I am trying to understand Lemma 2 in this paper.

Consider a state $\tau_{H^n}=\int \sigma^{\otimes n}_{H} \mu(\sigma)$ where $\mu(\sigma)$ is the measure on the space of density operators on a single subsystem induced by the Hilbert-Schmidt metric.

Let $\sigma_{HK}$ be the purification of the state $\sigma_{H}$. We then have $$\tau_{H^nK^n}=\int \sigma^{\otimes n}_{HK} d(\sigma)$$ where $d(\sigma)$ is the measure on the pure states induced by the Haar measure.

It is claimed that by Schur's lemma, $\tau_{H^nK^n}$ is proportional to the identity on the symmetric subspace $\text{Sym}^n(H\otimes K)$. I am not familiar with Schur's lemma and reading about it has me more confused about why this claim is true.

Question: What is Schur's lemma (in this context) and why does it prove that $\tau_{H^nK^n}$ is proportional to the identity on the symmetric subspace $\text{Sym}^n(H\otimes K)$?


1 Answer 1


Most generally, Schur's Lemma is used as a tool in representation theory. In this answer, I'll try to explain it without talking about said theory.

Preliminaries: Schur's Lemma

Let us consider a vector space $V$ on an algebraically closed field $\mathbb{K}$ (typically $\mathbb{C}$). We denote $\mathcal{L}(V)$ the space of endormophisms of $V$. Finally, let us consider a subspace $U$ of $\mathcal{L}(V)$.

$U$ is said to be irreducible if the only subspaces of $V$ that are left unchanged by every $u\in U$ are the nil subspace $\{0\}$ and $V$ itself. That is, we have: $$\forall F\text{ subspace of }V, [\forall u\in U, \forall x\in F, u(x)\in F]\iff[F=\{0\}\lor F=E]$$

What Schur's Lemma tells you is that if an endomorphism $v$ commutes with all elements of an irreducible $U$, then $v$ must be an homothety, that is $v$ is proportional to the identity. That is: $$[\forall u\in U, u\circ v=v\circ u]\implies[\exists\lambda\in\mathbb{K},v=\lambda\,\mathrm{id}]$$

Computing the integral

So now, let us consider your integral: $$\tau=\int\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)$$ where $\mu$ is the Haar-measure on $H\otimes K$. Let us decompose the reasoning in two steps.

Showing that $\tau$ is an operator on the symmetric subspace

First of all, note that $\tau$ is an operator on the symmetric subspace $\mathrm{Sym}^n(H\otimes K)$. In order to show this, let us define $P_\pi$ as the permutation of the tensor factors according to the permutation $\pi\in\mathfrak{S}_n$. That is, if we have a state $\left|x_1,\cdots,x_n\right\rangle\in(H\otimes K)^{\otimes n}$, then: $$P_\pi\left|x_1,\cdots,x_n\right\rangle=\left|x_{\pi^{-1}\left(1\right)},\cdots,x_{\pi^{-1}\left(n\right)}\right\rangle$$ Note that for any $\pi\in\mathfrak{S}_n$, we have: $$P_\pi\tau=P_\pi\int\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)=\int P_\pi\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)=\int\sigma^{\otimes n}\,\mathrm{d}\mu(\sigma)=\tau$$ And similarly we have $\tau P_\pi=\tau$. The penultimate equation is more easily seen when writing $\sigma$ as $|\psi\rangle\!\langle\psi|$, since $\sigma^{\otimes n}=|\psi\rangle^{\otimes n}\!\langle\psi|^{\otimes n}$ and $P_\pi|\psi\rangle^{\otimes n}=|\psi\rangle^{\otimes n}$.

As a recall, the symmetric subspace is defined by the fact that $|\psi\rangle\in\mathrm{Sym}^n(H\otimes K)$ if and only if $P_\pi|\psi\rangle=|\psi\rangle$ for every $\pi\in\mathfrak{S}_n$.

Let us consider $|\psi\rangle\in\mathrm{Sym}^n(H\otimes K)$. We have for any $\pi\in\mathfrak{S}_n$: $$P_\pi(\tau |\psi\rangle)=\left(P_\pi\tau\right)|\psi\rangle=\tau|\psi\rangle$$ Thus, $\mathrm{Sym}^n(H\otimes K)$ is stable by $\tau$. On the other hand, it is known that: $$P=\frac{1}{n!}\sum_{\pi\in\mathfrak{S}_n}P_\pi$$ is an orthogonal projector on $\mathrm{Sym}^n(H\otimes K)$. This means that: $$(H\otimes K)^{\otimes n}=\mathrm{Sym}^n(H\otimes K)\overset{\perp}{\oplus}\mathrm{Ker}(P)$$ So let us consider $|\psi\rangle\in\mathrm{Ker}(P)$. Note that $\tau=\tau P$. Thus: $$\tau|\psi\rangle=\tau P|\psi\rangle=\tau0=0$$ Thus, $\tau$ is an operator on $\mathrm{Sym}^n(H\otimes K)$: we only have to determine its behavior on this space to know it completly.

Using Schur's Lemma

We know want to find an irreducible subspace of operators on $\mathrm{Sym}^n(H\otimes K)$ such that $\tau$ commutes with every operator in this subspace. Note that, denoting $\mathcal{U}$ the space of unitaries on $H\otimes K$: $$\begin{align} \forall U\in\mathcal{U},U^{\otimes n}\tau&=\int(U|\psi\rangle)^{\otimes n}\langle\psi|^{\otimes n}\,\mathrm{d}\mu(|\psi\rangle)\\ &=\int|\psi\rangle^{\otimes n}(\langle\psi|U)^{\otimes n}\,\mathrm{d}\mu\left(U^\dagger|\psi\rangle\right)\\ &=\int|\psi\rangle^{\otimes n}(\langle\psi|U)^{\otimes n}\,\mathrm{d}\mu(|\psi\rangle)\\ &=\int|\psi\rangle^{\otimes n}\!\langle\psi|^{\otimes n}\,\mathrm{d}\mu(|\psi\rangle)U^{\otimes n}\\ &=\tau U^{\otimes n}\end{align}$$ Thus, $\tau$ commutes with every element of $\left\{U^{\otimes n}\middle|U\in\mathcal{U}\right\}$. It has been proven that this space is irreducible with respect to $\mathrm{Sym}^n(H\otimes K)$. We can thus apply Schur's Lemma, which gives us that $\tau$ acts as the identity (up to a multiplicative factor) on $\mathrm{Sym}^n(H\otimes K)$. Since it is nil outside of this space, we actually have: $$\exists\lambda\in\mathbb{C},\tau=\lambda P$$ We now only have to determine $\lambda$. But we know that $\mathrm{tr}(\tau)=1$ since it's a density matrix. Thus: $$1=\lambda\mathrm{tr}(P)$$ Since $P$ is a projector on $\mathrm{Sym}^n(H\otimes K)$, its trace is equal to the dimension of $\mathrm{Sym}^n(H\otimes K)$, which is known to be $\binom{\mathrm{dim}(H\otimes K) + n - 1}{n-1}$. This finally gives us: $$\tau=\frac{1}{\binom{\mathrm{dim}(H\otimes K) + n - 1}{n-1}}P$$

  • $\begingroup$ Thanks for a great answer. Just one question: How do we know that the set of i.i.d. unitaries is irreducible with respect to the symmetric subspace? You mention this after showing that $\tau U^{\otimes n} = U^{\otimes n}\tau$. Or did you mean that this is a known result but not shown in the answer? $\endgroup$ Commented Aug 11, 2023 at 1:59
  • 1
    $\begingroup$ @user1936752 The latter. It's generally proved using representation theory, but Harrow proved it using the Quantum formalism in this paper: arxiv.org/abs/1308.6595 (Theorem 5) $\endgroup$
    – Tristan Nemoz
    Commented Aug 11, 2023 at 6:59
  • $\begingroup$ I think you meant "the only subspaces of V that are left unchanged by every u" rather than "for any u∈U, the only subspaces of V that are left unchanged" $\endgroup$
    – EvgeniyZh
    Commented Sep 5, 2023 at 9:07
  • 1
    $\begingroup$ @EvgeniyZh Indeed, thanks for spotting it! $\endgroup$
    – Tristan Nemoz
    Commented Sep 5, 2023 at 11:57

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