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import mmf_setup;mmf_setup.nbinit()
import logging;logging.getLogger('matplotlib').setLevel(logging.CRITICAL)
%matplotlib inline
import numpy as np, matplotlib.pyplot as plt

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How to Renormalize The Schrödinger Equation

Here we work through the example discussed in [Lepage, 1997], which uses the example of bound states in a spherically symmetric potential, with Coulomb-like behaviour for large radii \(r\rightarrow \infty\), but with “unknown” corrections at short distances \(r\rightarrow 0\):

\[\begin{gather*} \left(\frac{-\hbar^2\nabla^2}{2m} + V(r) - E\right)\Psi(r, \Omega) = 0, \qquad \lim_{r\rightarrow \infty} V(r) \rightarrow \frac{-\alpha}{r}. \end{gather*}\]

To do this, I suggest you complete the following tasks:

1. Get a method to numerically solve the radial Schrödinger equation. You will use this to compute your “experimental” data. Note: this is a little challenging due to the nature of the Coulomb potential being singular at the origin, and long-ranged (falling off slowly \(V(r) \sim 1/r\) at long distances). The following notes have many hints and suggestions, but please try to use them only as needed.

   • S-Wave Scattering
   • How to Solve the (Radial) Schrödinger Equation
   • How to Solve the (Radial) Schrödinger Equation II
   • Diagonalizing the Radial Schrödinger Equation
   • Shooting the Radial Schrödinger Equation
   • My Approach to Renormalize The Schrödinger Equation

2. Perform the perturbative analysis discussed in [Lepage, 1997] but following with your own data. If there is anything you don’t know how to do, please ask on the Hypothes.is document so I can help fill in background in class or online.

How to Solve the Schrödinger Equation

To easily work through the discussion in [Lepage, 1997], we must be able to formulate and solve the Schrödinger equation for spherically symmetric potentials.

Radial Equation

Spherical symmetry allows use to express this as a simple 1D boundary value problem (BVP) for the radial equation:

With the usual spherical coordinates \(\phi\) and \(\theta\), and angular momentum quantum numbers \(l\) and \(m\), we have

\(d\)	\(\Omega\)	\(\lambda\)
\(2\)	\(\phi\)	\(\abs{m}\)
\(3\)	\((\phi, \theta)\)	\(l\)

for \(d=2\) dimensions (i.e. cylindrical coordinates, but no \(z\) dependence) and \(d=3\) dimensions respectively. The total orbital angular momentum operator is related to the Laplace-Beltrami operator by

\[\begin{gather*} \op{L}^2 = \hbar^2\op{\Delta}_{S^{d-1}} \end{gather*}\]

which has eigenvalues \(\hbar^2 m^2\) in \(d=2\) (i.e. \(\op{L}^2 \equiv \op{L}_z^2\) and \(\hbar^2 l (l - 1)\) in \(d=3\).

Our aim is to satisfy the Schrödinger equation for central potentials in \(d\)-dimensions, which we can do in the usual way by expressing the wavefunction \(\Psi(r, \Omega) = \psi(r)Y_{\lambda}(\Omega)\) in terms of the radial wavefunction \(u(r)\) and the appropriate generalized spherical harmonics \(Y_{\lambda}(\Omega)\):

\[\begin{gather*} \left(\frac{-\hbar^2\nabla^2}{2m} + V(r) - E\right)\Psi(r, \Omega) = 0, \\ \left[ \frac{\hbar^2}{2m} \underbrace{ \left(-\diff[2]{}{r} + \frac{\nu^2 - 1/4}{r^2}\right) }_{\op{K}} + V(r)\right]u(r) = E u(r),\\ u(r) = r^{(d-1)/2}\psi(r), \qquad \nu = \lambda + \frac{d}{2} - 1. \end{gather*}\]

Details

Here \(\Omega\) is the generalized solid angle. Rotational invariance implies that angular momentum is a good quantum number, so all eigenstates can be factored \(\Psi(r, \Omega) = \psi(r)Y_{\lambda}(\Omega)\) where \(Y_{\lambda}(\Omega)\) is the generalized spherical harmonic on the (\(d-1\))-dimensional sphere and \(\lambda \in \{0, 1, \dots\}\) is the generalized angular momentum:

\[\begin{gather*} \nabla^2 = \frac{1}{r^{d-1}}\diff{}{r}\left(r^{d-1} \diff{}{r}\right) + \frac{1}{r^2}\Delta_{S^{d-1}}, \\ \Delta_{S^{d-1}}Y_{\lambda}(\Omega) = \lambda (\lambda + d - 2)Y_{\lambda}(\Omega), \end{gather*}\]

where \(\Delta_{S^{d-1}}\) is the Laplace-Beltrami operator. Introducing the radial wavefunction \(u(r)\), this becomes quadratic:

\[\begin{gather*} u(r) = r^{(d-1)/2}\psi(r), \qquad \nu = \lambda + \frac{d}{2} - 1,\\ r^{(d-1)/2}\nabla^2 \psi(r) = \left(\diff[2]{}{r} - \frac{\nu^2 - 1/4}{r^2}\right)u(r). \end{gather*}\]

This follows after a little algebra from

\[\begin{gather*} \frac{1}{r^{d-1}}\diff{}{r}\left(r^{d-1} \diff{}{r}\right)r^{(1-d)/2}u(r)\\ = r^{(1-d)/2} \left( \diff[2]{}{r} - \frac{(d-3)(d-1)}{4} \right)u(r). \end{gather*}\]

This can be solved quite simply – but not very accurately – with finite differences. Highly accurate solutions can be obtained by shooting, and we present details in How to Solve the (Radial) Schrödinger Equation about how to do this.

The Essence

The essential idea is low-energy properties, like the bound state energies \(E_n\), should not be highly sensitive to details about the nature of the potential \(V(r)\) at short distances \(r \ll 1/\Lambda\) where \(\Lambda\) is a large momentum scale.