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I'm playing a bit with bosonization $ψ→:e^{-φ}:$ and $ψ^*→:e^{φ}:$ in the sense that

$$ \Bigg\langle 0_\mathrm{F} \Bigg|∏_{i=1}^nψ(z_i)ψ^*(w_i)\Bigg|0_\mathrm{F}\Bigg\rangle = \Bigg\langle 0_\mathrm{B}\Bigg|∏_{i=1}^n:e^{-φ(z_i)}::e^{φ(w_i)}:\Bigg|0_\mathrm{B}\Bigg\rangle $$

where the subscripts refer to fermionic/bosonic vacuum. I would like to know if it there is a way to recover $$ \left\{ψ(z),ψ^*(w) \right\}= δ(z-w) $$

in terms of bosons, something like

$$ \left\{:e^{-φ(z)}::e^{φ(w)}: \right\}= δ(z-w). $$

JamalS
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MaPo
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1 Answers1

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I have a recent discussion about this problem.

Let me assume you already have the expectation value of the equal time anti-commutator \begin{equation} \langle \{ \psi( x ) , \psi^{\dagger}( x' ) \} \rangle = \langle :e^{i \phi(x) }:, :e^{ -i \phi(x' )} :\rangle = i \delta( x - x' ) \end{equation} This can be established for example through analytically continuing the Euclidean boson correlator $\ln | z_1 - z_2|$ with $t = i \epsilon$ trick, or operator formalism.

For a single oscillator we have \begin{equation} :e^A: :e^B: = :e^{A+B}: e^{\langle A B \rangle } \end{equation} generalizing this to the vertex operator \begin{equation} \begin{aligned} \{:e^{i \phi(x) }:, :e^{ -i \phi(x' )} :\} &= (e^{\langle \phi(x) \phi(x') \rangle } + e^{\langle \phi(x') \phi(x) \rangle }):e^{i (\phi(x) - \phi( x' )) }: \\ &= \langle :e^{i \phi(x) }:, :e^{ -i \phi(x' )} :\rangle : e^{i (\phi(x) - \phi( x' )) }:\\ &= i \delta( x - x' ) e^{i (\phi(x) - \phi( x' )) }\\ &= i \delta( x - x' ) \end{aligned} \end{equation} The last line is an unexpected way to reduce an operator to a c-number.

anecdote
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