# Schur's lemma for quantum states

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 $$\text{Sym}^n(H\otimes K)$$?

• Aug 8 at 16:56

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: $$\tau=\frac{1}{\binom{\mathrm{dim}(H\otimes K) + n - 1}{n-1}}P$$

• 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? Aug 11 at 1:59
• @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) Aug 11 at 6:59
• 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" Sep 5 at 9:07
• @EvgeniyZh Indeed, thanks for spotting it! Sep 5 at 11:57