Using an operational definition we quantify the entanglement, EP, between two parties who share an arbitrary pure state of N indistinguishable particles. We show that EP ≤ EM, where EM is the bipartite entanglement calculated from the mode-occupation representation. Unlike EM, EP is super-additive. For example, EP = 0 for any single-particle state, but the state |1 |1 , where both modes are split between the two parties, has EP = 1/2. We discuss how this relates to quantum correlations between particles, for both fermions and bosons.PACS numbers: 03.65. Ta, Entanglement lies at the heart of quantum mechanics, and is profoundly important in quantum information (QI) [1]. It might be thought that there is nothing new to be said about bipartite entanglement if the shared state |Ψ AB is pure. In ebits, the entanglement is simply [2]Here S(ρ) is the binary von Neumann entropy −Tr [ρ log 2 ρ], and (since we will use unnormalized kets)However, in the context of indistinguishable particles, a little consideration reveals a less than clear situation, which has been the subject of recent controversy [3,4,5,6,7]. Consider, for example, a single particle in an equal superposition of being with Alice and with Bob. In the mode-occupation, or Fock, representation, the state isHere we are following the conventions of writing Alice's occupation number(s) followed by Bob's, separated by a comma, and of omitting any modes that are unoccupied. On the face of it, this is an entangled state with one ebit, and such a state has been argued to show nonlocality of a single photon [8,9]. However, the particle's wavefunction, in the co-ordinate representation, is of the formwhere the subscripts indicate where the wavepackets are localized in co-ordinate (x) space. In this representation, the above entanglement is not apparent, and indeed it has been argued that nonlocality cannot be a single-particle effect [10] (although see Ref.[11]). As a second example, consider a two-particle state where Alice has one particle and Bob the other. In the mode-occupation picture, the state is |1, 1 , which appears unentangled. But since these are identical particles, the wavefunction must be symmetrized asfor bosons and fermions respectively. This has the appearance of an entangled state. Finally, consider another two-particle state, but this time where the two particles are prepared and shared as in the first example, but in different modes. In the Fock representation, this state is entangled:(|0, 1 + |1, 0 )(|0, 1 + |1, 0 ) = |00, 11 + |01, 10 + |10, 01 + |11, 00 .The corresponding wavefunctionalso has the appearance of an entangled state. In this Letter we give an operational definition of entanglement between two parties who share an arbitrary pure state of indistinguishable particles. Applying this to the three states introduced above yields an entanglement (in ebits) of 0, 0, and 1/2 respectively.To justify these (non-obvious) answers, we proceed as follows. First we define precisely what we mean when we use the term particle. Next we review two pre...