A Bosonization

Demystifier

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Summary
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: "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?
 

A. Neumaier

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It must be understood in a weak sense (imposing proper boundary conditions), together with renormalization. The latter is in this case (1+1D) just done by normal ordering the exponential.
 
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Demystifier

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It must be understood in a weak sense (imposing proper boundary conditions), together with renormalization. The latter is in this case (1+1D) just done by normal ordering.
If we studied bosonization on a lattice, so that the number of degrees of freedom was finite, could it be an operator identity in this case?
 

A. Neumaier

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If we studied bosonization on a lattice, so that the number of degrees of freedom was finite, could it be an operator identity in this case?
If the mathematical arguments for the bosonization still go through on the lattice, yes. You'd need to check the derivation of the commutation rules.
 

Demystifier

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If the mathematical arguments for the bosonization still go through on the lattice, yes.
So, assuming that it is so, the Shankar's claim would not longer be true?
 

A. Neumaier

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So, assuming that it is so, the Shankar's claim would not longer be true?
The point of Shankar's argument is that one cannot sensibly exponentiate distributions, which quantum fields on a continuum are; the formula you quoted is strictly speaking only mnemonics for a complicated process.

But if a field is defined only at a finite number of points (e.g., on a bounded lattice) , the fields are functions, not distributions. Then there are no such problems, hence no associated claims.
 

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