# Problem Set 4


## 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.

## References

{% bibliography –cited %}

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