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Let us then consider a state 0 12 if we choose n0 e 2

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Unformatted text preview: ΨN has a well defined number of particles. Let us then consider a state |Ψ0 1/2 If we choose α = N0 = e− α2 2 † eαbp=0 |vac (3.8) we find that Ψ0 |Ψ(r)|Ψ0 = Ψ0 |Ψ† (r)|Ψ0 = ( N0 1/2 ) V (3.9) State (3.8) captures property (3.6) but does it in a more natural way. It has individual expectation values of Ψ(r) and Ψ† (r). One may be concerned by the fact that this state does not have a well defined number of particles, although Hamiltonian (3.1) commutes with the total number of particles N = p b† bp . p And according to the fundamental theorem in quantum mechanics, if some operator commutes with the Hamiltonian, then it can be made diagonal in the basis of energy eigenstates. This ”violation” of the basic theorem of quantum mechanics is the essence of the idea of spontaneous symmetry breaking. In state (3.8) we have a non-vanishing expectation value of the order parameter ΨN |Ψ(r)|ΨN and this automatically means that state |ΨN is not an eigenstate of the conserved operator N . Some justification for using wavefunction (3.8) i...
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