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 nonvanishing 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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 Fall '10
 EugeneDemler
 Physics, BoseEinstein condensation, Bogoliubov Theory, Bogoliubov quasiparticles

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