On the Exact Solution for a Luttinger Liquid with Repulsion and a Single Point Impurity

Abstract

The expression for the conductance of a 1D channel, which has been obtained using the well-known exact solution, is analyzed. It is shown that in the case of strong electron—electron interaction, the slowest (linear in frequency) asymptotics of the conductance is determined by the behavior of electron—electron interaction in the region of transition from 1D to 3D motion realized near the impurity.

REDUCTION OF THE HAMILTONIAN TO DIAGONAL FORM

1.1 A. “Schrödinger Equation” for Rotation Matrix

To calculate correlator $\mathcal{S}$(ω), we must pass from the Schrödinger to the Heisenberg representation. This can easily be done for free fermions, for which the differences between one representation from the other is reduced to the emergence of factor exp (–i${{\epsilon }_{n}}$t). For this purpose, expression (11) must be diagonalized. It should be noted that variable ζ (now, Φ_{n = 0}) appearing in the Hamiltonian describes the variation (and not total number) of chiral pairs in the ground state. It follows hence that the rotation matrix is determined only by the scattering process (i.e., transitions between states Φ_n ↔ Φ_n'; n, n' ≠ 0, and Φ_n ↔ Φ₀. Therefore, the matrix diagonalizing expression (11) has form

$$\begin{gathered} \lambda ({{\epsilon }_{n}}) = \frac{1}{L}\sum\limits_{m \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{m}})\Phi ({{p}_{m}}) + Z({{\epsilon }_{n}}){{\Phi }_{0}}} , \\ {{\lambda }_{0}} = \frac{1}{L}\sum\limits_{m \ne 0}^{} {T({{p}_{m}})\Phi ({{p}_{m}}).} \\ \end{gathered} $$

(19)

(The arguments of the elements of this matrix can be written as ${{\epsilon }_{n}}$, p_m for not to confuse the indices of rotated and nonrotated fields; in this notation, ${{\epsilon }_{n}}$ and ${{{v}}_{c}}$p_m are the emerges of “rotated” and “nonrotated” fields, respectively.)

The inverse matrix of the transition can be written as

$$\begin{gathered} \Phi ({{p}_{{{{m}_{1}}}}}) = \frac{1}{L}\sum\limits_{{{n}_{1}} \ne 0}^{} {{{S}^{{ - 1}}}({{\epsilon }_{{{{n}_{1}}}}},{{p}_{{{{m}_{1}}}}})\lambda ({{\epsilon }_{{{{n}_{1}}}}}) + {{Z}^{{ - 1}}}({{p}_{{{{m}_{1}}}}})} {{\lambda }_{0}}, \\ {{\Phi }_{0}} = \frac{1}{L}\sum\limits_{{{n}_{1}} \ne 0}^{} {{{T}^{{ - 1}}}({{\epsilon }_{{{{n}_{1}}}}})\lambda ({{\epsilon }_{{{{n}_{1}}}}}).} \\ \end{gathered} $$

(20)

To obtain expressions for the elements of the inverse matrix, we substitute Eqs. (20) into (19):

$$\begin{gathered} \lambda ({{\epsilon }_{n}}) = \frac{1}{L}\sum\limits_{m \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{m}})} \\ \times \,\left[ {\frac{1}{L}\sum\limits_{{{n}_{1}} \ne 0}^{} {{{S}^{{ - 1}}}({{\epsilon }_{{{{n}_{1}}}}},{{p}_{m}})\lambda ({{\epsilon }_{{{{n}_{1}}}}}) + {{Z}^{{ - 1}}}({{\epsilon }_{n}}){{\lambda }_{0}}} } \right] \\ \, + Z({{\epsilon }_{n}})\frac{1}{L}\sum\limits_{{{n}_{1}} \ne 0}^{} {{{T}^{{ - 1}}}({{\epsilon }_{{{{n}_{1}}}}})\lambda ({{\epsilon }_{{{{n}_{1}}}}}).} \\ \end{gathered} $$

By varying this expression with respect to λ_n, we obtain

$$L = \frac{1}{L}\sum\limits_{m \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{m}}){{S}^{{ - 1}}}({{\epsilon }_{n}},{{p}_{m}}) + Z({{\epsilon }_{n}}){{T}^{{ - 1}}}({{\epsilon }_{n}}).} $$

On the other hand,

$$\{ \lambda ({{\epsilon }_{n}}),\lambda ({{\epsilon }_{m}})\} = L{{\delta }_{{n, - m}}}.$$

Substituting Eq. (19) into this expression, we obtain

$$L = \frac{1}{L}\sum\limits_{m \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{m}})S({{\epsilon }_{{ - n}}},{{p}_{{ - m}}}) + Z({{\epsilon }_{{ - n}}})Z({{\epsilon }_{n}}).} $$

(21)

Comparing these two expressions, we see that

$${{S}^{{ - 1}}}({{\epsilon }_{n}},{{p}_{m}}) = S({{\epsilon }_{{ - n}}},{{p}_{{ - m}}}),\quad {{T}^{{ - 1}}}({{\epsilon }_{n}}) = Z({{\epsilon }_{{ - n}}}).$$

(22)

To obtain the equations for the rotation matrix, we must compare commutator

$${{[\lambda ({{\epsilon }_{n}}),{{\mathcal{H}}_{\lambda }}]}_{ - }} = {{\epsilon }_{n}}\lambda ({{\epsilon }_{n}})$$

with the same commutator written in terms of “nonrotated” fields Φ. Therefore, we must calculate commutator

$$\begin{gathered} \frac{{{{{v}}_{c}}}}{{2L}}\left[ {\sum\limits_{n \ne 0}^{} {{{p}_{n}}{{{\hat {\Phi }}}_{{ - n}}}{{{\hat {\Phi }}}_{n}}} + 2i{{\gamma }_{2}}} \right.{{\Phi }_{0}}\frac{1}{L}\sum\limits_{n \ne 0}^{} {{{{\hat {\Phi }}}_{n}}} , \\ {{\left. {\frac{1}{L}\sum\limits_{m \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{m}})\Phi ({{p}_{m}}) + Z({{\epsilon }_{n}}){{\Phi }_{0}}} } \right]}_{ - }}. \\ \end{gathered} $$

The term containing the kinetic energy gives

$$ - \frac{{{{{v}}_{c}}}}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}}){{p}_{{{{m}_{1}}}}}\Phi ({{p}_{{{{m}_{1}}}}}).} $$

It remains for us to consider commutator

$$\begin{gathered} \left[ {2i{{\gamma }_{2}}{{\Phi }_{0}}\frac{1}{L}\sum\limits_{n \ne 0}^{} {{{{\hat {\Phi }}}_{n}},} } \right. \\ {{\left. {Z({{\epsilon }_{n}}){{\Phi }_{0}} + \frac{1}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}})\Phi ({{p}_{{{{m}_{1}}}}})} } \right]}_{ - }}. \\ \end{gathered} $$

The first term gives

$$ - i{{\gamma }_{2}}Z({{\epsilon }_{n}})\frac{1}{L}\sum\limits_{n \ne 0}^{} {{{{\hat {\Phi }}}_{n}},} $$

while the second term yields

$$2i{{\gamma }_{2}}{{\Phi }_{0}}\frac{1}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}}).} $$

Ultimately, we obtain equality

$$\begin{gathered} - \frac{{{{{v}}_{c}}}}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}}){{p}_{{{{m}_{1}}}}}\Phi ({{p}_{{{{m}_{1}}}}})} \\ + 2i{{\gamma }_{2}}{{\Phi }_{0}}\frac{1}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}}) - i{{\gamma }_{2}}Z({{\epsilon }_{n}})\frac{1}{L}\sum\limits_{n \ne 0}^{} {{{{\hat {\Phi }}}_{n}}} } \\ = - {{\epsilon }_{n}}\left( {\frac{1}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}}){{p}_{{m1}}}\Phi ({{p}_{{{{m}_{1}}}}}) + Z({{\epsilon }_{n}}){{\Phi }_{0}}} } \right). \\ \end{gathered} $$

This relation must hold for any fields Φ; therefore,

$$\begin{gathered} ({{\epsilon }_{n}} - {{{v}}_{c}}{{p}_{m}})S({{\epsilon }_{n}},{{p}_{m}}) = i{{\gamma }_{2}}Z({{\epsilon }_{n}}), \\ {{\epsilon }_{n}}Z({{\epsilon }_{n}}) = - 2i{{\gamma }_{2}}\frac{1}{L}\sum\limits_{{{m}_{1}} \ne 0}^{} {S({{\epsilon }_{n}},{{p}_{{{{m}_{1}}}}})} . \\ \end{gathered} $$

(23)

Thus, we have obtained the following equation determining the spectrum:

$${{\epsilon }_{n}} = 2\gamma _{2}^{2}\frac{1}{L}\sum\limits_{m \ne 0}^{} {\frac{1}{{{{\epsilon }_{n}} - {{{v}}_{c}}{{p}_{m}}}}} .$$

(24)

Passing to dimensionless energy y_n = L${{\epsilon }_{n}}$/2π${{{v}}_{c}}$, we get

$${{y}_{n}} = {{d}^{2}}\left[ {\pi {\text{cot}}(\pi {{y}_{n}}) - \frac{1}{{{{y}_{n}}}}} \right],\quad {{d}^{2}}\, = \,2L{{\left( {\frac{{{{\gamma }_{2}}}}{{2\pi {{{v}}_{c}}}}} \right)}^{2}} \gg 1,$$

(25)

or

$$\tan (\pi {{y}_{n}}) = \frac{{\pi {{d}^{2}}{{y}_{n}}}}{{{{d}^{2}} + y_{n}^{2}}}.$$

It should be noted that [γ₂] ~ $\sqrt {\mathcal{M}}$; therefore, the thermodynamic transition in our case indicates the disregard corrections in 1/$\mathcal{M}L$ for $\mathcal{M}L$ $\gg$ 1, but not requires the passage to the limit L → ∞. In the zeroth approximation in this parameter, we seek the solution in form

$${{y}_{n}} = n + \delta {{y}_{k}},\quad n \gg \delta {{y}_{n}}.$$

Then the shift around each point y_n = n satisfies relation

$$\delta {{y}_{n}} = \frac{1}{\pi }\arctan \left( {\frac{{\pi n}}{{1 + {{{(n{\text{/}}d)}}^{2}}}}} \right),$$

where arctan is defined on interval (–π/2, π/2).

Ultimately, the spectrum is given by

$${{\epsilon }_{n}} = \frac{{2\pi {{{v}}_{c}}}}{L}\left( {n + \frac{1}{\pi }\arctan \left( {\frac{{\pi n}}{{1 + {{{(n{\text{/}}d)}}^{2}}}}} \right)} \right)$$

(26)

for n $\gg$ δy_n. However, for n $\gg$ 1, this condition is always satisfied because arctangent is defined on interval (–π/2, π/2), i.e., ~1. It follows hence that the level shift is small and can be ignored. We have the spectrum satisfying periodic boundary conditions: y_n = n. A transition to the thermodynamic limit is performed conventionally, $\mathcal{M}L$ $\gg$ 1, and 2πn/L = k is independent of L. Ultimately, the spectrum of the fields that are not scattered by the impurity is also equal to $\epsilon$(k) = ${{{v}}_{c}}$k.

1.2 B. Calculation of Rotation Matrix

Relations (20) and (22) make it possible to express the Guinier field in terms of the fields λ_n diagonalizing the Hamiltonian, while the normalization condition written in form (21) with account for the first equation of system (23) makes it possible to calculate the required matrix element of rotation matrix Z(${{\epsilon }_{n}}$):

$${{\left( {\frac{{{{\gamma }_{2}}}}{{2\pi {{{v}}_{c}}}}} \right)}^{2}}L\sum\limits_{{{m}_{1}} \ne 0}^{} {\frac{{Z({{\epsilon }_{n}})Z( - {{\epsilon }_{n}})}}{{{{{({{y}_{n}} - {{m}_{1}})}}^{2}}}}} + Z({{y}_{n}})Z( - {{y}_{m}}) = L.$$

(27)

To evaluate the sum on the left-hand side of this expression, we note that

$$\begin{gathered} - {{\partial }_{y}}\sum\limits_{{{m}_{1}} \ne 0}^{} {\frac{1}{{{{y}_{n}} - {{m}_{1}}}}} = - {{\partial }_{y}}\left( {\pi \cot (\pi y) - \frac{1}{y}} \right) \\ = {{\pi }^{2}}\left( {1 + \frac{{1 - {{{\sin }}^{2}}(\pi y)}}{{{{{\sin }}^{2}}(\pi y)}}} \right) - \frac{1}{{{{y}^{2}}}} \\ = {{\pi }^{2}} + {{\pi }^{2}}{{\cot }^{2}}(\pi y) - \frac{1}{{{{y}^{2}}}} = {{\pi }^{2}} + \frac{2}{{{{d}^{2}}}} + \frac{{y_{k}^{2}}}{{{{d}^{4}}}}. \\ \end{gathered} $$

Disregarding corrections in 1/L, we obtain the final expression for the transition of matrix elements to the diagonal form:

$$Z({{\epsilon }_{n}}) = \frac{{2\sqrt 2 i{{\gamma }_{2}}{\text{sgn}}({{\epsilon }_{n}})}}{{\sqrt {{{\mu }^{2}} + \epsilon _{n}^{2}} }},$$

(28)

$$S({{\epsilon }_{n}},{{p}_{m}}) = - \frac{{2\sqrt 2 \gamma _{2}^{2}{\text{sgn}}({{\epsilon }_{n}})}}{{\sqrt {{{\mu }^{2}} + \epsilon _{n}^{2}} ({{\epsilon }_{n}} - {{{v}}_{c}}{{p}_{m}})}},$$

(29)

where μ = $\gamma _{2}^{2}$/${{{v}}_{c}}$ ~ $\mathcal{M}$. This allows us to express the Guinier field in the Heisenberg representation in terms of fields λ(${{\epsilon }_{n}}$) of free quasiparticles:

$$\zeta (t) = - \frac{{4i{{\gamma }_{2}}}}{L}\sum\limits_{n \ne 0}^{} {\frac{{{\text{sgn}}({{\epsilon }_{n}})}}{{\sqrt {{{\mu }^{2}} + \epsilon _{n}^{2}} }}\lambda ({{\epsilon }_{n}})\exp ( - i{{\epsilon }_{n}}t).} $$

(30)