Orthogonality Catastrophe: More Truly is Different

This note originates from discussion with Jiaqi Cai, Ruihua Fan, Pavel Volkov, Jie Wang, and my recent project on $\textrm{MoTe}_2$. We note that the discovery of AIP's oral history interviews of P. W. Anderson made the writing process quite enjoyable.

Original Formulation: Anderson

Kohn and Majumdar confused Anderson with their paper in 1965 saying that if you turn on a scattering potential of arbitrary strength in the Fermi liquid it wouldn't create singularities at the Fermi surface. This was perplexing, because at some point, a bound state will form.

Let us instead think about the problem in the following way, following Anderson's masterpiece in^[1]. Let us calculate the overlap of the electronic ground state with no potential turned on and the ground state with the potential turned on: this will serve as a test for the validity of perturbation theory.

Before we turn on the interactions, the radial eigenstates are

$$\phi_0^n(r)=N_n\frac{\sin k_n r}{k_n r}$$

in the $\ell=0$ multiplet. Here $k_n=\frac{\pi n}{R}$ and $E_n=\frac{\hbar^2}{2m}k_n^2$.

After we turn on a scattering potential the outgoing scattering states will get a phase shift:

$$\psi_0^n(r)=N_n'\frac{\sin (k_nr-\delta(E_n)[1-r/R])}{k_nr}.$$

Why are these states morally justified? They satisfy two properties: They still vanish at $r=R$, so they satisfy the boundary conditions; in the region s.t. $a\ll r\ll R$ in which $a$ is the scattering length, they essentially have the form $e^{i(k_nr-\delta(E_n))}$.

We can calculate the overlap integral between typical members of states near the Fermi surface, in preparation to calculate the overlap between the Fermi seas:

$$A_{nn'}=\left\langle \phi^n|\psi^n \right\rangle\approx \frac{2\pi N_nN_{n'}}{k_nk_{n'}}\frac{\sin\delta_{n'}}{k_n-k_{n'}+\delta/R}.$$

(So far I haven't found a clean proof of this line – let me know if you have a clean way of proving it.) If we set $N_n=\frac{k_n}{\sqrt{2\pi R}}$ then

$$A_{nn'}=\frac{\sin\delta_{n'}}{\pi(n-n')+\delta}.$$

This matrix is unitary because of the identity

$$\csc^2\theta = \sum_n\frac{1}{(\pi n+\theta)^2}.$$

We can see this in the following way:

$$\sum_{n}\frac{1}{(\theta+\pi n)^2}=\sum_{n\in\mathbb{Z}}\mathrm{Res}\frac{1}{(\theta+z)^2}\cot(z).$$

now "integrate over a large loop around the poles" (this time the contour is nontrivial!) and the whole thing vanishes because of analyticity. However there is one pole that we have missed: the pole at $z=-\theta$, which gives $-\frac{d}{dz}\cot(z)|_{z=-\theta}=\csc(\theta)^2$.

With that, we can now compute the overlap between the Slater determinants:

$$S=|\left\langle \Phi|\Psi \right\rangle|=\det\limits_{E_n,E_{n'}\left.<\right. E_F} A_{nn'}.$$

We should think of this overlap as taking $N$ $\psi$ states and calculating the volume of the parallelepiped projected to the state generated by the occupied $\phi$ states. The volume is of course smaller than the norms of the projected vectors multiplied together:

$$S\leq \prod_{n}(1-\sum_{E_{n'}>E_F}|A_{nn'}|^2)^{1/2}.$$

Instead of the infinite product $S$ we should think about $\log S$ which is an infinite sum. Let $n_F$ be the critical $n$ s.t. if $n>n_F$ then it is unoccupied. We obtain

$$\log S = -\frac{1}{2}\sum_{n\left.<\right. n_F,m>n_F}|A_{mn}|^2=-\frac{1}{2}\sum_{n\left.<\right. n_F,m>n_F}\frac{\sin^2\delta}{(\pi(n-m)+\delta)^2}\sim\frac{\sin^2\delta}{\pi^2}\log M.$$

where $E_M$ is where $\delta$ takes a small and constant value (i.e. around the Fermi surface). Crudely we assume $\log M\sim\log N$ with some proportionality constant. Then, we have the final bound

$$S\leq N^{-\sin^2\delta/\pi^2}.$$

This tells us that if we are at resonance – $\delta\sim\pi/2$ – then the overlap between the two states is strongly reduced. This is especially the case where $N$ is rather large.

Lessons learned:

1. Fock space has a huge dimension. Take two Slater determinants that are made of almost the same single particle states, and the overlap is vastly reduced.

2. From the form of the answer $S\leq N^{-\sin^2\delta/\pi^2}$ one should be made aware that this answer is deeply RG like. It almost hints at us that a phase transition is occuring.

3. Adiabaticity matters. One can imagine turning on the scattering potential super slowly, which is exactly what Kohn and Majumdar did - and although one needs to be careful about what one means by "slowly" in a gapless system, it is not totally crazy that the Fermi surface is still regular.

Alternative derivation that is less model dependent. Consider

$$\log S = -\frac{1}{2}\sum_{n\left.<\right. n_F,m>n_F}|A_{mn}|^2.$$

Consider a (potentially weak) potential $V(\boldsymbol{ r})$. Do a first order perturbation:

$$A_{mn}=\frac{\left\langle m|V|n \right\rangle}{E_n-E_m}.$$

Now let's say $\left\langle m|V|n \right\rangle$ can be approximated by a constant $V$ ($s$-wave scattering?). Then, we can do the integral:

$$\log S = -\frac{V^2}{2}\sum_{n\left.<\right. n_F,m>n_F}\frac{1}{(E_m-E_n)^2}=-\frac{V^2}{2}\int_0^{E_F}d\omega\int_{E_F}^\infty d\omega' \frac{\rho(\omega)\rho(\omega')}{(\omega-\omega')^2}.$$

we again approximate the density of states $\rho(\omega)$ is more or less a constant $\rho_0$. Now perform an integration and take a cutoff that is a power in $N$. This means that $\log S\propto (V\rho)^2\log N$.

Bosonization from Schotte and Schotte

After brewing for a year after the groundbreaking work by Anderson, a classic trilogy emerged from the writings of Nozieres and de Dominicis and friends^[2][3][4]. They studied implications of the orthogonality catastrophe in the context of X-ray absorption. Instead of going through their monolithic calculations, I choose the easier path of Klaus-Dieter Schotte and Ursula Schotte^[5] – they were a couple, by the way – and apply the Tomonaga-Luttinger(-Lieb-Mattis, but that's another story) liquid formalism. We will try to keep our discussion as broad as possible, to which we owe our knowledge to^[6], sections 3.6 and 9.3.

The statement of the problem is the following. Consider hitting your favorite metal with an X-ray photon. By energy conservation, the most common scenario is that a core electron is hit out of the core.

We consider the Green's function of the core electron, created by $d^\dagger$. Basically, let's calculate

$$G(t)=\left\langle d(t)d^\dagger(0) \right\rangle.$$

Why is this related to the orthogonality catastrophe above? Note that this is truly a many-body system. Apart from the core electrons there are also itinerant electrons that make up the Fermi surface. The core electron that absorbs the X-ray leaves a hole behind, creating an attractive potential. This potential renormalizes the Fermi surface, strongly reducing the matrix element – and thus creating singularities at the edge of the absorption spectrum.

Particularly, the Green's function above models the following process for the Fermi liquid: an attractive potential is turned up for $0\left.<\right. t'\left.<\right. t$. Calculate the overlap between the original Fermi liquid and the Fermi liquid after experiencing the attractive potential for time $t$.

We want to highlight the Fourier transform of this Green's function. $G(\omega)$ really represents the spectral function of the core electron, which will correspond directly to the X-ray absorption amplitude.

Now let's calculate $G(t)$. We will calculate it from a purely Fermi liquid point of view. We assume that the potential Hamiltonian is written as

$$V(t)=\int d\boldsymbol{ r} V(\boldsymbol{ r})n(\boldsymbol{ r},t)=\int d\boldsymbol{ q} V(\boldsymbol{ q})\rho(\boldsymbol{ q},t).$$

The time evolution operator can be written as

$$U(t)=T \exp({i\int_0^t dt' V(t')})$$

and thus

$$G(t) = \left\langle U(t) \right\rangle = \exp[\sum F_l(t)]$$

in which $F_l$ is the connected diagram of $l$ interaction terms:

$$F_l(t)=\frac{(-1)^l}{l}\int_0^t dt_1\dots \int_0^t dt_l T\left\langle V(t_1)\dots V(t_l) \right\rangle_{\textrm{connected}}$$

Arguably, this has flavors of the "effective action" expansion, where the log of the partition function is given by the sum of all connected diagrams.

As usual we will perturbatively expand. The first order contribution $F_1$ will give a constant energy shift, due to Fumi's theorem (see Mahan). Already at second order do we get orthogonality catastrophe:

$$\begin{aligned} F_2(t)&=\frac{1}{2}\int d\boldsymbol{ q} \int_0^t dt_1\int_0^t dt_2|V(\boldsymbol{ q})|^2T\left\langle \rho(\boldsymbol{ q},t_1)\rho(\boldsymbol{ q},t_2) \right\rangle\\ &=\int d\boldsymbol{ q}\int_0^\infty d\omega \int_0^{t_2} dt_1\int_0^t dt_2 |V(\boldsymbol{ q})|^2\left\langle \rho(\boldsymbol{ q},\omega)\rho(-\boldsymbol{ q},-\omega) \right\rangle e^{i\omega(t_1-t_2)}\\ &=\int d\boldsymbol{ q}\int_0^\infty d\omega |V(\boldsymbol{ q})|^2\left\langle \rho(\boldsymbol{ q},\omega)\rho(-\boldsymbol{ q},-\omega) \right\rangle\frac{1-e^{-i\omega t}-i\omega t}{\omega^2} \end{aligned}$$

For reasons unknown, the author claimed that only the real part contributes to the term linear in $t$ (a constant energy shift), and only the imaginary part contributes to the interesting correction. Let's see if we can figure it out later...

Now comes the wisdom of the Schotte duo. They figured that they can use bosonization to describe the density operators $\rho(\boldsymbol{ q})$. In the bosonization language one should write

$$b_{\boldsymbol{ k}}=\frac{1}{\sqrt{k}}\rho(\boldsymbol{ k}),\quad b_{\boldsymbol{ k}}^\dagger=\frac{1}{\sqrt{k}}\rho(-\boldsymbol{ k})$$

and

$$[b_{\boldsymbol{ k}},b_{\boldsymbol{ k}'}^\dagger]=\delta_{\boldsymbol{ k}\boldsymbol{ k}'}.$$

We have to make sure that the bosonization procedure is actually well defined. We know it's not, in dimensions greater than $1$, which is why so many people are working on it (for example, Umang's thesis Postmodern Fermi Liquids is a great review). However, from Anderson's exposition above, we know that the relevant scattering is $s$-wave, so the higher-dimensionality dependence is reduced.

We take contact interaction $V(\boldsymbol{ q})=V$. We may thus evaluate the integral

$$\begin{aligned} F_2(t)&=V^2\int d\boldsymbol{ q} \left\langle b_{\boldsymbol{ q}}b^\dagger_{\boldsymbol{ q}} \right\rangle\frac{1-e^{-iq t}}{q}\\ & \propto V^2\int dq \frac{1-e^{-iq t}}{q}\\ & \propto -V^2\log(1+it\Lambda) \end{aligned}$$

in which $\Lambda$ is a cutoff and the last line comes from Mahan's (9.198). We thus read off

$$G(t)\propto(1+it\Lambda)^{-V^2}.$$

Its Fourier transform gives the spectral weight. We can guess that its long time behavior $t^{-V^2}$ gives its low energy behavior at the edge of the spectrum:

$$G(\omega)\propto (\omega-\omega_c)^{V^2-1}.$$

We note that Joaquin Luttinger himself was involved in related topics as well: he pointed out the Cauchy's determinant in Anderson and Yuval's paper^[7], which greatly simplifies some of Nozieres and de Dominicis' derivations. Anderson's interview and his article on Luttinger's biographical memoir has some hilarious accounts on this matter.

"Fermi liquid" and Alien Particles - Just Not Electrons

The calculation above not only applies to electronic Fermi liquids. We can also have composite Fermi liquids occuring at half filled Landau levels^[8]. Composite fermions are not electrons: they are gauge fluxes attached to physical electrons, and they are neutral. They are also responsible for most of the fractional quantum Hall states we know.

Suppose that these pretty exotic particles indeed forms a Fermi liquid. Now suppose you try to tunnel an electron into this system. Note that the electron is charged, whereas these composite fermions are not. Thus the electron is an alien particle that has its own dynamics and is pretty much decoupled from the active degrees of freedom – the gapless composite fermions. It only adds a potential term for the composite fermions. This scenario is reminiscent of the scenario above: a core electron is excited, and it only adds a potential term for the electrons.

In a short paper by Bert and friends^[9] they used exactly the formalism above to calculate the tunneling density of states. They use the fact that that the density-density response function $\chi(\boldsymbol{ q},\omega)$ is dominated by a diffusive mode

$$\chi(\boldsymbol{ q},\omega)=\left\langle \rho(\boldsymbol{ q},\omega)\rho(-\boldsymbol{ q},-\omega) \right\rangle\sim\frac{1}{1-i\omega/\beta q^2}.$$

Substituting into the result above, they find that $G(\tau)\sim\sqrt{\tau}$ and $G(\omega)\sim e^{-\omega_0/\omega}$.

References

[1]	Infrared Catastrophe in Fermi Gases with Local Scattering Potentials (1967). DOI

[2]	Singularities in the X-Ray Absorption and Emission of Metals. I. First-Order Parquet Calculation (1969). DOI

[3]	Singularities in the X-Ray Absorption and Emission of Metals. II. Self-Consistent Treatment of Divergences (1969). DOI

[4]	Singularities in the X-Ray Absorption and Emission of Metals. III. One-Body Theory Exact Solution (1969). DOI

[5]	Tomonaga's Model and the Threshold Singularity of X-Ray Spectra of Metals (1969). DOI

[6]	Many-Particle Physics (2000). DOI

[7]	Exact Results in the Kondo Problem: Equivalence to a Classical One-Dimensional Coulomb Gas (1969). DOI

[8]	Theory of the half-filled Landau level (1993). DOI

[9]	Tunneling into a Two-Dimensional Electron System in a Strong Magnetic Field (1993). DOI