Abstract

In contrast to normal Fermi liquids, where the momentum distribution n(k) has a discontinuous jump at k=${\mathit{k}}_{\mathit{F}}$, in Luttinger liquids a power-law singularity appears in n(k). It was argued that this singularity shows up not only at ${\mathit{k}}_{\mathit{F}}$ but at 3${\mathit{k}}_{\mathit{F}}$,5${\mathit{k}}_{\mathit{F}}$, . . . as well, and the anomaly exponent has been detemined for the one-dimensional Hubbard model using numerical methods or assuming conformal invariance. We present an exact calculation of the anomaly exponent for the Tomonaga-Luttinger model and compare it with the results for the Hubbard model.

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