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Chemistry as we know it would not exist if electrons were Bosons, because many of the details of the chemical "bonding" of atoms are strongly affected by the requirement that the overall wavefunction of all the electrons involved be anti-symmetric. This anti-symmetrization principle (which is the basis of the "Pauli Exclusion Principle") along with the fact ...

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It seems to be referring to the fact that bosons don't have to obey the Pauli exclusion principle, whereas fermions do. With fermionic electrons, only one can occupy each atomic energy state at a time, so as you add more electrons, they have to form sequential shells, which makes different atoms behave differently chemically. If electrons were bosons, then ...

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Using the rule (3) you can sort the four operators in the order you prefer, by swapping terms. After each swapping, you obtain a piece with four operators, plus an additional one with two operators. And so on. In your case, by swapping the second and the third term, $$a_j a_i a_k^+ a_l^+ = - a_j a_k^+ a_i a_l^+ + \delta_{ki} a_j a_l^+$$ and by swapping ...

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Introducing: $a_ia_k^+=\delta_{ik}-a_k^+a_i$ Replacing 3 in 2 and solving: $\langle K|L\rangle=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+|\rangle=\delta_{ik}\langle|a_ja_l^+|\rangle-\langle|a_ja_k^+a_ia_l^+|\rangle$ $\langle K|L\rangle=\langle|a_ja_ia_k^+a_l^+|\rangle$ $$=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+\rangle$$ $$=\langle ... 2 In the context of ultracold Fermi gases, a BEC-BCS crossover means that by tuning the interaction strength (the s-wave scattering length), one goes from a BEC state to a BCS state without encountering a phase transition (thus the word "crossover"). It is also useful to know that the BEC state is a Bose-Einstein condensate of two-atom molecules, while the ... 0 Regarding question 2, in Weinberg's The Quantum Theory of Fields page 295, the author defines the functional derivative of an arbitrary bosonic functional F[q,p]:$$\frac{\delta F[q,p]}{\delta q}=i[p,F[q,p]]\frac{\delta F[q,p]}{\delta p}=i[F[q,p],q]$$where q,p are a pair of operators that satisfy canonical conmutation relations. If that is indeed ... 0 I hear that analogy too. Spin 0: any rotation left the "object" invariant, like a circle who rotates. Spin 1/2: half rotation to het the initial state of the object, and here we are: any figure of the playing card "has spin 1/2". Spin 1: any non figure card, like who knows, the ace of clubs. One integer rotation to get is as it was initially. Spin 2: no ... 4 Comment to the question (v4): Classically, the Lagrangian for a fermion system reads$$ L ~=~ \int\! d^3x~ i\psi^{\dagger}\dot{\psi}-H.\tag{A} The Legendre transformation from the Lagrangian to the Hamiltonian formalism is tricky for at least three reasons: The traditional Dirac-Bergmann analysis leads to constraints. See e.g. my Phys.SE answers here ...

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