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To provide context for my question, I'll simplify it to consider three symmetric particles moving in one dimension. In my actual case, I'm dealing with $N$ generic particles in three dimensions.

Let $| \psi \rangle$ represent a state composed of three bosonic particles moving in one dimension, which can be expressed as follows: $$ | \psi \rangle = \int dx dy dz \ \phi(x,y,z) \ | x, y, z \rangle, $$ where $\phi(x,y,z)$ is the wave function of the state and $$ | x, y, z \rangle = \hat{a}^\dagger(x) \hat{a}^\dagger(y) \hat{a}^\dagger(z) \ |0 \rangle, $$ with the creation (or annihilation) operator given by the following commutation relations: $$ [\hat{a}(x), \hat{a}^\dagger(y)] = \delta (x-y), \\ [\hat{a}^\dagger(x), \hat{a}^\dagger(y)] = [\hat{a}(x), \hat{a}(y)] = 0. $$

My question is the following:

Using the commutation relations, the orthogonal relation of the basis $| x, y, z \rangle$, should be given by: $$ \langle x', y', z' | x, y, z \rangle = \delta (x-x') \delta (y-y') \delta (z-z') + \rm{permutations}. $$ However, I have seen in some papers the orthogonal relation of the Fock state given by $$ \langle x', y', z' | x, y, z \rangle = \delta (x-x') \delta (y-y') \delta (z-z'), $$ that is, without the permutations. I suspect that this simplification may be justified when the particles are distinguishable and independent from each other (in which case, the wave function can be factorized as $\phi(x,y,z) = \phi_1(x) \phi_2(y) \phi_3(z)$). However, I haven't been able to find any references or notes that specifically address this point.

Could someone please provide me with relevant bibliography or a better clarification on this matter?

Qmechanic
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aruera
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  • "However, I have seen in some papers" - Please give references. This issue depends on some conventions. But in general there should appear permutation terms (however, one often wants to construct a basis set, so one has to take care that the vectors of this set are linearly independent, and then there are no permutation terms left)-- any good book on second quantization should explain this properly. Which one do you follow? – Tobias Fünke Oct 11 '23 at 11:48
  • @TobiasFünke Thanks for the answer. I think citing the papers will not help because this assumption is implicit and is never written explicitly in the papers. I don't follow any second quantization book. Can you give me an example of a good book , please? – aruera Oct 11 '23 at 12:07
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    Have a look here. There are also plenty of free online sources, such as lecture notes. – Tobias Fünke Oct 11 '23 at 12:10
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    @aruera the significance of citing papers here is that the same, or very similar notation is often used for slightly (or indeed wildly) different things. It is important to check that the cases omitting those don't simply mean something else. You say "it is never written explicitly" but it may be implicit somewhere you didn't think to look – By Symmetry Oct 11 '23 at 12:55
  • @TobiasFünke I'm not sure there should be permutation terms. Second quantization should take care of this under the hood. We explicitly have three bosonic systems, so the inner product really is $\langle x^\prime|x\rangle \times\langle y^\prime|y\rangle \times\langle z^\prime|z\rangle$; it doesn't make sense to take the ket from one boson and the bra from another – Quantum Mechanic Oct 11 '23 at 13:39
  • @QuantumMechanic No. The inner product of e.g. two such symmetric states is a permanent. Likewise, the inner product of two such anti-symmetrized states is a determinant. I don't know how to understand your last sentence; we do consider indistinguishable particles, where the single-particle states are from a (single-particle) Hilbert space $\mathfrak h$. – Tobias Fünke Oct 11 '23 at 13:48
  • The paper in question is this one in, e.g., equation (28). The source of my question is because the expectation value of the density operator in those coordinates is written as (for 2 particles to make it easier) $\int_{r_1, r_2} \langle b_1-r_1/2, b_2-r_2/2| a^\dagger(x) a(x) |b_1+r_1/2, b_2+r_2/2 \rangle = \delta(x - b_1) + \delta(x - b_2)$. This only makes sense for me if you don't have into account the permutation terms since you should also have a term $\delta(x-b_1-b_2)$. – aruera Oct 11 '23 at 14:23
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    Due to the lack of a better reference, cf. this theorem 3.1. See also eq. 3.17 and 18. ---this summarizes my first comment. As I said above, you should really take a text book and work through the algebra. And as already mentioned, you should make sure that two sources use the same conventions. – Tobias Fünke Oct 11 '23 at 14:31
  • @TobiasFünke oh I see - I thought we had three bosonic modes, but it's one mode with three bosons – Quantum Mechanic Oct 11 '23 at 17:01

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