# Notes on Ghirardi/Marineto (2004)

$$\newcommand{\lt}{\left} \newcommand{\me}{\middle} \newcommand{\rt}{\right} \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\ua}{|\hspace{-3 pt}\uparrow\ra} \newcommand{\da}{|\hspace{-3 pt}\downarrow\ra} \newcommand{\uabra}{\la\uparrow\hspace{-3 pt}|} \newcommand{\dabra}{\la\downarrow\hspace{-3 pt}|}$$

In this paper Ghirardi and Marineto offer conditions under which bipartite states of indistinguishable particles should be considered entangled. These conditions are not the same as for distinguishable particles. In the distinguishable case, it’s easy: two particles are non-entangled of and only if they can be written as a product state. I.e. the Schmidt number is 1 (number of terms in Schmidt decomposition). In the indistinguishable case, this would mean that fermions are always entangled, and that bosons are only non-entangled if we have a state that can be written as $$| \phi \ra_1 | \phi \ra_2$$. But, there is no way to measure ‘particle 1’ and ‘particle 2’ seperately (unlike the distinguishable case). Instead, measurement operators are symmetric.

G&M start with a definition for entanglement: that to each particle we can describe a full set of definite properties. They then formalize this and come up with the criterion that the state satisfies

$$\la\phi|E_P|\phi\ra = 1,$$

where

\begin{align} E_P &= P_1 \otimes (\mathbb{1} – P_2) + (\mathbb{1} – P_1) \otimes P_2 + P_1 \otimes P_2\\ &= \mathbb{1} – (\mathbb{1} – P_1) \otimes (\mathbb{1} – P_2) \end{align}

(Is $$E_P$$ a projector itself? Yes it is. A sum of orthogonal projectors is itself a projector. This projector corresponds to the question: “is the state corresponing to P occupied at least once?” or “Is there at least one particle in the state $$|\phi\ra$$?”). This in turn implies that for fermions, a state is non-entangled iff it is the antisymmetrization of a product state. For bosons, it implies that a state is non-entangled iff it is the symmetrization of a product state of two orthogonal states, or it is the product of two identical states.

This in turn can be linked to the Schmidt number/Slater number of the bipartite state, and to the Von Neumann entropy of the reduced 1-particle states (the reduced states are identical for ‘particles’ 1&2. The Slater number is for antisymmetrized states, and is just half the Schmidt number (because of the antisymmetrization, the Schmidt terms come in pairs, so it is always an even number and we can rewrite the sum using those antisymmetrical pairs, and have half the number of terms in the sum, i.e. a Slater number amount of terms). The Von Neumann entropy is defined as $$-Tr \rho \log \rho$$ (log base 2). For a pure state, this equals 0, for a mixed state, it is greater than 0. For a perfectly entangled bipartite state, it is $$-2*1/2 \log 1/2 = 1$$

For fermions, a state is nonentangled iff the Slater number of the state is 1 (Schmidt number 2). The (Von Neumann) entropy is 1. If the state is entangled, both the Slater number and the entropy are higher. G&M interpret a entropy of 1 because of the unavoidable uncertainty ‘which particle has which properties’, and a higher entropy signifies an additional uncertainty because of the entanglement. (I’m not so certain of this). Note that because of degeneracy in the Schmidt decomposition a nonentangled state can also be written as the antisymmetrization of two different orthogonal states (spanned by the other two states).

For bosons, a state is nonentangled iff…

• the Schmidt number of the state is 1 (which is equivalent to the Von Neumann entropy being 0) OR
• the Schmidt number is 2 and the Von Neumann entropy is 1 (then it is a symmetrization of two orthogonal states)

Note that two bosons are entangled if the state is a symmetrization of two nonorthogonal states. Then it can also be written as a|1>|1> + b|2>|2>. Also note that a non-entangled states that is the symmetrization of two orthogonal states can again be rewritten as that of two other states (spanned by the original states).

G&M state, but do not proof, that entanglement coincides with being able to violate Bell inequalities and perform teleportation. I beg to differ.

Some points I want to investigate and write about… (I have now included comments after writing more)

• The Schmidt decomposition has a particular form for bosons and fermions because of the (anti-)symmetrization. G&M allude to theorems in matrix theory about that. I wonder if it can be shown simpler. Also in the bosons case the coefficients are real, in the fermion case, they are complex (conjugates). I wonder why, since in the Schmidt decomposition, the coefficients are always real.

It turns out it’s not so easy to make it simpler. The reason is that, even for (anti-)symmetrical states, the two Schmidt bases need not be identical. The Slater decomposition has in fact different basis states than the Schmidt one (you need those for the minus signs)
• It seems a bit weird that in the boson case, the state is entangled if it is the symmetrization of two non-orthogonal states. Can I show that a Bell inequality can be violated in that way?

No, and this is criticism to G&M, what do they mean with violating a Bell inequality in this case?
• I have a counterexample: a G&M-nonentangled state which violates a Bell inequality. I wonder if there is also a counterexample of a G&M-entangled state that cannot violate a Bell inequality?

I don’t think so, it seems that for every state a Bell inequality can be violated, if you pick your measurement operators weird enough!
• I think the G&M notion of entanglement only works if the measurement locations coincide with properties that distinguish the particles (projectors P and Q). This can be made more precise? (maybe a project for later)

Yes I think so. But for non-orthogonal boson states, such projectors do not exist.
• Is (1) a projector, and what happens when applying it to a symmetrization of non-orthogonal states?

Yes it is. When applying it to such a state, you get 1 for the non-orthogonal states. When applying the counting operator which also has a 2 in it, you get an expectation value between 1 and 2.

## Schmidt decomposition

Ok, it’s two days later, and I figured out that the Schmidt decomposition thing is in fact more subtle than I initially thought. This is mainly because the Schmidt bases do not have to be the same in general, even for (anti-)symmetric states. For fermions, they can not even be the same, if the Schmidt coefficients are all positive, the negative sign of the antisymmetry has to come from somewhere, more on that later. So we are just going to repeat the decompositions from G&M and observe some relations to the Schmidt decomposition.

For fermions, there is the decomposition

$$\sum_{I=1}^N a_i \frac{1}{\sqrt{2}} (|2i-1\ra_1|2i\ra_2 – |2i-1\ra_1|2i\ra_2)$$

where $$a_i$$ are complex and $$\sum_i |a_i|^2 = 1$$. Now define

\begin{align} |\bar{1}\rangle &:= \alpha^*|1\rangle – \beta|2\rangle & |\bar{2}\rangle &:= \beta^*|1\rangle + \alpha|2\rangle\\\ |1\rangle &= \alpha|\bar{1}\rangle + \beta|\bar{2}\rangle & |2\rangle &= -\beta^*|\bar{1}\rangle + \alpha^*|\bar{2}\rangle \end{align}

where $$|\alpha|^2 + |\beta|^2 = 1$$. Then $$|1\ra|2\ra – |1\ra|2\ra = |\bar1\ra|\bar2\ra – |\bar1\ra|\bar2\ra$$. The unitary operator that takes non-barred to barred vectors is

$$U = \begin{pmatrix} \alpha^* & \beta* \\ -\beta & \alpha \end{pmatrix}; U^{-1} = U^\dagger = \begin{pmatrix} \alpha & -\beta^* \\ \beta & \alpha^* \end{pmatrix}$$

This is not the most general unitary operator; it cannot add separate phase factors to basis states. For example, there is also the unitary matrix

$$U = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix}$$

The state can also be Schmidt decomposed:

$$|\psi\ra = \sum_i c_i |a_i\rangle |b_i\rangle$$

where all $$c_i$$ are positive. The first decomposition can be related to the Schmidt decomposition by letting $$c_{2i} = c_{2i-1}$$ and $$|b_{2i}\ra = |a_{2i-1}\ra, |b_{2i-1}\ra = -|a_{2i}\ra$$.

Let’s turn to bosons. Apparently any symmetric state can be written

$$\sum_i a_i |i\ra |i\ra$$

where the a_i are positive. This is already a Schmidt decomposition, where the Schmidt bases are identical. Define

\begin{align} |\bar1\ra &= \alpha|1\ra + \beta|2\ra \\ |\bar2\ra &= -\beta|1\ra + \alpha|2\ra \end{align} U = \begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \end{pmatrix}; U^{-1} = U^\dagger = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}

with $$\alpha, \beta$$ real! Now $$|\bar1\ra|\bar1\ra + |\bar2\ra|\bar2\ra = |1\ra|1\ra + |2\ra|2\ra$$.

Suppose $$|\psi\ra = \frac{1}{\sqrt{2}}(|\phi\ra|\xi\ra + |\xi\ra|\phi\ra)$$. Define

\begin{align} |1\ra &= \frac{1}{\sqrt{2}}(|\phi\ra + |\xi\ra) & |\phi\ra &= \frac{1}{\sqrt{2}}(|1\ra – i|2\ra)\\ |2\ra &= \frac{i}{\sqrt{2}}(|\phi\ra – |\xi\ra) & |\xi\ra &= \frac{1}{\sqrt{2}}(|1\ra + i|2\ra) \end{align}$$U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -i & +i \end{pmatrix}; U^{-1} = U^\dagger = \begin{pmatrix} 1 & +i \\ 1 & -i \end{pmatrix}$$

Then $$|\psi\ra = \frac{1}{\sqrt{2}}(|1\ra|1\ra + |2\ra|2\ra)$$.

## Counting operators

A counting operator is an operator which ‘counts’ the number of particles that are in a certain state, or in a subspace. A state might be in an eigenstate of a counting operator, in which case the number of particles in that state/subspace is fixed, but it might also be in a superposition. The expectation value gives the averaged number of particles. Let’s start simple: two particles in total. The counting operator for state $$|\psi\ra$$ is then

$$1 \cdot ([\phi] \otimes (\mathbb{1} – [\phi]) + (\mathbb{1} – [\phi]) \otimes [\phi]) + 2 \cdot [\phi] \otimes [\phi]$$

where $$[\phi] := |\phi\ra\la\phi|$$. More generally, for a subspace on which the projector $$P$$ projects, we have

$$1 \cdot (P \otimes (\mathbb{1} – P) + (\mathbb{1} – P) \otimes P) + 2 \cdot P \otimes P$$

The question: “Is there 0, 1 or 2 particles in some state” must always be answered with YES:

$$(\mathbb{1} – P) \otimes (\mathbb{1} – P) + P \otimes (\mathbb{1} – P) + (\mathbb{1} – P) \otimes P + P \otimes P = \mathbb{1}$$

Also, summing all particles in a complete set of orthogonal states/subspaces must yield the total number of particles:

$$\sum_i 1 \cdot (P_i \otimes (\mathbb{1} – P_i) + (\mathbb{1} – P_i) \otimes P_i) + 2 \cdot P_i \otimes P_i = 2\cdot\mathbb{1}$$

For more particles, the operator becomes a bit cumbersome because it involves a lot of permutations. But we can write it like this:

$$\sum_{i=1}^N i \cdot \mathrm{Comb}(P^i \otimes (\mathbb{1} – P)^{N-i})$$

where Comb() gives all the combinations of tensor product orderings (not permutations because we don’t want double counting). We might even construct a counting operator that works for all N simultaneously, i.e. in a Fock-like Hilbert space to be defined later:

$$\sum_{N=1}^\infty \sum_{i=1}^N i \cdot \mathrm{Comb}(P^i \otimes (\mathbb{1} – P)^{N-i})$$

## Non-orthogonal boson states

G&M devote considerable attention to two-boson states that are the result of anti-symmetrising a product state of two non-orthonal states (we ignore normalisation again):

$$|\psi\ra = |\phi\ra |\xi\ra +|\xi\ra |\phi\ra$$

By decomposing $$|\xi\ra$$ into $$|\phi\ra, |\phi_\perp\ra$$, where the latter state is perpendicular to $$|\phi\ra$$, the state can be written

$$|\psi\ra = a \cdot |\phi\ra |\phi \ra + b \cdot (|\phi\ra |\phi_\perp\ra + |\phi_\perp\ra |\phi\ra)$$

This state is an eigenstate of $$E_P$$ both for $$|\xi\ra$$ and $$|\phi\ra$$. Meaning that for both states there is at least one particle occupying it. In terms of counting operators, the state is in a superposition of 1 and 2 particles occupying each state. If one would measure the number of particles in one of the states, the outcome would be 1 or 2. The expectation value is somewhere inbetween 1 and 2.

According to G&M, such a state is entangled. This follows from their definition of nonentanglement, which to them is equivalent to each constituent possessing a complete set of properties. G&M state that in this case the statement “there is exactly one particle possessing properties P” is not true. Correct, because it can be 1 or 2 particles. Still it is not entirely clear for me why this should count as entangled. The state $$|\phi\ra|\phi\ra$$ does count as nonentangled to them, although then the above statement is also false, since there are 2 particles possessing those properties then. Maybe it has to do with the fact that with the latter case or with orthogonal states, measuring $$[\phi]$$ does not disturb the state, while with nonorthogonal states, it does.

It is also not clear to be how a Bell inequality could be violated with such states. G&M state that this is a crucial aspect of entanglement. But i would say BIV involves two distant parties, and if there are two particles separated, then they must be orthogonal because of non-overlapping wave packets in the spatial domain. OK, we could allow for the possibility of two particles at one side, but if we then allow this also for orthogonal states, then we get the counterexample I came up with before.

## Counterexample

Just to be complete, I’ll repeat the counterexample here, both for bosons and for fermions:

\begin{align} |\psi\ra &= (|L\ra + |R\ra)\ua(|L\ra + |R\ra)\da \pm (|L\ra + |R\ra)\da(|L\ra + |R\ra)\ua\\ &= (|L\ra|L\ra + |R\ra|R\ra) (\ua\da \pm \da\ua)\\ &+ |L\ra|R\ra (\ua\da \pm \da \ua) \pm |R\ra|L\ra (\da\ua \pm \ua\da) \end{align}

Now, it can be that two particles are found on the left and two to the right. But IF one particle is found at each location, then the spins are maximally entangled (singlet for fermions, with a plus sign for bosons). So a Bell inequality can be violated (one can also write down symmetric operators for this, no need to explicitly talk about postselection).