Problem Set 4

\[ \nonumber \newcommand{\cN}{\mathcal{N}} \newcommand{\br}{\mathbf{r}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bk}{\mathbf{k}} \newcommand{\bq}{\mathbf{q}} \newcommand{\bv}{\mathbf{v}} \newcommand{\pop}{\psi^{\vphantom{\dagger}}} \newcommand{\pdop}{\psi^\dagger} \newcommand{\Pop}{\Psi^{\vphantom{\dagger}}} \newcommand{\Pdop}{\Psi^\dagger} \newcommand{\Phop}{\Phi^{\vphantom{\dagger}}} \newcommand{\Phdop}{\Phi^\dagger} \newcommand{\phop}{\phi^{\vphantom{\dagger}}} \newcommand{\phdop}{\phi^\dagger} \newcommand{\aop}{a^{\vphantom{\dagger}}} \newcommand{\adop}{a^\dagger} \newcommand{\bop}{b^{\vphantom{\dagger}}} \newcommand{\bdop}{b^\dagger} \newcommand{\cop}{c^{\vphantom{\dagger}}} \newcommand{\cdop}{c^\dagger} \newcommand{\Nop}{\mathsf{N}^{\vphantom{\dagger}}} \newcommand{\bra}[1]{\langle{#1}\rvert} \newcommand{\ket}[1]{\lvert{#1}\rangle} \newcommand{\inner}[2]{\langle{#1}\rvert #2 \rangle} \newcommand{\braket}[3]{\langle{#1}\rvert #2 \lvert #3 \rangle} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} \newcommand{\abs}[1]{\lvert{#1}\rvert} \newcommand{\brN}{\br_1, \ldots, \br_N} \newcommand{\xN}{x_1, \ldots, x_N} \newcommand{\zN}{z_1, \ldots, z_N} \]

Anomalous Commutators for 1D Fermions

To describe the physics at low energies, it is convenient to write the fermion field operator in one dimension as

\[ \pop(x) = e^{ik_\text{F}x}\pop_R(x)+e^{-ik_\text{F}x}\pop_L(x), \]

where the modes contributing to \(\pop_{L/R}(x)\) are restricted to a momentum shell \([-\Lambda,\Lambda]\) around the Fermi points, with \(\Lambda\ll k_\text{F}\). Thus

\[ \pop_R(x)=\sum_{k=-\Lambda}^\Lambda \aop_{k+k_\text{F}}e^{ikx}. \]

In this way we hope to affect a clean separation of low energy excitations into right and left movers. This has some slightly surprising consequences, however.

We define the density operators for right movers by

\[ \rho^R_q = \sum_{p=-\Lambda+\abs{q}/2}^{\Lambda-\abs{q}/2}\adop_{p-q/2+k_\text{F}}\aop_{p+q/2+k_\text{F}}, \label{rhodef} \]

where we have chosen the labels so that the hermiticity condition \(\left(\rho^R_q\right)^\dagger=\rho^R_{-q}\) is respected.

  1. Show that as a result

    \[ \left[\rho^R_{q},\rho^R_{-q}\right] = \frac{qL}{2\pi},\quad \abs{q}<2\Lambda \label{Anom} \]

    leading to the real space commutator

    \[ \left[\rho_R(x),\rho_R(x')\right] =-\frac{i}{2\pi}\delta'(x-x'). \]

  2. Show that the left moving commutator has the opposite sign, so the total density \(\rho^R(x)+\rho^L(x)\) still commutes at different points.

  3. \(\eqref{Anom}\) tells us that \(\rho^R_q\) can be written

    \[ \begin{align} \rho^R_q &= \sqrt{\frac{qL}{2\pi}}r_q, \\ \rho^R_{-q} &= \sqrt{\frac{qL}{2\pi}}r^\dagger_q, \quad q>0, \end{align} \]

    where \(\left[r_q,r_{q'}^\dagger\right]=\delta_{q,q'}\). From the definition \(\eqref{rhodef}\) we see that \(\rho_q^R\) annihilates the ground state, while \(\rho^R_{-q}\) creates an excitation with momentum \(q\).

  4. Show that the structure factor of the density \(\rho_q = \rho^R_q+\rho^L_q\) saturates the Onsager bound discussed in Problem Set 3.

Bosonized Form of the Fermion Operator

Using the density operator in the previous question, define

\[ \begin{align} \phi^R(x) = 2\pi\int^x \rho^R(x')dx' = -i\sum_{q>0}\sqrt{\frac{2\pi}{qL}}\left[r_q e^{iqx}-r^\dagger_q e^{-iqx}\right]. \end{align} \]

Show that \(e^{\pm i\phi^R(x)}\) anticommute at different positions, and further that

\[ \left\{e^{i\phi^R(x)},e^{-i\phi^R(x')}\right\}\propto \delta(x-x'). \]

The overall normalization is cut-off dependent.

From the Anderson Model to the Kondo Model

One route to understanding the origin of the Kondo coupling is to start from the Anderson impurity model.

\[ \begin{align} H = \sum_{\bk,s}\xi(\bk)\adop_{\bk,s}\aop_{\bk,s}+\overbrace{\sum_s\epsilon_b N_{b,s} + U N_{b,\uparrow}N_{b,\downarrow}}^{\equiv H_b} + \overbrace{\frac{t}{\sqrt{V}}\sum_{\bk,s}\left[\adop_{\bk,s}\bop_s+\text{h.c.}\right]}^{\equiv H_t} \end{align} \]

here the \(\bop\) fermion describes the impurity site: \(N_{b,s}=\bdop_{s}\bop_s\).

For \(t=0\) and \(\epsilon_b=-U/2\) the singly occupied, doubly degenerate impurity state has energy \(-U/2\), with the empty and double occupied states having larger energy \(0\) (this choice is made for simplicity). We can then ask for the effective Hamiltonian describing the singly occupied site. We’ve seen one way to do this in Lecture 7. A second way is to use the Schrieffer–Wolff transformation {%cite Schrieffer:1966aa %}. The idea is to perform a unitary transformation on the Hamiltonian that removes the coupling in the last term to lowest order.

We write

\[ H\to H'\equiv e^S H e^{-S}, \]

where \(S\) is antihermitian.

  1. Show that, if we ignore the kinetic energy, \(S\) must be chosen so that

    \[ \left[S, H_b\right] = - H_t. \]

  2. Show that the transformed Hamiltonian \(H'\) contains the term

    \[ H^{(2)} = \frac{1}{2}\left[S,H_t\right] \]

    of order \(t^2\).

  3. Try \(S\) of the form

    \[ S = \sum_{k,s} f(N_{b,\bar s})\adop_{\bk,s}\bop_s - \text{h.c.}, \]

    where \(\bar\uparrow=\downarrow\), \(\bar\downarrow=\uparrow\). Find the form of \(f(N)\) and compute \(H^{(2)}\), retaining only those parts that keep the occupancy of the impurity fixed. You should be able to write \(H^{(2)}\) as a Kondo Hamiltonian and identify \(J\). What sign does it have?

    Hint: \(\bdop_s f(N_b) = f(N_b-1)\bdop_s\), \(\bop_s f(N_b) = f(N_b+1)\bop_s\)

Spin-Flip Mediated Inelastic Scattering

{% cite Kaminski:2001aa %}

Consider the scattering of two fermions in the Kondo problem

\[ \ket{\bk_1,s_1,\bk_2,s_2} = \adop_{\bk_1,s_1}\adop_{\bk_2,s_2}\ket{\text{FS}} \]

Find at second order in \(J\) the amplitude for the process

\[ \bk'_1,s'_1,\bk'_2,s'_2\longrightarrow \bk_1,s_1,\bk_2,s_2, \]

and show that the total scattering rate, averaged over initial spin states of the fermions and the impurity spin \(S\), is

\[ \begin{align} \Gamma(\epsilon_1',\epsilon_2',\epsilon_1,\epsilon_2)&=\frac{\pi}{4}S(S+1)\left(\frac{J}{V}\right)^4\\ &\times\left[\frac{1}{(\epsilon'_1-\epsilon_1)^2}+\frac{1}{(\epsilon'_1-\epsilon_2)^2}\right]\delta(\epsilon_1+\epsilon_2-\epsilon_1'-\epsilon_2'). \end{align} \]

Making sense of Golden Rule calculations is always tricky. To convert this into a total rate at which a particle scatters from state \(1\) requires us to multiply by a factor

\[ N_\text{imp} n_\text{F}(\epsilon_2)\left[1-n_\text{F}(\epsilon_3)\right]\left[1-n_\text{F}(\epsilon_4)\right]\nu^3 V^3 d\xi_2 d\xi_3 d\xi_4, \]

where \(\nu\) is the density of states per spin per unit volume. \(n_\text{F}(\epsilon)\) is the Fermi distribution, as states \(2\) must be occupied, while \(3\) and \(4\) must be empty. This rate is volume independent, as it should be. Note that if we wanted the total rate for all particles in state 1 we’d need another \(n_\text{F}(\epsilon_1)\nu Vd\xi_1\), which would be proportional to volume.

Kondo Effect Without Spin

{% cite Matveev:1990aa %}

A model for (spinless) electrons hopping on and off a metal grain or quantum dot (\(a\) fermions) to a lead (\(b\) fermions) is

\[ \begin{align} H &= \sum_{\bk} \xi(\bk)\left[\adop_\bk\aop_\bk+\bdop_\bk\bop_\bk \right] + \frac{Q^2}{2C}+\varphi Q\\ &\quad+\overbrace{\frac{1}{\sqrt{V_aV_b}}\sum_{\bk,\bk'} \left[t\,\adop_\bk\bop_{\bk'}+\text{h.c.}\right]}^{\equiv H_t}. \end{align} \]

Here \(C\) is the capacitance, and \(Q\) is the charge of the dot

\[ Q = e\sum_\bk \left[\adop_\bk\aop_\bk - \theta(-\xi(\bk))\right], \]

defined so that the dot is neutral when filled to \(\xi(\bk)=0\). \(\varphi\) is a gate voltage. In the region

\[ -\frac{e}{2C}< \varphi < \frac{e}{2C}, \]

the ground state of the decoupled system (\(t=0\)) has \(Q=0\).

  1. Show that at second order in the hopping, the ground state expectation of the charge on the dot is

    \[ \langle Q\rangle \sim eg \log\left(\frac{e/2C-\varphi}{e/2C+\varphi}\right), \]

    where \(g \equiv \abs{t}^2\nu_a(0)\nu_b(0)\).

  2. What has this got to do with the Kondo effect? Try to establish a dictionary between the two phenomena.

Ground State of the Attractive Lieb–Liniger Model

For attractive interactions \(c<0\) the ground state of \(N\) bosons on the infinite line is

\[ \Psi(x_1,\ldots,x_N) = \prod_{j<k}\exp\left(-\frac{\abs{c}}{2}\abs{x_j-x_k}\right),\quad x_1<x_2<\cdots <x_N \]

Show that this is a Bethe state, find the Bethe roots, and calculate the energy. Compare with the prediction of the Gross–Pitaevskii equation found in Problem Set 2.

Density Correlations in Lieb–Liniger

If \(E_0(c)\) is the ground state energy of the Lieb–Liniger model as a function of \(c\), show that

\[ \frac{dE_0}{dc} = \bra{\Psi_0}\sum_{j<k}\delta(x_j-x_k)\ket{\Psi_0}, \]

where \(\ket{\Psi_0}\) is the ground state. By solving the Bethe equations numerically (see the IPython notebook), find how the RHS varies with \(c\), and interpret your results.


{% bibliography –cited %}