Chapter 9: Translational and Rotational Symmetry in the Two-Body Problem

Note

The atoms or elementary particles themselves are not real; they form a world of potentialities or possibilities rather than one of things or facts. ― Werner Heisenberg

Summary

Attention

Up to now, we have tiptoed our way in quantum mechanics, first introducing single particles that can only assume two states, then introducing the combination of two such particles. Next we introduced the position-space (and momentum-space) representations of the quantum state vector, using the eigenstates of the $\hat{x}$ operator as a complete basis. What we did may seem like a far cry from an understanding of quantum mechanics. Do not despair! In spite of the simplicity of the approach, we have learned a great deal about the formalism and this chapter is mostly an extension of what we learned but now considering the three-dimensional nature of space.

In particular, we find that the probability of finding a particle in the volume $d r^{3}=dx dy dz$ at position ${\mathbf{r}}$ is now given by:

$d^{3} r|\langle\mathbf{r} \mid \psi\rangle|^{2}$

Similar to Chapter 6: Wave mechanics in one-dimension, we introduce translation operators, this time along three different directions:

$\begin{array}{l} \hat{T}\left(a_{x} \mathbf{i}\right)|x, y, z\rangle=\left|x+a_{x}, y, z\right\rangle \\ \hat{T}\left(a_{y} \mathbf{j}\right)|x, y, z\rangle=\left|x, y+a_{y}, z\right\rangle \\ \hat{T}\left(a_{z} \mathbf{k}\right)|x, y, z\rangle=\left|x, y, z+a_{z}\right\rangle \end{array}$

Again, using our usual approach, this allows us to introduce the components of the linear momentum (vector) operator, as generators of each of those translations (see box below). The momentum operators are Hermitian and we can use their eigenstates to expand any states (we say that the eigenstates constitute a complete basis).

We also introduce a two-body Hamiltonian with potential energy that only depends on distance between the two bodies:

$\hat{H}=\frac{\hat{\mathbf{p}}_{1}^{2}}{2 m_{1}}+\frac{\hat{\mathbf{p}}_{2}^{2}}{2 m_{2}}+V\left(\left|\hat{\mathbf{r}}_{1}-\hat{\mathbf{r}}_{2}\right|\right)$

We then realize that we are now working in a quantum space of two particles, as we have already seen in Chapter 5: Combining two spin-1/2 particles for the combination of two spin-1/2 particles. Here, we combine two particles that are represented by a position space vector. Naturally, we combine the space spanning the possible state of each particle using the direct (or tensor) product:

$\left|\mathbf{r}_{1}, \mathbf{r}_{2}\right\rangle=\left|\mathbf{r}_{1}\right\rangle_{1} \otimes\left|\mathbf{r}_{2}\right\rangle_{2}$

We also remember that the operators that operate in different individual space commute with one another (since they operate on different things). This leads us to introduce a new operator that operates on the combined space where the two particles live. This operator is the total linear momentum and it corresponds to the generation of translations on the two-body space:

$\hat{\mathbf{P}}=\hat{\mathbf{p}}_{1}+\hat{\mathbf{p}}_{2}$

We demonstrate mathematically an important result: the Hamiltonian mentioned above commutes with the operator $\hat{\mathbf{P}}$ . This is not surprising since translating rigidly the two-body system does not change its energy (this is because the potential only depends on the distance between the particles!):

$[\hat{H}, \hat{\mathbf{P}}]=0$

Students by now should be conditioned! We have two operators that commute and it follows that they share the same eigenstates (who would have believed this?). This also means that because the Hamiltonian is invariant under translation, we find to conservation of total momentum of the system (using results derived in Chapter 4: Time evolution):

$\frac{d\langle\mathbf{P}\rangle}{d t}=\frac{i}{\hbar}\langle\psi|[\hat{H}, \hat{\mathbf{P}}]| \psi\rangle=0$

Next, we develop the classical argument of changing variables from the position each particles to variables corresponding to the position of the center of mass and of the position relative to the center of mass:

$\begin{array}{l} \hat{\mathbf{r}}=\hat{\mathbf{r}}_{1}-\hat{\mathbf{r}}_{2} \\ \hat{\mathbf{R}}=\frac{m_{1} \hat{\mathbf{r}}_{1}+m_{2} \hat{\mathbf{r}}_{2}}{m_{1}+m_{2}} \end{array}$

Finally, this yields, using the total momentum operator and the momentum operator of the relative motion:

$\hat{H}=\frac{\hat{\mathbf{P}}^{2}}{2 M}+\frac{\hat{\mathbf{p}}^{2}}{2 \mu}+V(|\hat{\mathbf{r}}|)$

The two variables can then be separated and we solve two separate problems (the first one being, trivially, the motion of a free particle that students typically solve during kindergarten):

$\hat{H}_{\mathrm{cm}}=\frac{\hat{\mathbf{P}}^{2}}{2 M} \quad\textrm{and}\quad \hat{H}_{\mathrm{rel}}=\frac{\hat{\mathbf{p}}^{2}}{2 \mu}+V(|\hat{\mathbf{r}}|)$

We just end up with solving a one-body (like) problem corresponding to the motion relative to the CM:

$\hat{H}=\frac{\hat{\mathbf{p}}^{2}}{2 \mu}+V(|\hat{\mathbf{r}}|)$

This problem is called the single-body system in a central potential.

One of the main concepts we introduce in this chapter is related to the orbital angular momentum operators. See box below about the angular momentum operators we have studied so far. Similar to classical mechanics, the orbital angular momentum operators are noted $(\hat{L}_{x},\hat{L}_{y},\hat{L}_{z})$ and are defined as $\hat{\mathbf{L}}=\hat{\mathbf{r}} \times \hat{\mathbf{p}}$ (note that this equation looks like what you have studied in classical mechanics but, here, this is an operator equation!). The problem we are interested is rotationally invariant (we see this because the Hamiltonian $\hat{H}_{\mathrm{rel}}$ only includes distances, in position and momentum spaces – as we formally proved in the course), this can be translated as:

$\left[\hat{H}, \hat{L}_{z}\right]=0$

and

$\left[\hat{H}, \hat{L}^2\right]=0$

in addition, we established that

$\left[\hat{L}^2, \hat{L}_{z}\right]=0$

These last three commutation relations consitute a central result of this chapter. We have three operators that commute with each other and it follows that one can find eigenstates that are common to all the three operators $\hat{H}$ , $\hat{L}^2$ , and $\hat{L}_{z}$ . In other words, we can write the eigenvalue problem as (we also remember that all three operators are Hermitian and thus have real eigenvalues):

$\begin{aligned} \hat{H}|E, l, m\rangle &=E|E, l, m\rangle \\ \hat{\mathbf{L}}^{2}|E, l, m\rangle &=l(l+1) \hbar^{2}|E, l, m\rangle \\ \hat{L}_{z}|E, l, m\rangle &=m \hbar|E, l, m\rangle \end{aligned}$

It is a good time to take a minute and understand the significance of the equations above. What they really mean is that we now know it is possible to know the energy, the total momentum, and the projection of the angular momentum on the z-axis at the same time.

In this chapter, we also work out the Schrodinger equation in position space (see box below on how this is done):

$\left\langle\mathbf{r}\left|\frac{\hat{\mathbf{p}}^{2}}{2 \mu}\right| \psi\right\rangle+\langle\mathbf{r}|V(|\hat{\mathbf{r}}|)| \psi\rangle=-\frac{\hbar^{2}}{2 \mu}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)\langle\mathbf{r} \mid \psi\rangle+\frac{\left\langle\mathbf{r}\left|\hat{\mathbf{L}}^{2}\right| \psi\right\rangle}{2 \mu r^{2}}+V(r)\langle\mathbf{r} \mid \psi\rangle=E\langle\mathbf{r} \mid \psi\rangle$

Solving this equation is a bit intimidating but it is mostly a matter of remembering how to represent operators into their position-space representations. In addition, using spherical coordinates, we can separate the radial, azimuthal, and polar parts as:

$\langle\mathbf{r} \mid E, l, m\rangle=R(r) \Theta(\theta) \Phi(\phi)$

The radial part of the Schrodinger equation becomes (we will come back to the azimuthal and polar parts later):

$\left[-\frac{\hbar^{2}}{2 \mu}\left(\frac{d^{2}}{d r^{2}}+\frac{2}{r} \frac{d}{d r}\right)+\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}+V(r)\right] R(r)=E R(r)$

We tranform this equation by writing $R(r)=\frac{u(r)}{r}$ to find:

$\left[-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}+V(x)\right]\langle x \mid E\rangle=E\langle x \mid E\rangle$

Where the effective potential is:

$V_{\mathrm{eff}}(r)=\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}+V(r)$

It is informative to understand the origin of this effective potential. The second part is just the original central potential (for example, the electrostatic potential). The first part corresponds to a centrifugal barrier (“barrier” because it is a repulsive part, away from the center). However, it is important to understand that this “potential” comes originally from a kinetic energy term (it originates from the fact the system rotates). This has far reaching consequences since that term depends on l (and is, in fact, zero for $l=0$ ). We will see the consequence on the ground-state of the hydrogen atom in the next chapter.

Solving the radial equation requires more information about the system (that is: the function $V(r)$ ) as we shall see in the next chapter. For now, let’s conclude this chapter with a study of the solution of the angular part of the equation. In spherical coordinates, we write:

$\langle r, \theta, \phi \mid E, l, m\rangle=R(r) Y_{l, m}(\theta, \phi)$

Where $Y_{l, m}(\theta, \phi)$ are known as the spherical harmonics:

$Y_{l, m}(\theta, \phi)=\frac{(-1)^{l}}{2^{l} l !} \sqrt{\frac{(2 l+1)(l+m) !}{4 \pi(l-m) !}} e^{i m \phi} \frac{1}{\sin ^{m} \theta} \frac{d^{l-m}}{d(\cos \theta)^{l-m}} \sin ^{2 l} \theta$

This last equation is one of the reasons professors are happy to let students use cribsheet. However, students should be aware of the general shape of those functions (as we describe in Special Mathematical Functions.)

Note

From operators to position-space representations

Starting from the operator version of the Schroginder equation

$\left\langle\mathbf{r}\left|\frac{\hat{\mathbf{p}}^{2}}{2 \mu}\right| \psi\right\rangle+\langle\mathbf{r}|V(|\hat{\mathbf{r}}|)| \psi\rangle = E\langle\mathbf{r} \mid \psi\rangle$

Linear momentum (radial coordinate):

We use:

$\hat{p}_{r} \rightarrow \frac{\hbar}{i}\left(\frac{\partial}{\partial r}+\frac{1}{r}\right)$

to find

$-\frac{\hbar^{2}}{2 \mu}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)\langle\mathbf{r} \mid \psi\rangle=\frac{\left\langle\mathbf{r}\left|\hat{p}_{r}^{2}\right| \psi\right\rangle}{2 \mu}$

This allows us to write:

$-\frac{\hbar^{2}}{2 \mu}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r} \frac{\partial}{\partial r}\right)\langle\mathbf{r} \mid \psi\rangle+\frac{\left\langle\mathbf{r}\left|\hat{\mathbf{L}}^{2}\right| \psi\right\rangle}{2 \mu r^{2}}+V(r)\langle\mathbf{r} \mid \psi\rangle=E\langle\mathbf{r} \mid \psi\rangle$
Since we can choose eigenstates that are eigenstates of $\hat{\mathbf{L}}^{2}$ , we have, for the eigenstates:

$\frac{\left\langle\mathbf{r}\left|\hat{\mathbf{L}}^{2}\right| \psi\right\rangle}{2 \mu r^{2}} =\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}} \langle\mathbf{r} \mid \psi\rangle$

Finally, we get, for the radial part:

$\left[-\frac{\hbar^{2}}{2 \mu}\left(\frac{d^{2}}{d r^{2}}+\frac{2}{r} \frac{d}{d r}\right)+\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}+V(r)\right] R(r)=E R(r)$

Note

Angular solution

Because we could separate angular and radial coordinates, we find the angular dependence as solutions to

$\hat{\mathbf{L}}^{2} \rightarrow-\hbar^{2}\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right]$

and

$\left\langle r, \theta, \phi\left|\hat{L}_{z}\right| \psi\right\rangle=\frac{\hbar}{i} \frac{\partial}{\partial \phi}\langle r, \theta, \phi \mid \psi\rangle$

The solutions of those equations are the Spherical Harmonics, as described in Special Mathematical Functions.

Learning Material

Copy of Slides

The slides for Chapter 9 are available in pdf format here: 📂.

Screencast

Key Learning Points

Position-space representation in 3D:

$\hat{x}|\mathbf{r}\rangle=x|\mathbf{r}\rangle \quad \hat{y}|\mathbf{r}\rangle=y|\mathbf{r}\rangle \quad \hat{z}|\mathbf{r}\rangle=z|\mathbf{r}\rangle$

where

$|\mathbf{r}\rangle=|x, y, z\rangle$
Completeness relation in 3D (position-space representation):

$|\psi\rangle=\iiint d x d y d z|x, y, z\rangle\langle x, y, z \mid \psi\rangle=\int d^{3} r|\mathbf{r}\rangle\langle\mathbf{r} \mid \psi\rangle$
Orthogonality (with: $|\mathbf{r}\rangle=|x, y, z\rangle$

$\left\langle x, y, z \mid x^{\prime}, y^{\prime}, z^{\prime}\right\rangle=\delta\left(x-x^{\prime}\right) \delta\left(y-y^{\prime}\right) \delta\left(z-z^{\prime}\right)$

or, more succinctly:

$\left\langle\mathbf{r} \mid \mathbf{r}^{\prime}\right\rangle=\delta^{3}\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$
Generators of translation in 3D:

$\begin{array}{l} \hat{T}\left(a_{x} \mathbf{i}\right)=e^{-i \hat{p}_{x} a_{x} / \hbar} \\ \hat{T}\left(a_{y} \mathbf{j}\right)=e^{-i \hat{p}_{y} a_{y} / \hbar} \\ \hat{T}\left(a_{z} \mathbf{k}\right)=e^{-i \hat{p}_{z} a_{z} / \hbar} \end{array}$

and because the different generators of translation commute, we have:

$\hat{T}(\mathbf{a})=e^{-i \hat{p}_{x} a_{x} / \hbar} e^{-i \hat{p}_{y} a_{y} / \hbar} e^{-i \hat{p}_{z} a_{z} / \hbar}=e^{-i \hat{\mathbf{p}} \cdot \mathbf{a} / \hbar}$
Commutation relationship (using Kronecker delta):

$\left[\hat{x}_{i}, \hat{p}_{j}\right]=i \hbar \delta_{i j}$
Position-space representation of the momentum operator vector:

$\langle\mathbf{r}|\hat{\mathbf{p}}| \psi\rangle=\frac{\hbar}{i} \nabla\langle\mathbf{r} \mid \psi\rangle$
position-space representation of eigenstates of the momentum operator vector:

$\langle\mathbf{r} \mid \mathbf{p}\rangle =\frac{1}{(2 \pi \hbar)^{3 / 2}} e^{i \mathbf{p} \cdot \mathbf{r} / \hbar}$
Angular momentum operators, commutation relationships (using the Levi-Civita symbol :

$\left[\hat{L}_{i}, \hat{L}_{j}\right]=i \hbar \sum_{k=1}^{3} \varepsilon_{i j k} \hat{L}_{k}$
Position-space representation of angular momentum operator (in spherical coordinates):

$\hat{L}_{z} \rightarrow \frac{\hbar}{i} \frac{\partial}{\partial \phi}$

and

$\hat{\mathbf{L}}^{2} \rightarrow-\hbar^{2}\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right]$

Test your knowledge

What pairs of operators below commute?
1. $\hat{x}$ and $\hat{p}_y$
2. $\hat{x}$ and $\hat{p}_x$
What pairs of operators below commute?
1. $\hat{S}_x$ and $\hat{\boldsymbol{S}}^2$
2. $\hat{S}_x$ and $\hat{S}_y$
Among these vector states, which ones are not eigenvectors of operator $\hat{r}=(\hat{x}, \hat{y}, \hat{z})$ ?
1. $\bra{x,y,z}$
2. $\ket{x-\delta,y,z}$
3. $1/\sqrt{2} (\ket{x,y,z}-\ket{-x,y,z})$
Among the possibilities below, select all correct normalization relationships.
1. $\langle p_x, p_y, p_z | p'_x, p'_y, p'_z \rangle = \delta(p_x-p'_x)\delta(p_y-p'_y)\delta(p_z-p'_z)$
2. $\langle p_x, p_y, p_z | p'_x, p'_y, p'_z \rangle = \delta(x'-x)\delta(y'-y)\delta(z'-z)$
3. $\langle p_x, p_y, p_z | p'_x, p'_y, p'_z \rangle = \delta_{p'_x,p_x}\delta_{p'_y,p_y}\delta_{p'_z,p_z}$ (where $\delta_{ij}$ is Kronecker’s delta)
Imagine a two-body system where the interaction potential between the two bodies only depends on their separation distance. What are the correct statements from the list below?
1. We can rewrite the Hamiltonian in the center-of-mass frame so that it becomes effectively a one-body problem. This is an approximation.
2. The system is invariant under translation. This is translated mathematically as $\comm{\hat{H}}{\hat{\boldsymbol{P}}}=0$ (where $\hat{\boldsymbol{P}}$ is the total linear momentum of the system).
3. The system is invariant under rotation. However, because IQM is so weird, one cannot claim that $\comm{\hat{H}}{\hat{\boldsymbol{L}^2}}=0$ (where $\hat{\boldsymbol{L}}$ is the total angular momentum of the system).
4. None of the other claims is correct.
We know that a system is invariant under rotation. How do you translate this mathematically?
1. $\comm{\hat{H}}{\hat{\boldsymbol{L}}^2}=\comm{\hat{H}}{\hat{{L}}_x}=\comm{\hat{H}}{\hat{{L}}_y}=\comm{\hat{H}}{\hat{{L}}_z}=0$ .
2. $\comm{\hat{H}}{\hat{\boldsymbol{L}}^2}=\comm{\hat{H}}{\hat{L}_z}=0$ ; and $\comm{\hat{L}_x}{\hat{{L}}_y}=0$ .
3. $\comm{\hat{\boldsymbol{L}}^2}{L_z}=0$ .
4. None of the other mathematical descriptions corresponds to a system that is invariant under rotation.
(select the most accurate answer) The orbital momentum operator $\hat{L}_z$
1. is the generator of rotations around the $z$ axis.
2. can be expressed as $\hat{L}_z=\hat{x}\hat{p}_y-\hat{y}\hat{p}_x$ .
3. Both answers are correct
Consider a system of two bodies whose interaction only depends on the separation between the two particles
1. The system is invariant under translation and but not under rotation of both particles
2. The system is invariant under translation but not under rotation because $\comm{\hat{L}_z}{\hat{L}_x} \neq 0$
3. The Hamiltonian of the system can be written such that the rotational energy of the entire system appears as an effective potential, in a way similar to the existence of a fictitious centrifugal force in a classical system
4. None of the other claims is correct
What are the spherical harmonics?
1. Spherical harmonic oscillators
2. Solutions of the angular part of the general solution of the 2-body problem in the presence of a spherically symmetric potential
3. Spherical model explaining Bohr’s model of the hydrogen atom.
4. Functions whose Fourier transforms are obtained by direct product of vector spaces of two coupled harmonic oscillators.

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 following problems from the textbook: 9.4, 9.7, 9.13, 9.17, 9.23

Recitation Assignment

Solve the following problems from the textbook: 9.3, 9.8, 9.12, 9.16, 9.20