Let's have two orthogonal n-particle quantum states: $|\psi \rangle$ and $|\phi \rangle$. In theory it is always possible to make an unambiguous measurement.

However, things get complicated when one restricts oneself to a certain class of measurements. With so called LOCC (Local Operations and Classical Communication, that is, we have to measure particles separately, but we are allowed to communicate and to have dependence for measurements on the outcomes of previous measurement) still it is possible to unambiguously distinguish any two states (see: Walgate et al., Local Distinguishability of Multipartite Orthogonal Quantum States, Phys. Rev. Lett. 85, 23: 4972-4975 (2000) arXiv:quant-ph/0007098).

With fixed local operations (and thus classical communications only after all measurements are done) sometimes we can't unambiguously distinguish between $|\psi \rangle$ and $|\phi \rangle$.

  • Is there any simple argument why?
  • Are there any simple criteria which says which orthogonal states can be unambiguously distinguished with local measurements and communication only after them?

To keep things simple, let's talk about two-qubit states.

A single qubit could have an orthonormal basis $\{|0\rangle, |1\rangle\}$. But it could also have a different orthonormal basis $\{|+\rangle,|-\rangle\}$, where $$|+\rangle = \large(\normalsize|0\rangle \small+\normalsize |1\rangle\large)\normalsize / \sqrt{2}$$ $$|-\rangle = \large(\normalsize|0\rangle \small-\normalsize |1\rangle\large)\normalsize / \sqrt{2}$$

No suppose the two states you're trying to distinguish are: $$|\psi\rangle = \large(\normalsize|0\rangle|0\rangle \small+\normalsize |1\rangle|+\rangle\large)\normalsize / \sqrt{2}$$ and $$|\phi\rangle = \large(\normalsize|0\rangle|1\rangle \small+\normalsize |1\rangle|-\rangle\large)\normalsize / \sqrt{2}$$

Here the first qubit in each term is Alice's and the second qubit in each term is Bob's.

Let's say Alice does a measurement in the $\{|0\rangle, |1\rangle\}$ basis. After the measurement the state will be projected into a new state, but which new state depends on the result of Alice's measurement.

If Alice observes state $|0\rangle$: $$|\psi\rangle \rightarrow |0\rangle|0\rangle$$ $$|\phi\rangle \rightarrow |0\rangle|1\rangle$$

If Alice observes state $|1\rangle$:$$|\psi\rangle \rightarrow |1\rangle|+\rangle$$ $$|\phi\rangle \rightarrow |1\rangle|-\rangle$$

If Bob knows Alice observes $|0\rangle$, then he can distinguish the two states by making a measurement in the $\{|0\rangle,|1\rangle\}$ basis. Likewise, if Bob knows Alice observes $|1\rangle$, then he can distinguish the two states by making a measurement in the $\{|+\rangle,|-\rangle\}$ basis.

But if Bob doesn't know the result of Alice's measurement until after he makes his measurement, then he doesn't know which measurement to make.

If the true state is $|1\rangle|+\rangle$ but Bob guesses it's either $|0\rangle|0\rangle$ or $|0\rangle|1\rangle$, then he's going to make a measurement in the $\{|0\rangle,|1\rangle\}$ basis. This means he's equally likely to observe $|0\rangle$ or $|1\rangle$, since the true state of his qubit, $|+\rangle$, is an equal superposition of $|0\rangle$ and $|1\rangle$. Note that even if the true state of his qubit were $|-\rangle$ (the only other possibility after Alice's measurement), Bob would still have been equally likely to observe $|0\rangle$ or $|1\rangle$.

So after comparing results, Alice and Bob realize that Bob's measurement tells them nothing, and moreover the result of Alice's measurement by itself does nothing to distinguish the two initial states.


I'm going to try to give a more general rule for which states Alice and Bob can distinguish with local measurements only (no communication until afterwards.)

If Alice and Bob aren't going to communicate until their measurements are done, then the combined measurement is effectively chosen before hand. As Frédéric Grosshans points out in his answer, this measurement will be a projection into the eigenbasis of $A \otimes B$, where $A$ is a local observable that Alice can measure and $B$ is a local observable that Bob can measure.

If $S_A$ is the set of eigenstates of $A$, and $S_B$ is the set of eigenstates of $B$, then the set of eigenstates of $A \otimes B$ is the direct product (the Cartesian product) of $S_A$ and $S_B$.

As an example, if $S_A = S_B = \{|0\rangle, |1\rangle\}$, Alice and Bob will only distinguish $\{|0\rangle|0\rangle, |0\rangle|1\rangle, |1\rangle|0\rangle, |1\rangle|1\rangle\}$. If $S_A = \{|0\rangle,|1\rangle\}$ and $S_B = \{|+\rangle,|-\rangle\}$, Alice and Bob will only distinguish $\{|0\rangle|+\rangle, |0\rangle|-\rangle, |1\rangle|+\rangle, |1\rangle|-\rangle\}$.

If you have a set like my two entangled states above, or Frédéric Grosshans' four states $\{|0\rangle|0\rangle, |0\rangle|1\rangle, |1\rangle|+\rangle, |1\rangle|-\rangle\}$, then there's no choice of $A$ and $B$ that will suffice, because these states aren't in the direct product of the eigenstates of two local measurements.

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  • $\begingroup$ Thanks. However, you constrained Alice to measure in $\{|0\rangle, |1\rangle\}$ basis. I can numerically check that there are no bases for Alice and Bob in which they can perform an unambiguous measurement. So it do not give me any further insight in 'why' such phenomena has place and how to check if such unambiguous measurement is possible. $\endgroup$ – Piotr Migdal Nov 12 '10 at 17:40
  • $\begingroup$ @Piotr, certainly the above is not a rigorous proof, even for the two qubit case. But I think that it does capture the basic idea of "why" the communication can't be deferred to the end. In order for Bob to distinguish the states, they must be eigenstates of some observable (after projection by Alice's measurement). And moreover, he has to be able to know what measurement it is (and thus, what eigenbasis) without knowing Alice's result. There are some states for which those requirements can't be met. $\endgroup$ – Tim Goodman Nov 14 '10 at 23:47
  • $\begingroup$ Admittedly, I didn't prove this, just gave an example of how an attempt might fail. But you did ask for a "simple argument" after all. $\endgroup$ – Tim Goodman Nov 14 '10 at 23:49
  • $\begingroup$ @Piotr, on further consideration, I think perhaps what you're wanting is a more general condition for when local measurements alone won't suffice. I've added an edit to try to address this. $\endgroup$ – Tim Goodman Nov 15 '10 at 2:30
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    $\begingroup$ When it comes ${|00\rangle,|1+\rangle}$ there is $\sigma_z \otimes \mathbb{1}$ measurement, so you do not need to add anything about single local measurement. BTW: $A\otimes B$ is called tensor product. $\endgroup$ – Piotr Migdal Nov 16 '10 at 15:54

(Edited to correct TeX typo)

I complete Tim Goodman's answer in answer to get something more systematic. A local measurement has to be written as a tensorial product of two observables A⊗B. And it can only distinguish (with probability 1) its eigenstates. The states |ϕ⟩ and |ψ⟩ of Tim's example cannot be written as eigenstates of a tensorial product.

Note that this does not correspond exactly to states which cannot be written as tensorial product of states. For example, if let's use the following 4 states : $|\psi_0\rangle=|00\rangle$ , $|\psi_1\rangle=|1+\rangle$,
$|\phi_0\rangle=|01\rangle$ , $|\phi_1\rangle=|1-\rangle$ and let's try to distinguish the ψs from the ϕs.Tim's arguments are still valid, even if each the 4 states is a product state and is orthogonal to all the 3 others. Furthermore, each pair of state is locally distinguishable.

I think locally distinguishable subspaces has something to do with to the direct sum of locally orthogonal subspaces, but I don't exactly know how to write it.

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  • $\begingroup$ +1. Good point about the measurement distinguishing eigenstates of the $A \otimes B$ for local observables $A$ and $B$. $\endgroup$ – Tim Goodman Nov 15 '10 at 2:42
  • $\begingroup$ Well, saying I need measurement in the form $A \otimes B$ is only a reformulation of the problem, with two drawback: 1) (Major) I do not know if its analogue works for $n>2$ qubits (I guess it does not). 2) (Minor) You cannot extend it to measuring 3 different outcomes with one setting (let Alice use $0/1$ and Bob $+/-$ basis then the subset $\{|00\rangle, |1+\rangle, |1-\rangle \}$ ) is distinguishable (but there is no $A\otimes B$ operator, if I am right). $\endgroup$ – Piotr Migdal Nov 16 '10 at 16:03
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    $\begingroup$ About point 2) (Nice example, btw) Take the operators $A=|1\rangle\langle1|$ and $B=|+\rangle\langle+|-|-\rangle\langle-|$, then your 3 states are eignestates of $A\otimes B$ of eigenvalues 0,±1. But I guess that on could find another example (with high dimensions $d$ and less than $d^2$ states) where it is not eigenstates, but I'm pretty sure it can be seen as a decomposition on loval eigenstates (where $|00\rangle$ would be seen as a subspace of the space spanned by $|0\pm\rangle$.) $\endgroup$ – Frédéric Grosshans Nov 16 '10 at 19:29
  • $\begingroup$ About point 1): lot of things are more complicated for three party (ore more) settings. But I would be surprised if the property would not be rephrased as $A\otimes B\otimes C$. After all, it's just a reformulation of the problem ;-) $\endgroup$ – Frédéric Grosshans Nov 16 '10 at 19:33

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