# Friedel Sum Rule and Phase Shifts

When I took undergraduate quantum mechanics, one of the most painful subjects to study was scattering theory, due to the usage of special functions, phase shifts and partial waves. To be honest, the sight of those words still makes me shudder a little.

If you have felt like that at some point, I hope that this post will help alleviate some fear of phase shifts. Phase shifts can be seen in many classical contexts, and I think that it is best to start thinking about them in that setting. Consider the following scenarios: a wave pulse on a rope is incident on a (1) fixed boundary and (2) a movable boundary. See below for a sketch, which was taken from here.

Fixed Boundary Reflection

Movable Boundary Reflection

Notice that in the fixed boundary case, one gets a phase shift of $\pi$, while in the movable boundary case, there is no phase shift. The reason that there is a phase shift of $\pi$ in the former case is that the wave amplitude must be zero at the boundary. Therefore, when the wave first comes in and reflects, the only way to enforce the zero is to have the wave reflect with a $\pi$ phase shift and interfere destructively with the incident pulse, cancelling it out perfectly.

The important thing to note is that for elastic scattering, the case we will be considering in this post, the amplitude of the reflected (or scattered) pulse is the same as the incident pulse. All that has changed is the phase.

Let’s now switch to quantum mechanics. If we consider the same setup, where an incident wave hits an infinitely high wall at $x=0$, we basically get the same result as in the classical case.

Elastic scattering from an infinite barrier

If the incident and scattered wavefunctions are:

$\psi_i = Ae^{ikx}$      and      $\psi_s=Be^{-ikx}$

then $B = -A = e^{i\pi}A$ because, as for the fixed boundary case above, the incident and scattered waves destructively interfere (i.e. $\psi_i(0) + \psi_s(0) =0$). The full wavefunction is then:

$\psi(x) = A(e^{ikx}-e^{-ikx}) \sim \textrm{sin}(kx)$

The last equality is a little misleading since the wavefunction is not normalizable, but let’s just pretend we have an infinite barrier at large but not quite infinite $(-x)$. Now consider a similar-looking, but pretty arbitrary potential:

Elastic scattering from an arbitrary potential

What happens in this case? Well, again, the scattering is elastic, so the incident and reflected amplitudes must be the same away from the region of the potential. All that can change, therefore, is the phase of the reflected (scattered) wavefunction. We can therefore write, similar to the case above:

$\psi(x) = A(e^{ikx}-e^{i(2\delta-kx)}) \sim \textrm{sin}(kx+\delta)$

Notice that the sine term has now acquired a phase. What does this mean? It means that the energy of the wavefunction has changed, as would be expected for a different potential. If we had used box normalization for the infinite barrier case, $kL=n\pi$, then the energy eigenvalues would have been:

$E_n = \hbar^2n^2\pi^2/2mL^2$

Now, with the newly introduced potential, however, our box normalization leads to the condition, $kL+\delta(k)=n\pi$ so that the new energies are:

$E_n = \hbar^2n^2\pi^2/2mL^2-\hbar^2\delta(k)^2/2mL^2$

The energy eigenvalues move around, but since $\delta(k)$ can be a pretty complicated function of $k$, we don’t actually know how they move. What’s clear, though, is that the number of energy eigenvalues are going to be the same — we didn’t make or destroy any new eigenvalues or energy eigenstates.

Let’s now move onto some solid state physics. In a metal, one usually fills up $N$ states in accordance with the Pauli principle up to $k_F$. If we introduce an impurity with a different number of valence electrons into the metal, we have effectively created a potential where the electrons of the Fermi gas/liquid can scatter. Just like in the cases above, this potential will cause a phase shift in the electron wavefunctions present in the metal, changing their energies. The amplitudes for the incoming and outgoing electrons again will be the same far from the scattering impurity.

This time, though, there is something to worry about — the phase shift and the corresponding energy shift can potentially move states from above the Fermi energy to below the Fermi energy, or vice versa. Suppose I introduced an impurity with an extra valence electron compared to that of the host metal. For instance, suppose I introduce a Zn impurity into Cu. Since Zn has an extra electron, the Fermi sea has to accommodate an extra energy state. I can illustrate the scenario away from, but in the neighborhood of, the Zn impurity schematically like so:

E=0 represents the Fermi Energy. An extra state below the Fermi energy because of the addition of a Zn impurity

It seems quite clear from the description above, that the phase shift must be related somehow to the valence difference between the impurity and the host metal. Without the impurity, we fill up the available states up to the Fermi wavevector, $k_F=N_{max}\pi/L$, where $N_{max}$ is the highest occupied state. In the presence of the impurity, we now have $k_F=(N'_{max}\pi-\delta(k_F))/L$. Because the Fermi wavevector does not change (the density of the metal does not change), we have that:

$N'_{max} = N_{max} + \delta(k_F)/\pi$

Therefore, the number of extra states needed now to fill up the states to the Fermi momentum is:

$N'_{max}-N_{max}=Z = \delta(k_F)/\pi$,

where $Z$ is the valence difference between the impurity and the host metal. Notice that in this case, each extra state that moves below the Fermi level gives rise to a phase shift of $\pi$. This is actually a simplified version of the Friedel sum rule. It means that the electrons at the Fermi energy have changed the structure of their wavefunctions, by acquiring a phase shift, to exactly screen out the impurity at large distances.

There is just one thing we are missing. I didn’t take into account degeneracy of the energy levels of the Fermi sea electrons. If I do this, as Friedel did assuming a spherically symmetric potential in three dimensions, we get a degeneracy of $2(2l+1)$ for each $l$ where the factor of 2 comes from spin and $(2l+1)$ comes from the fact that we have azimuthal symmetry. We can write the Friedel sum rule more precisely, which states:

$Z = \frac{2}{\pi} \sum_l (2l+1)\delta_l(k_F)$,

We just had to take into consideration the fact that there is a high degeneracy of energy states in this system of spherical symmetry. What this means, informally, is that each energy level that moves below the Fermi energy has the $\pm\pi$ phase shift distributed across each relevant angular momentum channel. They all get a little slice of some phase shift.

An example: If I put Ni  (which is primarily a $d$-wave $l=2$ scatterer in this context) in an Al host, we get that $Z=-1$. This is because Ni has valence $3d^94s^1$.  Now, we can obtain from the Friedel sum rule that the phase shift will be $\sim -\pi/10$. If we move onto Co where $Z=-2$, we get $\sim -2\pi/10$, and so forth for Fe, Mn, Cr, etc. Only after all ten $d$-states shift above the Fermi energy do we acquire a phase shift of $-\pi$.

Note that when the phase shift is $\sim\pm\pi/2$ the impurity will scatter strongly, since the scattering cross section $\sigma \propto |\textrm{sin}(\delta_l)|^2$. This is referred to as resonance scattering from an impurity, and again bears a striking resemblance to the classical driven harmonic oscillator. In the case above, it would correspond to Cr impurities in the Al host, which has phase shift of $\sim -5\pi/10$. Indeed, the resistivity of Al with Cr impurities is the highest among the first row transition metals, as shown below:

Hence, just by knowing the valence difference, we can get out a non-trivial phase shift! This is a pretty remarkable result, because we don’t have to use the (inexact and perturbative) Born approximation. And it comes from (I hope!) some pretty intuitive physics.