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Let us consider \begin{equation} |T\rangle = |\psi \rangle^{\otimes m} \end{equation} for an $n$-qubit quantum state $|\psi\rangle$. Let $\mathcal{V}$ be the space of all $(m + 1)$-partite states that are invariant under permutations. Consider $|\psi\rangle|T\rangle$, which belongs to the space $\mathcal{V}$. I am trying to prove:

  1. For $|\phi\rangle$ orthogonal to $|\psi\rangle$, $|\phi\rangle |T\rangle$ is almost orthogonal to $\mathcal{V}$, upto an error $\mathcal{O}\big(\frac{1}{m}\big)$.
  2. There exists a measurement to determine whether $|\phi\rangle |T\rangle$ is in $\mathcal{V}$, upto an error $\mathcal{O}\big(\frac{1}{m}\big)$, and when doing so, the state $|T\rangle$ is only disturbed by $\mathcal{O}\big(\frac{1}{m}\big)$.

(For reference, these statements are in this paper, in the last paragraph of page 13).

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    $\begingroup$ Take a look at page 41 in the paper. $\endgroup$ – tsgeorgios Oct 18 '20 at 10:43
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Let $P$ be the projector onto the symmetric subspace. We want to find $$ \max|\langle\gamma|\phi,T\rangle| $$ for $|\gamma\rangle\in \mathcal{V}$. This is equivalent to $$ \max|\langle\gamma|P|\phi,T\rangle|, $$ so if we calculate $P|\phi\rangle|T\rangle$, then $|\gamma\rangle$ will be the normalised state parallel to that.

Now, if we project $|\phi\rangle|T\rangle$ onto the fully symmetric subspace, we get $$ |\gamma\rangle=\frac{1}{\sqrt{m+1}}\sum_{i=1}^{m+1}|\psi\rangle^{\otimes (i-1)}|\phi\rangle|\psi\rangle^{\otimes(m+1-i)}. $$ Hence, the maximum value of the overlap is $1/\sqrt{m+1}$. If you're talking about an error, you'r probably talking about a measurement, and hence need the mod-square, so the probability of finding the state in the symmetric subspace is $1/(m+1)$.

How to perform the measurement? I could just specify measurement projectors $\{P,I-P\}$. However, you may subsequently ask how you perform that measurement using standard gate elements. I've never looked at this in great detail, but assume it is related to topics such as this. Equally, for this specific form of state, I imagine you could do something generalising the swap-test (which projects two systems onto their symmetric subspace).


Further explanation of the claimed result:

Let $$ |\omega_j\rangle=\frac{1}{\sqrt{m+1}}\sum_{k=1}^{m+1}\omega^{kj}|\psi\rangle^{\otimes (k-1)}|\phi\rangle|\psi\rangle^{\otimes(m+1-k)} $$ where $\omega=e^{2\pi i/(m+1)}$. Each of the $|\omega_j\rangle$ is an eigenvector, eigenvalue $\omega^j$, of the cyclic permutation operator. The cyclic permutation operator and the projector onto the symmetric subspace commute (because the symmetric subspace is invariant under permutations), and hence they share a common eigenbasis and, in particular, only $|\omega_0\rangle$ is in the symmetric subspace.

We have $$ |\phi\rangle|T\rangle=\frac{1}{\sqrt{m+1}}\sum_{j=0}^m|\omega_j\rangle. $$ Thus, $$ P|\phi\rangle|T\rangle=\frac{1}{\sqrt{m+1}}|\omega_0\rangle. $$ The normalised version of this is hence $$ |\gamma\rangle=|\omega_0\rangle. $$

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  • $\begingroup$ Could you comment on why the projection of $\vert\phi\rangle\vert T\rangle$ gives the state you got? And why it is also denoted as $\vert\gamma\rangle$? $\endgroup$ – user1936752 Oct 18 '20 at 17:18
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Hope this helps you get started, If two vectors are orthogonal, then their inner product is zero. The inner product is a measure of how similar two vectors are; or, the degree to which the first vector lies along the second. So you know that |𝜙⟩|𝑇⟩~ 0

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