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- Is the bosonization formula an operator identity?

Shankar, in the book "Quantum Field Theory and Condensed Matter", at page 328 writes the famous bosonization formula in the form

$$\psi_{\pm}(x)=\frac{1}{\sqrt{2\pi\alpha}} e^{\pm i \sqrt{4\pi} \phi_{\pm}(x)}$$

and then writes:

I don't understand this statement. ##\psi_{\pm}(x)## satisfies anticommutation relations, so why can't it change the fermion number the way ψ can?

$$\psi_{\pm}(x)=\frac{1}{\sqrt{2\pi\alpha}} e^{\pm i \sqrt{4\pi} \phi_{\pm}(x)}$$

and then writes:

*"This is not an operator identity: no combination of boson operators can change the fermion number the way ψ can."*I don't understand this statement. ##\psi_{\pm}(x)## satisfies anticommutation relations, so why can't it change the fermion number the way ψ can?