analytical expression for I_{d=4}

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.