Consider two states $\sigma_0,\sigma_1\in\text{L}(\mathcal{H}_{AB})$, and suppose $\sigma_0,\sigma_1$ are separable and orthogonal. Is it possible to distinguish between $\sigma_0,\sigma_1$ through LOCC?

My approach so far har been to write out $$ \sigma_0 = \sum_{i=1}^n p_i |a_ib_i\rangle\langle a_ib_i| \quad\text{and}\quad \sigma_1 = \sum_{j=1}^n q_j |a_jb_j\rangle\langle a_jb_j|, \quad \text{where } p_i,q_j\geq 0, $$ and since $$ 0 = \text{Tr}(\sigma_0^\dagger\sigma_1) = \sum_{i,j}^n p_iq_j \lvert\langle a_ib_i | a_jb_j\rangle\rvert^2, $$ it follows that all terms in the decomposition of $\sigma_0$ are orthogonal to all terms in the decomposition of $\sigma_1$. My idea was now to measure using projections onto the two subspaces spanned by the terms in each decomposition, and these are separable projections. I am stuck at implementing this as an LOCC protocol, so any help with this or giving an alternative approach is appreciated!

  • $\begingroup$ One of the problems that you have is that while the vectors $|a_i\rangle|b_i\rangle$ may be orthonormal and separable, that does not guarantee that the sets $\{|a_i\rangle\}$ and $\{|b_i\rangle\}$ form orthonormal bases, so you may not be able to set them up as local measurements. It doesn't mean that there isn't an LOCC protocol that distinguishes them, but it does suggest that a general protocol is, at best, problematic. $\endgroup$
    – DaftWullie
    Jun 25, 2020 at 8:49
  • $\begingroup$ Does "distinguish" mean "with certainty", or "with some chance"? That's a major difference. $\endgroup$ Jun 25, 2020 at 9:59
  • $\begingroup$ @NorbertSchuch Good point, I mean with certainty. $\endgroup$
    – user114158
    Jun 25, 2020 at 17:27

2 Answers 2


Consider these two states $$ \sigma_0 = \frac{1}{2}(|11\rangle\langle 11| + |++\rangle\langle ++|) $$ $$ \sigma_1 = \frac{1}{2}(|0-\rangle\langle 0-| + |-0\rangle\langle -0|) $$

I believe they are indistinguishable (with certainty), though to be sure it's better to find the exact proof.

Also check this paper https://arxiv.org/abs/quant-ph/9804053.
It's impossible to distinguish a set of product states in general, though this is not directly answers your question.

As John Watrous explained in the comments, $\sigma_0$,$\sigma_1$ are indeed indistinguishable. They have orthogonal images that span the whole space. So, the only way to distinguish them with certainty is to use the two-outcome projective measurement where projections correspond to the images. But these projections are not separable, we can use PPT criterion to check this. For example, the projection on $\text{Im}(\sigma_0)$ is $$ P_0 = \frac{1}{3}\begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 3 \\ \end{pmatrix} $$ and you can check that the partial transpose $$ P_0^{T_2} = \frac{1}{3}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 3 \\ \end{pmatrix} $$ has a negative eigenvalue.

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    $\begingroup$ Regarding the quoted paper: What does it say? (And if it is about a larger set, then even is a strange word: Is is clearly harder to distinguish more states.) $\endgroup$ Jun 25, 2020 at 9:58
  • $\begingroup$ Yes, it's not exactly comparable. On the one hand product state is more restrictive condition than separable, but on the other hand it is harder to distinguish more states. This is why I presented a different example. $\endgroup$
    – Danylo Y
    Jun 25, 2020 at 10:21
  • $\begingroup$ I have to admit that "I don't see a way to distinguish them" is not a very compelling argument. (To quote Reinhard Werner: Lack of imagination is not an argument.) For sure, you can probabilistically distinguish them. Also, note that LOCC protocols can be very complicated (multi-round protocols etc.!), so I'm not sure whether "some calculations" are sufficient. $\endgroup$ Jun 25, 2020 at 10:33
  • $\begingroup$ I'm not very familiar with such kind of proofs. But the question is asking about some help. Maybe I'll find the exact proof later. $\endgroup$
    – Danylo Y
    Jun 25, 2020 at 10:48
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    $\begingroup$ This is s great guess. Indeed these two states cannot be perfectly discriminated by an LOCC measurement, or even by a separable measurement, which is surprising given that the states themselves are separable. The reason is that the states have orthogonal images that together span the entire two-qubit space, so the only measurement that can perfectly discriminate them is the two-outcome projective measurement corresponding to the projections onto their images. These projections, however, are not separable, as the PPT test reveals. $\endgroup$ Jun 30, 2020 at 1:28

If you restrict to pure states, all orthogonal multipartite states are locally distinguishable. By "local" I mean here a LOCC measurement where you measure the different parties in sequence and the each measure can to be conditioned to the previous measurement results — it's thus a relatively simple kind of LOCC measurement that doesn't require any back and forth. This was originally shown in Walgate et al. 2000.

To show this, let $|\Psi\rangle,|\Phi\rangle\in \mathbb{C}^n\otimes\mathbb{C}^m$ be arbitrary pure bipartite orthogonal states, $\langle\Psi|\Phi\rangle=0$. For locally distinguishability it is enough to find an ON basis $\{|u_a\rangle\}\subset\mathbb{C}^n$ such that $\langle\Psi|(\mathbb{P}_{u_a}\otimes I)|\Phi\rangle=0$ for all $a$, using the notation $\mathbb{P}_{u_a}\equiv|u_a\rangle\!\langle u_a|$. Indeed, this would imply that, performing this projective measurement, regardless of which outcome $a$ is found, the resulting conditional states on $\mathbb{C}^m$ remain orthogonal, and therefore deterministically distinguishable.

First attempt at a proof — We want to show that there is always such a basis. To this end, observe that $$\langle \Psi|(\mathbb{P}_{u_a}\otimes I)|\Phi\rangle = \langle u_a|\underbrace{\operatorname{tr}_2(|\Phi\rangle\!\langle\Psi|)}_{\equiv \Phi\Psi^\dagger}|u_a\rangle = \langle u_a|M|u_a\rangle,$$ where $2M= \Phi\Psi^\dagger+\Psi\Phi^\dagger$, and I used the fact that $\Phi\Psi^\dagger-\Psi\Phi^\dagger$ is skew-Hermitian and thus has vanishing expectation values. Note that I'm denoting with $\Psi$ the matrix whose elements are those of the vector $|\Psi\rangle$, that is, $(\Psi)_{ij}=\langle i,j|\Psi\rangle$. The matrix $M$ is Hermitian and thus admits an eigendecomposition of the form $M=\sum_k\lambda_k \mathbb{P}_k$. If you then define $|u_a\rangle\equiv \frac{1}{\sqrt n}\sum_b \omega^{ab} |b\rangle$, with $\omega\equiv e^{2\pi i/n}$, you get $$\langle u_a|M|u_a\rangle = \frac1n\sum_{bc} \omega^{-ab+ac} \langle b|M|c\rangle = \frac1d \sum_b \lambda_b =\frac1n \operatorname{tr}(M)=0.$$ We thus conclude that the orthonormal basis $\{|u_a\rangle\}\subset\mathbb{C}^n$ satisfies the constraints, i.e. performing a projective measurement on it preserves orthogonality on the conditional states on $\mathbb{C}^m$, and thus one can always deterministically distinguish any pair of orthogonal pure states via local measurements. Note that I'm showing that the basis to use is the Fourier transform of the eigenbasis of $M$. Though I think you don't need to use this specific basis: any other orthonormal basis that is unbiased wrt the eigenbasis of $M$ should work just the same — it just so happens that the QFT is a good standard way to get an unbiased basis.

IMPORTANT EDIT: I'm leaving the above paragraph for reference, but I later found out that this proof is wrong. It only works as presented in some cases, such as when $\Psi\Phi^\dagger$ is normal, or when it's 2x2. The issue is that it uses $\langle u_a|B|u_a\rangle=0$ for skew-Hermitian $B$, which is simply wrong in general. Nevertheless, the statement still holds: one can prove in general that any traceless matrix has zero diagonal in some basis, as discussed in this math.SE post. It seems the proof isn't constructive in the general case though (and note that Walgate et al's proof also requires some enlarging of the space shenanigans for the general case).

Toy example 1

Consider the bipartite example with $$\sqrt2|\Psi\rangle = |00\rangle+|11\rangle, \qquad \sqrt2|\Phi\rangle = |00\rangle-|11\rangle.$$ Note that measuring in the computational basis wouldn't work: if you for example measured $|0\rangle$ on the first qubit, the second qubit would be $|0\rangle$ regardless of which of the two states you're given.

To find the right measurement basis, we can follow the procedure outlined above: compute $$\Psi\Phi^\dagger = \frac12\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ In this case this is Hermitian so we don't need to introduce the $M$ operator. Clearly the eigenbasis of this is the computational basis, and therefore the correct basis to measure in is $|\pm\rangle$, that is, the eigenbasis of the Pauli X matrix.

You can verify that this works because measuring $|+\rangle$ on the first qubit the second qubit would transform as $|\Psi\rangle \to |+\rangle$, $|\Phi\rangle \to |-\rangle,$ while if you get $|-\rangle$ on the first qubit you have $|\Psi\rangle \to |-\rangle$, $|\Phi\rangle \to |+\rangle$. You therefore clearly always have orthogonal states on the second qubit regardless of the first measurement result. Furthermore, you don't even need to change the computational basis in this case: you just measure both qubits in the $|\pm\rangle$ basis.

It's however worth noting that the resulting protocol still involves some classical communication, because to figure out which bipartite state you've been given from the second measurement result, you need to know the first measurement result. Namely, you'll overall (correctly) guess the input was $|\Psi\rangle$ if you measure (+,+) or (-,-), and you'll guess $|\Phi\rangle$ if you measure (+,-) or (-,+).

Toy example 2

Say now $$\sqrt2|\Psi\rangle = |00\rangle+|11\rangle, \qquad \sqrt2|\Phi\rangle = |01\rangle-|10\rangle.$$ This gives $\Psi\Phi^\dagger=\frac12\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, which is interesting because it's not Hermitian, and if you were to define the $M$ operator you'd just get $M=0$. This tells you that it doesn't matter what measurement basis you use for the first qubit, because the leftover second qubit will remain orthogonal no matter what. You can also see this as direct consequence of $\Psi\Phi^\dagger$ being skew-Hermitian, and thus its diagonal is always made of zeros.

Toy example 3

Let's ease into the multipartite case. Say we want to distinguish between two GHZ-like states: $$\sqrt2|\Psi\rangle = |000\rangle+|111\rangle, \qquad \sqrt2|\Phi\rangle = |000\rangle-|111\rangle.$$ In such cases we start showing how to measure the first qubit leaving the rest orthogonal. In other words, we first treat the states as bipartite in a $\mathbb{C}^2\otimes\mathbb{C}^4$ space. Note how this means that $\Psi,\Phi$ are not square matrices anymore: we have $$\Psi= \frac{1}{\sqrt2}\begin{pmatrix}1&0&0&0\\0&0&0&1\end{pmatrix}, \qquad \Phi= \frac{1}{\sqrt2}\begin{pmatrix}1&0&0&0\\0&0&0&-1\end{pmatrix},\\ \Psi\Phi^\dagger = \begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ Following the usual recipe we conclude that the first qubit needs to be measured in the $|\pm\rangle$ basis. The leftover two-qubit states will be the Bell states $|00\rangle\pm|11\rangle$, and thus we already know how to continue the procedure from the first example.

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    $\begingroup$ Nice. You also could mention that the basis $\{|u_a\rangle\}$ is the Fourier transform of the basis $\{|a\rangle\}$ which is the eigenbasis of $M$. $\endgroup$
    – Danylo Y
    Mar 5 at 12:35
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    $\begingroup$ In essence you've made $0$ the real part of diagonal of $\Phi\Psi^\dagger$. To make $0$ the imaginary part of that diagonal too you can use a sequence of two-level unitaries. If $(d_1,d_2)$ is a diagonal of some 2x2 complex matrix where $d_1,d_2$ are real and $d_1 \le 0 \le d_2$ then there should be a unitary that transforms it to a matrix with the diagonal $(0, d_1+d_2)$, it seems (I'm not sure how to prove it constructively). Similarly, we could from $(id_1,id_2)$ to $(0, i(d_1+d_2))$. And then by recursion when dimension is more than 2. $\endgroup$
    – Danylo Y
    Mar 6 at 14:20

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