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I have asked the question in math.stackexchange, but perhaps it should be more relevant here. Hence I am re-posting it with necessary reediting.

Let $\mathcal{H}=\mathbb{C}^n$ be a Hilbert space. A state $\rho\in\mathcal{B(H)}$ is a positive semi-definite operator with unit trace. $\rho\in \mathcal{B(H)}$, where $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2=\mathbb{C}^{n_1}\otimes\mathbb{C}^{n_2}$, is called entangled, if it can not be written as convex sum of one dimensional projections - like $P_x\otimes P_y$, where $|x\rangle\in\mathcal{H}_1$ and $|y\rangle\in\mathcal{H}_2$.

In a similar spirit can we define entanglement in the symmetric space $\mathcal{H}\bigvee\mathcal{H}$ and antisymmetric space $\mathcal{H}\bigwedge\mathcal{H}$?

As you have already understood, I am looking for entanglement in the indistinguishable particles. Hence the above definition for the entanglement of distinguishable particles does not work. A quick googling gave me a few papers, which refer to different definitions (for multipartite systems, and some are based on certain entropic conditions). Hence I want to know whether there is mathematical definitions for entanglement in such systems which is closest to the spirit of the definition mentioned above (for distinguishable particle). Advanced thanks for any help in this direction.

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What properties do you want entangled/unentangled systems of bosons or fermions to have? – Peter Shor Aug 30 '12 at 5:25
@PeterShor It may sound stupid, but actually I did not consider any particular property of bosons and fermions. I wanted to see the problem of entanglement witness (for mixed states) for the indistinguishable particles and construct an approach by using positive but not completely positive maps. – RSG Aug 31 '12 at 5:14

1 Answer 1

up vote 2 down vote accepted

For symmetric or antisymmetric tensor product, the most useful definition of an unentangled state is a state of the form $$ |\psi\rangle = \sum_{p} (-1)^{2J\cdot |p|} \prod^\otimes_i |\psi_{p(i)}\rangle $$ which is just a convoluted symbolic expression for the totally symmetrized (for bosons) or totally antisymmetrized (for fermions) tensor product. The sum goes over all permutations $i\mapsto p(i)$. The power of $(-1)$ is $1$ for bosons i.e. $j\in {\mathbb Z}$ while it is $(-1)^{|p|}$, i.e. the sign of the permutation, for fermions with $j\in {\mathbb Z}+1/2$. The product is a tensor product.

All states that can't be written in this form are entangled.

If we kept the usual definition of unentangled states as "strict tensor products", there would be too few unentangled states because most multi-boson states and almost all multi-fermion states refuse to be strict tensor products.

So the definition above makes the form of unentangled states more general. However, it may be too general for various purposes. In general, the tensor factors $|\psi_{p(i)}\rangle$ should correspond to states where objects are localized in different regions of space, otherwise the state that is unentangled according to the definition above could be entangled "morally".

In particular, in quantum field theory, unentangled states may be obtained as $$ A^\dagger B^\dagger C^\dagger \dots |0\rangle $$ where $A^\dagger$ and others are polynomials in creation operators but $A^\dagger,B^\dagger,C^\dagger$ are only made of creation operators acting on disjoint, non-overlapping regions. The actual wave function of the state will have to be symmetrized or antisymmetrized so it won't be a strict tensor product but the state of the form above will still enjoy many features of unentangled states.

The discussion about the non-overlapping regions may get important and subtle because, for example, the simple singlet state of two spins, $|up\rangle |down\rangle - |down\rangle |up\rangle$, is entangled according to the normal definition but it could end up being unentangled because it's an antisymmetrized tensor product. However, it usually only makes sense to call such a state unentangled if the up-spinning and down-spinning electron are also located at different positions. If they share the same location, the strict singlet state should be called entangled in all conventions.

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Lubos: why should this be called an unentangled state? From general principles, you would expect a thermal state to be unentangled. According to your definition, it's not. – Peter Shor Aug 30 '12 at 20:59
Dear @Peter Shor, I only discussed pure states. Thermal states are mixed states and this answer of mine hasn't attempted to discuss which of those are entangled. It could actually be easier and more natural to define entangled states - even in the presence of identical particles - for mixed states. The cleanest general definition would be that unentangled are those whose degrees of freedom may be divided to regions and all the probabilities $P(A=A_i,B=B_i)$ of combined properties referring to regions a,b factorize. – Luboš Motl Aug 31 '12 at 5:47

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