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It is known that the maximally entangled qubit states form a basis (the Bell basis). Let $\Phi$ be the canonical maximally entangled state i.e.

$$\Phi = \left(\frac{\vert 00\rangle + \vert 11\rangle}{\sqrt{2}}\right)\left(\frac{\langle 00\vert + \langle 11\vert}{\sqrt{2}}\right)$$

Let $\Phi^\perp = \mathbb{1} - \Phi$, where $\mathbb{1}$ is the maximally mixed state on two qubits.

Let $P_{k}^{n}\left(\Phi, \Phi^{\perp}\right)$ denotes the summation of $n$-fold tensor products of $\Phi$ and $\Phi^{\perp}$ with exactly $k$ factors of $\Phi$. For example, $P_{1}^{3}\left(\Phi, \Phi^{\perp}\right)=\Phi^{\perp} \otimes \Phi^{\perp} \otimes \Phi+\Phi^{\perp} \otimes \Phi \otimes \Phi^{\perp}+\Phi \otimes \Phi^{\perp} \otimes \Phi^{\perp}$.

Do the terms $P_{k}^{n}\left(\Phi, \Phi^{\perp}\right)$ where $1\leq k\leq n$ form a basis for states that are invariant under permutation of the $n$ registers? That is, can any permutation invariant state of dimension $4^n$ be uniquely written as

$$\Psi = \sum_k c_k P^n_k(\Phi, \Phi^\perp)$$

If yes, how do I see why this is the case?

EDIT: Corrected dimension based on answer

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2 Answers 2

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No. Take the case $n=k=1$. Then your basis contains a single element, $\Phi$, whereas the permutation-invariant subspace of two qubits has dimension 10.

More generally, there's no need to reinvent the wheel, it's easy to construct an explicit basis for the permutation-invariant subspace.

EDIT: I don't know a reference for the construction, but it's easy to to it yourself. Let $\{\sigma_i\}_{i=0}^{d^2-1}$ be a Hermitian basis for the Hilbert space (e.g. the usual generalized Gell-Mann matrices). You need to generate the list of indices that are equivalent under permutations. There are $\binom{n+d^2-1}{n}$ non-equivalent indices. For example, for $n=2$ and $d=2$ the list is

00

01 10

02 20

03 30

11

12 21

13 31

22

23 32

33

Then each element of your basis will be the sum of the tensor product of the Gell-Mann matrices of a given line, e.g., $$\sigma_0 \otimes \sigma_0$$ $$\sigma_0\otimes\sigma_1 + \sigma_1\otimes\sigma_0$$ $$\sigma_0\otimes\sigma_2 + \sigma_2\otimes\sigma_0$$ and so on.

Generating this list of indices is a straightforward but tedious exercise in combinatorics. QETLAB can automate it for you.

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  • $\begingroup$ Thank you - is there a good reference to learn about this construction? $\endgroup$ Jun 30, 2022 at 7:11
  • $\begingroup$ I've edited my answer to explain the construction. $\endgroup$ Jun 30, 2022 at 8:51
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Your dimensions don't appear to work out. You're considering states of dimension $2^n$, but when $n=1$ you've written down a state for a Hilbert space of dimension $4$.

Anyway, I don't think $\Phi$ and $\Phi^\perp$ are good states to consider, even if we just consider $n=1$. Consider $(|01\rangle+|10\rangle)(\langle 01|+\langle 10|)$. There's no way to write this as a combination of $\Phi$ and $\Phi^\perp$.

In general, it's hard to see how this construction could get you all pure symmetric states, since you're looking at linear combinations of a pure state $\Phi$ and a mixed state $\mathbb{1}$.

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