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I am interested in the following Haar average over the unitary group

$D(x) = \int d\mathscr{U} ~(\mathscr{U})^{\otimes 2}(|\tilde{x}_{\mathscr{U}}\rangle\langle \tilde{x}_{\mathscr{U}}|)^{\otimes 2} (\mathscr{U}^\dagger)^{\otimes 2}$, where

$|\tilde{x}_{\mathscr{U}}\rangle = \frac{|x\rangle}{\sqrt{p_{\mathscr{U}}(x)}},$ $ p_{\mathscr{U}}(x) = \sum_i p_i ~\textrm{tr}(\mathscr{U}|x\rangle\langle x|\mathscr{U}^\dagger \rho_i),$ and $\{p_i, \rho_i\}$ is an ensemble of quantum states.

Now, following the standard approach, we consider

$V^{\otimes 2} D(x) (V^\dagger)^{\otimes 2} = \int d\mathscr{U} ~(\mathscr{U})^{\otimes 2}(|\tilde{x}_{V^\dagger\mathscr{U}}\rangle\langle \tilde{x}_{V^\dagger\mathscr{U}}|)^{\otimes 2} (\mathscr{U}^\dagger)^{\otimes 2}$, which follows from the invariance of Haar measure.

And so we find that

$(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2}D(x)(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2} = \int d\mathscr{U} ~(V^\dagger\mathscr{U})^{\otimes 2}(|\tilde{x}_{V^\dagger\mathscr{U}}\rangle\langle \tilde{x}_{V^\dagger\mathscr{U}}|)^{\otimes 2}((V^\dagger\mathscr{U})^{\dagger})^{\otimes 2}=D(x)$.

This means

$[D(x), (V^\dagger \otimes V)^{\otimes 2}] =0, ~~~~\forall V$. Operators that satisfy this relation are simply

$\mathbb{1}_4, ~\mathbb{1}_2 \otimes SWAP, ~SWAP \otimes \mathbb{1}_2, ~SWAP\otimes SWAP$. Taking linear combinations would allow us to work out $D(x)$.

I'm wondering if this is correct? And if this means that what we have is not a unitary design but a tensor product of two 2-designs. Note that $D(x)$ is not a homogenous polynomial in degrees of $(\mathscr{U}, \mathscr{U}^\dagger)$.

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    $\begingroup$ This has already been answered many times, see e.g. quantumcomputing.stackexchange.com/questions/21803/… quantumcomputing.stackexchange.com/questions/29671/… $\endgroup$ Jun 22, 2023 at 10:07
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    $\begingroup$ Hi Marcus. Thanks for pointing out the answers. I hadn't seen the second one. But they are not quite what I was looking for. I was mainly wondering if one can we say that $[D(x), (V^\dagger)^{\otimes 2}\otimes V^{\otimes 2}]=0$ is equivalent to sampling from the set $\{U_i \otimes U_j\}$, where $\{U_i\}$ form a 2-design, and taking expectation with respect to this set to do the Haar average... $\endgroup$ Jun 22, 2023 at 10:41

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You have a syntactic mistake in your argument. The dimensions in the equation

$$(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2}D(x)(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2} = D(x) \,,$$

don't match. I am not sure what you intended to do, but you cannot conclude anything from this.

To address the other question independently, the vector space of linear operators which satisfy $[D, (V^\dagger \otimes V)^{\otimes 2}] =0, \forall V \in U(d)$ should be much larger. This subspace is isomorphic to the commutant of $V^{\otimes 4}$ and thus has dimension $4!=32$ (if $d\geq 4$). This follows by computing the following using suitable partial transposes: $$ \mathbb{E}_{V\sim U(d)} (V^\dagger \otimes V)^{\otimes 2} (\cdot) (V^\dagger \otimes V)^{\otimes 2\dagger} \simeq \mathbb{E}_{V} (V^\dagger \otimes V^\dagger \otimes V\otimes V)\otimes (V^\top \otimes V^\top \otimes \bar V\otimes \bar V) \\ \simeq \mathbb{E}_{V} (\bar V \otimes \bar V \otimes V\otimes V)\otimes (V \otimes V \otimes \bar V\otimes \bar V) \\ \simeq \mathbb{E}_{V} (V\otimes \bar V)^{\otimes 4} \simeq \mathbb{E}_{V} V^{\otimes 4}(\cdot) (V^{\otimes 4})^{\dagger} $$

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  • $\begingroup$ I'm not sure what dimension argument you are referring to; if that were true it would be true for the 4-design case as well (or any t-design!). The commutator $[D, V^{\otimes 4}]=0$ is the same as $V^{\otimes 4} D (V^\dagger)^{\otimes 4} = D$, and is another way of saying $V^{\otimes 4} D = D V^{\otimes 4}$. About the isomorphism between the subspace in question and $V^{\otimes 4}$, I am not entirely convinced since the SWAP operator $W$ acting on a bipartition of $V^{\otimes 4}$ will not commute with $(V^\dagger \otimes V)^{\otimes 2}$ but would commute with $V^{\otimes 4}$... $\endgroup$ Jun 26, 2023 at 10:09
  • $\begingroup$ To add: If we were to get a 4-design it would mean the Haar average is over a polynomial function of degree 4 in $(U, U^\dagger)$ which is obviously not true. $\endgroup$ Jun 26, 2023 at 10:17
  • $\begingroup$ See cheat sheet (pg 10 of pdf) in this old QIP tutorial by Patrick Hayden: qipconference.org/2011/images/QIPtutorial1.pdf $\endgroup$ Jun 26, 2023 at 10:24
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    $\begingroup$ @Ghost-of-PPPF $D(x)$ is a $d^2\times d^2$ matrix (assuming your unitaries are from $U(d)$). $(V^\dagger)^{\otimes 2} \otimes V^{\otimes 2}$ ia a $d^4\times d^4$ matrix. The equation does not make sense to me. Judging by the middle equation, you just conjugated by $(V^\dagger)^{\otimes 2}$ which obviously undoes the first step. $\endgroup$ Jun 26, 2023 at 12:00
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    $\begingroup$ Hey Markus. I understand your point about the syntactic mistake which for some reason I couldn't appreciate earlier. Sorry about this. :) $\endgroup$ Sep 2, 2023 at 7:45

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