Consider the following density matrix over $n$ qubits, with $C$ being a single qubit operator: $$ \rho_{n} = \int_{C \sim \text{Haar}} \big(C|0\rangle\langle0|C^\dagger\big)^{\otimes n} dC. $$

Let's say I measure the qubit with respect to orthogonal measurement operators $\{M_i : I \in [k] \}$. Then, for each $M_i$:

$$\text{Tr}\big(M_i \rho_n\big) = \int_{C \sim \text{Haar}}\text{Tr}\bigg(M_i \big(C|0\rangle\langle0|C^\dagger\big)^{\otimes n} \bigg) dC $$

$$= \int_{C \sim \text{Haar}} \text{Tr}\bigg(C^{\dagger \otimes n} M_i C^{\otimes n} \left(|0^{\otimes n}\rangle\langle 0^{\otimes n}|\right)\bigg) dC $$

$$= \text{Tr} \bigg(\overline{M}_i |0^{\otimes n}\rangle\langle 0^{\otimes n}| \bigg),$$

where we have

\begin{equation} \overline{M}_i = \int_{C \sim \text{Haar}} C^{\dagger \otimes n} M C^{\otimes n}. \end{equation}

Now note that by the left and right invariance of the Haar measure, for any one qubit unitary $U$,

$$U^{\otimes n} \overline{M}_i = \overline{M}_i U^{\otimes n}.$$

Then, by the Schur Weyl duality, with each distinct $\pi$ being a distinct permutation operator acting on the $n$ registers and for some choices of $a_{\pi} \in \mathbb{C}$

$$\overline{M}_i = \sum_{\pi \in S_{n}} a_{\pi} ~\pi.$$

Note that for any choice of $\pi$,

$$\pi |0^{n}\rangle = |0^{n}\rangle. $$


$$\text{Tr}\big(M_i \rho_{n}\big) = \text{Tr} \bigg(\overline{M}_i |0^{\otimes n}\rangle\langle 0^{\otimes n}| \bigg).$$

What is the relation between $a_{\pi}$ and irreducible representations of symmetric subspaces and weak Schur sampling? For example, on page 42 of this link (https://arxiv.org/pdf/1310.2035.pdf), it is written that we can write an operator like $\overline{M_i}$ as

$$ \overline{M_i} = \sum_{\lambda} a_{\lambda} P_{\lambda},$$

where $\lambda$ is a partition of $n$. How do I see the relation between these two representations?

Additionally, can we exploit symmetry properties, like the fact that

$$ \pi \overline{M_i} = \overline{M_i},$$

for any $\pi$, to say anything more about $a_{\pi}$ and $\overline{M_i}$?


Note that the quoted relation $$ \bar M_i = \sum_\lambda a_\lambda P_\lambda, $$ only holds if the $M_i$ also commute with the representation of the symmetric group! Otherwise this can obviously not be true by a simple counting argument: The dimension of the commutant $U\mapsto U^{\otimes n}$ is $n!$ but there are certainly less partitions of $n$. Thus, the mutually orthogonal projectors $P_\lambda$ only span a subspace of the commutant, which is exactly its intersection with the commutant of the symmetric group $S_n$.

The other formula, $$ A = \sum_{\pi\in S_n} a_\pi \rho_d(\pi), $$ holds for any element $A$ in the unitary commutant and $\rho_d$ is the representation of $S_n$ on $(\mathbb C^d)^{\otimes n}$. The fact the permutations form a (non-orthogonal) basis of the commutant is a consequence of Schur-Weyl duality.

I think there is no real relation between these representations. The expansion into projectors uses that any element in the unitary commutant has the form $$ A = \bigoplus_{\lambda\vdash n} \mathrm{id} \otimes A_\lambda, $$ where the orthogonal decomposition is the same as in Schur-Weyl duality, i.e. $A_\lambda$ acts on the multiplicity space of the unitary irrep labelled by $\lambda$. This fact follows from Schur's lemma. If, on top, $A$ commutes with $S_n$, then it $A_\lambda$ has to be proportional to the identity, again by Schur's lemma. But then $$ A = \bigoplus_{\lambda\vdash n} a_\lambda \mathrm{id} \otimes \mathrm{id} = \sum_{\lambda} a_\lambda P_\lambda, $$ since these blocks are exactly the ranges of the projectors.

As the basis of permutations is non-orthogonal, computing the expansion is in practice not so simple. However, there is a remarkable formula which goes back to works of Collins and Sniady (see e.g. http://arxiv.org/abs/math-ph/0402073): $$ A = d^{-n} \big(\sum_{\pi\in S_n} \mathrm{tr}( \rho_d(\pi)^\dagger A ) \rho_d(\pi)\big) W, $$ where $W$ is the Weingarten function defined as the inverse of $W^{-1}:= \sum_{\pi\in S_n} \bar\chi_d(\pi) \rho_d(\pi)$ and $\chi_d = \mathrm{tr} \rho_d$ is the character of the representation. The factor comes from the fact that the permutations are not normalised.

$W$ has a small norm such that the expansion
$$ d^{-n} \sum_{\pi\in S_n} \mathrm{tr}( \rho_d(\pi)^\dagger A $$ is often a good approximation to $A$ (the error is suppressed by $d^{-1}$). For exact calculations of integrals of the unitary group one can use the Weingarten calculus (see https://arxiv.org/abs/2109.14890).

  • $\begingroup$ What is meant by representations of a symmetric group and the commuting of $\overline{M_i}$ with the same? What is a unitary commutant? And what might be some examples of $\overline{M_i}$ that commutes with the representations of a symmetric group? $\endgroup$
    – BlackHat18
    Nov 5 at 14:48
  • $\begingroup$ Also, $\pi_i$ — the permutation matrices — are orthogonal. So, why do they form a “non orthogonal basis of the commutant”? $\endgroup$
    – BlackHat18
    Nov 5 at 14:52
  • 1
    $\begingroup$ @BlackHat18 you might want to read up a bit on Schur-Weyl duality ... Here are some short answers: The symmetric group acts by as you assumed in your post, i.e. by permuting the systems, $\rho_d(\pi)(\psi_1\otimes\dots\otimes\psi_n)= \psi_{\pi(1)}\otimes\dots\otimes\psi_{\pi(n)}$. An operator $A$ commutes with a representation of a group if it commutes with each element, in this case $[A,\rho_d(\pi)]=0$ for all $\pi\in S_n$. With "unitary commutant" I meant the commutant of the unitary group, i.e. the operators for which $[A,U^{\otimes n}]=0$ for all $U\in U(d)$. $\endgroup$ Nov 8 at 9:07
  • $\begingroup$ @BlackHat18 They are an operator basis but not orthogonal with respect to the trace / Hilbert-Schmidt inner product $(A,B):=\mathrm{tr}(A^\dagger B)$. $\endgroup$ Nov 8 at 9:09
  • $\begingroup$ @MarkusHeinrich not to be nit-picky but the dimensionality of the unitary commutant being $n!$ is only true if the dimension $d$ of the space $U$ acts on satisfies $d \geq n$, right? $\endgroup$
    – nervxxx
    Nov 8 at 13:27

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