Chapter 10: Bound states of central potential

Note

We would like to mention one particular feature of the wave functions [of the hydrogen atom] for higher l: for l>0 the amplitudes are zero at the center. That is not surprising, since it’s hard for an electron to have angular momentum when its radius arm is very small. For this reason, the higher the l, the more the amplitudes are “pushed away” from the center. ― Richard Feynman, Lectures on physics.

Summary

Attention

We can summarize Chapter 9: Translational and Rotational Symmetry in the Two-Body Problem with just one equation and a comment:

$\langle\mathbf{r}|\hat{H}| E, l, m\rangle=\left[-\frac{\hbar^{2}}{2 \mu}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)+\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}+V(r)\right]\langle\mathbf{r} \mid E, l, m\rangle$

We were able to write this equation after we established that the eigenstates $|E, l, m\rangle$ are simultaneous eigenstates of $\hat{H}$ , $\hat{\mathbf{L}}^{2}$ , and $\hat{L}_{z}$ since those three operators commute in pairs for a potential energy operator that only depends on the distance between two particles. When doing so, we realized that the solution of the equation (in spherical coordinates) can be separated for the radial and angular (i.e., azimuthal and polar angles) solutions. For all problems of this type, the angular solutions are eigenstates of the angular momentum operators and they are called spherical harmonics. What is left to do now: find the solution of the radial part of the equation above for a given interacting potential $V(\hat{r})$ .

In this chapter, we study various examples of such potentials and discuss the solution of the radial part. Students will not forget that the angular solutions are already known (see Special Mathematical Functions).

Before moving to specific situations, we realize that regardless of the details of the potential, we can describe some general properties of the radial wavefunction near the origin ( $r=0$ ). These properties are related to the effective potential $\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}$ (students will remember that the origin of his term has to be found in the rotational kinetic energy). We find that, close to the origin:

$R_{E, l} \underset{r \rightarrow 0}{\longrightarrow} r^{l}$

This means that with the notable exception of the $l=0$ solution, the wavefunction vanishes at the origin.

We are now summarizing our findings for four different examples.

Hydrogen atom:

The first example is that of an electrostatic potential between an electron and $Z$ protons:

$-\frac{Z e^{2}}{|\hat{\mathbf{r}}|}$

After introducing dimensionless variables, the Schrodinger equation becomes (with $R(r)=u(r)/r$ ):

$\frac{d^{2} u}{d \rho^{2}}-\frac{l(l+1)}{\rho^{2}} u+\left(\frac{\lambda}{\rho}-\frac{1}{4}\right) u=0$

with

$\rho=\sqrt{\frac{8 \mu|E|}{\hbar^{2}}} r$

and

$\lambda=\frac{Z e^{2}}{\hbar} \sqrt{\frac{\mu}{2|E|}}$

Here, just like we did in Chapter 7: The Linear Harmonic Oscillator, we first consider the solution of the equation far from $r=0$ . This leads us to write the solution as:

$u(\rho)=\rho^{l+1} e^{-\rho / 2} F(\rho)$

We understand that the first factor is used for the $r\to0$ solution and the second factor accounts for the $r\to\infty$ solution. On a side note, we remember that the exponential function decays faster than any polynomial and the asymptotic solution is indeed dominated by the second factor (relative to the first one).

We show that $F(r)$ is a polynomial, known as Laguerre polynomial of degree $n_r$ and the energy eigenvalues are:

$E_{n}=-\frac{\mu Z^{2} e^{4}}{2 \hbar^{2} n^{2}}$

where the principal quantum number is

$n=1+l+n_r$

The radial solutions for the first few $n$ and $l$ values are provided in the box below.

Degeneracy:

For each value of $n$ , there are $n$ values of $l$ and for each $l$ , there are $2l+1$ values of $m$ that give the same energy (remember the energy only depends on $n$ ). All-in-all, we have a total degeneracy of $n^2$ .
Deuteron:

A deuteron is a bound state between a proton and a neutron. The interaction is governed by a strong force that appears when the particles touch each other.

Here we build a simple system corresponding to a spherical well (this means the potential is a square potential when plotted as a function of radial coordinate). We also use the experimental fact that such a potential has only one bound state (in other words, any solution of the Schrodinger equation that is not the ground state would not be normalizable).

In this study, we have two parameters: the potential well size and its depth. We are able to estimate the properties of the system on the basis on information stemming from experimental observations: the size of the well and the fact we only have one bound state. We conclude that the potential is quite deep (about 35 MeV) while the deuteron ground state is about 2.2 Mev. This “barely bound state” extends pretty far in the classically forbidden region (tunneling)
Infinite well:

Based on the information we obtained for the deuteron system (that is: the interaction between two nucleons), we build a model of the nucleus:

$V=\left\{\begin{array}{ll} 0 & r<a \\ \infty & r>a \end{array}\right.$

The radial equation we have been using throughout this chapter becomes (inside the well):

$\frac{d^{2} R}{d r^{2}}+\frac{2}{r} \frac{d R}{d r}-\frac{l(l+1)}{r^{2}} R+k^{2} R=0$

where $k=\sqrt{\frac{2 \mu E}{\hbar^{2}}}$ .

This equation is well-known to mathematicians and its solutions are either the spherical Bessel functions or the spherical Neumann functions (see Special Mathematical Functions). Only the former are acceptable solutions since the latter are irregular at the origin. Since the potential is infinite, we must impose that the radial solution vanishes at $r=a$ . This condition leads to a quantization of the energy, corresponding to the roots of the Spherical Bessel functions.

The resulting energy spectrum allows us to build a nucleus (by ignoring Coulomb repulsion between protons, spin-effects, etc) and we obtain an estimate of the energy of the nuclei. This is a crude approximation but it allows us to already highlight the existence of magic numbers (i.e., those nuclei feature closed shells).
Three-dimensional harmonic oscillator:

This is a particularly illuminating example of a three-dimensional problem featuring spherical symmetry. It is also important as it highlights this notion of labels and degeneracy we have encountered frequently in this course.

To summarize, we realize that because we are looking at an isotropic harmonic oscillator, we can treat it equality well using Cartesian or spherical coordinates. Formally:

$V(r)=\frac{1}{2} \mu \omega^{2} r^{2}=\frac{1}{2} \mu \omega^{2}\left(x^{2}+y^{2}+z^{2}\right)$

Using the first expression, one can solve the Schrodinger equation we have examined in this chapter. Alternatively, one can use Cartesian coordinates and realize that, mathematically, we can separate the three spatial coordinates so that the problem becomes equivalent to the problem of 3 uncoupled one-dimensional harmonic oscillators (the good news is that we know the solution to that problem from Chapter 7: The Linear Harmonic Oscillator.). Of course the solutions have to be the same using either approach. Let’s examine the eigen-states and eigenenergies of the Hamiltonian. Using Cartesian coordinates, we use the set of labels corresponding to each of the uncoupled harmonic oscillator:

$|E\rangle=\left|E_{x}, E_{y}, E_{z}\right\rangle$

with energy ( $n=n_x+n_y+n_z$ ):

$E_{n}=\left(n+\frac{3}{2}\right) \hbar \omega$

When using spherical coordinates, we have:

$|E\rangle=\left|n,l,m\right\rangle$

where $n=2n_r+l$ ( $n_r$ being the order of the polynomial solution of the Schrodinger equation).

It is most instructive to look at the degeneracy obtained in both methods and compare them:
1. For the ground state: $n=0$ and the degeneracy is one.
  1. Cartesian coordinates: $n=n_x=n_y=n_z=0$
  2. Spherical coordinates: $n=n_r=l=0$ and also $m=0$ since it varies from $-l$ to $l$ .
2. For the first excited state: $n=1$ and the degeneracy is three.
  1. Cartesian coordinates: $n_x=1, n_y=0, n_z=0$ ; $n_x=0, n_y=1; n_z=0$ ; and $n_x=0, n_y=0, n_z=1$
  2. Spherical coordinates: $l=1, m=-1$ ; $l=1, m=0$ ; and $l=1, m=1$
3. For the second excited state: $n=2$ and the degeneracy is 6.
  1. Cartesian coordinates: $n_x=2, n_y=0, n_z=0$ ; $n_x=0, n_y=2; n_z=0$ ; $n_x=0; n_y=0, n_z=2$ ; $n_x=1, n_y=1, n_z=0$ ; $n_x=1, n_y=0, n_z=1$ ; and $n_x=0, n_y=1, n_z=1$
  2. Spherical coordinates: $n_r=0, l=2, m=-2$ ; $n_r=0, l=2, m=-1$ ; $n_r=0, l=2, m=0$ ; $n_r=0, l=2, m=1$ ; $n_r=0, l=2, m=2$ ; and $n_r=1, l=0, m=0$
Of course, the degeneracy is the same regardless which point of view one adopts. This example also shows that quantum labels are useful but are not unique!

Note

What do we mean by bound state?

These are states that are solutions of the Schrodinger equation and that can be normalized. Their energy eigenvalue is lower than the top of the potential energy so that the particle cannot escape the potential. Of course, this is a classical picture since the particle can tunnel outside the potential but the majority of the probability of finding the particle is inside the potential well.

Warning

What is a potential that is less singular than $r^{-2}$ at the origin?

Mathematically, such a potential obeys the following conditions:

$r^{2} V(r) \underset{r \rightarrow 0}{\longrightarrow} 0$

This means the potential is dominated by the $r^{-2}$ term at the origin. This is important for the behavior of the wavefunction of the origin since this means that the centrifugal barrier dominates at the origin.

Learning Material

Copy of Slides

The slides for Chapter 10 are available in pdf format here: 📂.

Screencast

Key Learning Points

Radial wavefunction close to the origin:

$R_{E, l} \underset{r \rightarrow 0}{\longrightarrow} r^{l}$

Hydrogen atom’s wavefunction:

Ground state ( $n=1$ ): $R_{1,0}=2\left(\frac{Z}{a_{0}}\right)^{3 / 2} e^{-Z r / a_{0}}$
First excited states ( $n=2$ ):

$\begin{array}{l} R_{2,0}=2\left(\frac{Z}{2 a_{0}}\right)^{3 / 2}\left(1-\frac{Z r}{2 a_{0}}\right) e^{-Z r / 2 a_{0}} \\ R_{2,1}=\frac{1}{\sqrt{3}}\left(\frac{Z}{2 a_{0}}\right)^{3 / 2} \frac{Z r}{a_{0}} e^{-Z r / 2 a_{0}} \end{array}$

The second function has a degeneracy of 3 ( $m=1, 0, -1$ ).

Bound state of deuteron:

Solutions of the transcendental equation:

$\tan k_{0} a=-\frac{k_{0}}{q}$

where

$\begin{array}{l} k_{0}=\sqrt{\frac{2 \mu}{\hbar^{2}}\left(V_{0}+E\right)} \\ q=\sqrt{-\frac{2 \mu E}{\hbar^{2}}} \end{array}$
Bound energy: -2.2 Mev, potential depth: 35Mev and potential size: $1.710^{-15}$ m

Infinite spherical well:

Solutions are the Spherical Bessel Functions
Contains “magic angles” structure of nuclei (though not accurately given the rough approximation)

3D Harmonic Oscillator

Energy: $E_{n}=\left(n+\frac{3}{2}\right) \hbar \omega$
Can be solved in Cartesian or Spherial coordinates

Test your knowledge

True or False? None of the solutions of the Schrodinger equation for the hydrogen atom allows for a non-zero wavefunction at $r=0$ .
1. True
2. False
What is the expectation value $\expval{L_z}$ for the lowest energy state of the hydrogen atom?
1. $-\hbar/2$
2. 0
3. $+\hbar/2$
What do we mean when we talk about the degeneracy of an eigenvalue?
1. We mean something deep but I would need to study more to answer this!
2. We are talking about many different eigenvectors have the same eigenvalue.
3. We refer to the number of eigenvalues that are not real.
What is the degeneracy of the energy eigenvalue corresponding to the principal number $n$ ?
1. 0
2. $n$
3. $n^2$
4. $n^3$
We spent a lot of time discussing the radial part of the eigenstates of the hydrogen atom. What about the angular part?
1. There is no angular part: the problem is spherically symmetric!
2. The angular dependence is irrelevant since the hydrogen atom is a one-dimensional problem.
3. The angular dependence is found by searching for the eigenstates of the orbital angular momentum operators: these are known as the spherical harmonics.
Consider the solution(s) of the Schrodinger equation of the hydrogen atom for $n=3$ . What are the possible degree(s) of the corresponding Laguerre polynomial?
1. 0
2. 0 and 1
3. 0, 1, and 2
An eigenstate of the Hamiltonian of a finite-depth spherical well is a bound state of that system. Its energy is not larger than the top of the potential value.
1. True
2. False
A bound state can only be found in an infinite well.
1. True
2. False
In a finite well, a bound state decays exponentially in the region of space where the potential value is larger than its energy.
1. True
2. False
The experimentally proven existence of magic number of nucleons in nature can be qualitatively explained using an infinite well potential based off knowledge acquired when treating the deuteron system.
1. True
2. False
It is possible to tune the properties of a fine-well spherical well so that it only accepts one bound state. The bound state is usually shallow.
1. True
2. False
The ground state of an Hamiltonian with a finite spherical-well potential is necessarily equal to the depth of the potential (both the eigenstate and potential are measured using the same asymptotic reference).
1. True
2. False
The three-dimensional spherical harmonic oscillator can be solved using either the Cartesian coordinates or the Spherical coordinates, as we can use separation of variables in both cases.
1. True
2. False
For the 3D spherical harmonic oscillator, the notations $\ket{n_x,n_y,n_z}$ and $\ket{n,l,m}$ for the eigenstates are equivalent: one can find a one-to-one correspondence between kets of each set. (Hint: Students may want to think about degeneracy before answering too fast).
1. True
2. False
The ground state of a central potential must have a $l>0$ ground state due to the presence of a centrifugal barrier.
1. True
2. False
When solving the Schrodinger equation for a central potential, one needs to know the details of the potential in order to solve for the angular part of the eigenstates.
1. True
2. False

Hint

Find the answer keys on this page: Answers to selected test your knowledge questions. Don’t cheat! Try solving the problems on your own first!

Homework Assignment

Solve the step-by-step problem described here: pdf

Recitation Assignment

Solve the following problems from the textbook: 10.9, 10.10