Abstract
We calculate zero temperature Green’s function, the density–density correlations and expectation values of a one-dimensional quantum particle which interacts with a Fermi-sea via a δ-potential. The eigenfunctions of the Bethe-Ansatz solvable model can be expressed as a determinant. This allows us to obtain a compact expression for the Green’s function of the extra particle. In the hardcore limit the resulting expression can be analyzed further using Painlevé V transcendents. It is found that depending on the extra particles momentum its Green’s function undergoes a transition of that for hardcore Bosons to that of free Fermions.
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Notes
There is a typo in Eq. (1.29c) of Ref. [41]. The square bracket has to read \([\theta+\frac{A^{2}}{4}-n^{2}-1]\).
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Acknowledgements
We acknowledge fruitful discussion with B. Gutkin, A.M. Lunde and V. Osipov. For helpful comments we thank V. Leiss. CH acknowledges financial support by Studienstifutung des deutschen Volkes. HK acknowledges financial support by the CSIC through the JAE program.
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Appendix: Representation as Töplitz Determinant
Appendix: Representation as Töplitz Determinant
For finite interaction strength the determinant in Eq. (40) representing the Green’s function is not a Töplitz determinant due to the non constant diagonal entries. However, in the limit c→∞ the Green’s function has a representation as Töplitz determinant. To reveal it we write the quantities g jl in Eq. (42) as
Here we use the notations
and furthermore assume that the quantum number n j are given by the set in Eq. (15). Now the diagonal entries are constant, i.e. independent of k j . For the linear combination of the quantities g nm as it appears in Eq. (41) this yields
where we have defined
The full Green’s function (see Eq. (41)) then reads
Employing the normalization condition G(0)=1/L this evaluates to
Correspondingly the interaction part G I(t) reads
For λ=0 Eq. (96) is equivalent to the representation of the Green’s function for hardcore Bosons as Töplitz-determinant [27, 30]. If on the other hand λ→∞ we obtain from Eq. (96)
Expanding the determinant in the equation above yields the following recursion
With help of the relation 2cos(x)sin(x)=sin(x−y)+sin(x+y) we see that its solution is given by
Consequently Eq. (98) acquires the form
Analogously one obtains for the interaction part
Equations (101) and (102) correspond to the free Fermion result.
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Recher, C., Kohler, H. From Hardcore Bosons to Free Fermions with Painlevé V. J Stat Phys 147, 542–564 (2012). https://doi.org/10.1007/s10955-012-0482-1
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DOI: https://doi.org/10.1007/s10955-012-0482-1