Lecture 6 Jan 23#

Last time we discussed the symmetries of the two-body Hamiltonian:

  • translational symmetry when we move both particles together. The generator is the total momentum \(\hat{\mathbf{P}}\). The total momentum is conserved

\[{d\langle{\mathbf{P}}\rangle\over dt} = {i\over \hbar}\langle[\hat{H},\hat{\mathbf{P}}]\rangle = 0.\]

  • rotational symmetry. The generator is the component of orbital angular momentum about some (arbitrarily) chosen axis, \(\hat{\mathbf{L}}_z\). Angular momentum is conserved:

\[{d\langle L_z\rangle\over dt} = {i\over \hbar}\langle[\hat{H},\hat{L}_z]\rangle = 0\]

Another symmetry is parity which takes \(\mathbf{r}\rightarrow -\mathbf{r}\), i.e. \(\hat{P}\ket{\mathbf{r}} = \ket{-\mathbf{r}}\). The eigenvalues of \(\hat{P}\) are either +1 or -1 since \(\hat{P}^2\ket{\mathbf{r}} = \ket{\mathbf{r}}\). We will see that, because \([\hat{H},\hat{P}] = 0\), the stationary states of the two-body problem have a definite parity.

Today, we want to focus on the angular momentum eigenstates and in particular derive position-representations of the angular momentum operators and from there obtain the angular wavefunctions.

Reminder of angular momentum operators#

First some results you will have seen already in the context of spin, but now we can apply them to orbital angular momentum:

  • Commutation relations. Unlike translations, rotations do not commute, so the angular momentum operators in different directions also do not commute:

\[[\hat{L}_i, \hat{L}_j] = i\hbar \epsilon_{ijk} \hat{L}_k\]

(You will prove this specifically for orbital angular momentum in HW2). Each \(\hat{L}_i\) does commute with the total angular momentum \([\hat{L}^2,\hat{L}_i]=0\) where

\[\hat{L}^2 = \hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2.\]

So in quantum mechanics, we can specify the total angular momentum and one component of the angular momentum simultaneously.

  • Eigenstates. The angular momentum eigenstates are \(\ket{\ell,m}\) which satisfy

\[\hat{L}^2\ket{\ell, m} = \ell(\ell+1)\hbar^2 \ket{\ell, m}\]
\[\hat{L}_z\ket{\ell, m} = m\hbar \ket{\ell, m}\]

\(m\) ranges from \(-\ell\) to \(+\ell\) in integer steps.

  • Ladder operators. \(\hat{L}_{+-} = \hat{L}_x\pm i\hat{L}_y\)

\[\hat{L}_+\ket{\ell,m} = \hbar\sqrt{\ell(\ell+1)-m(m+1)}\ket{\ell,m+1}\]
\[\hat{L}_-\ket{\ell,m} = \hbar\sqrt{\ell(\ell+1)-m(m-1)}\ket{\ell,m-1}\]

Note that \(\hat{L}_+\ket{\ell, \ell} = 0\) and \(\hat{L}_-\ket{\ell, -\ell} = 0\) (why?). These are useful for computing for example \(\braket{\hat{L}_x}\) when you know the state in the \(\hat{L}_z\) basis because you can express \(\hat{L}_x\) in terms of the ladder operators that act on the \(\hat{L}_z\) basis states.

Orbital angular momentum has integer values#

One important difference between spin angular momentum and orbital angular momentum is that \(m\) can take only integer values for orbital angular momentum; half-integer values are not allowed. To see this (this argument is from Townsend 9.8), go back to the rotation operator \(\hat{R}(d\phi\,\mathbf{k})\) and write the position eigenstates now in spherical coordinates, so

\[\hat{R}(d\phi\,\mathbf{k})\, \ket{r, \theta, \phi} = \ket{r, \theta, \phi+d\phi}\]
\[\Rightarrow \bra{r, \theta, \phi}\,\hat{R}^\dagger(d\phi\,\mathbf{k}) = \bra{r, \theta, \phi+d\phi}\]
\[\Rightarrow \bra{r, \theta, \phi}\,\hat{R}(d\phi\,\mathbf{k}) = \bra{r, \theta, \phi-d\phi}\]

(since \(\hat{R}^\dagger\hat{R}=1\) so \(\hat{R}^\dagger\) is the inverse of \(\hat{R}\)). This allows us to see how the rotation operator acts on the wavefunction:

\[\braket{r, \theta, \phi|\hat{R}(d\phi\,\mathbf{k})|\Psi} = \braket{r, \theta, \phi-d\phi|\Psi}= \braket{r, \theta, \phi|\Psi} - d\phi {\partial\over\partial \phi}\braket{r, \theta, \phi|\Psi}\]

(keeping first order terms only for infinitesimal \(d\phi\)). We can also write this in terms of the generator

\[\braket{r, \theta, \phi|\hat{R}(d\phi\,\mathbf{k})|\Psi} = \braket{r, \theta, \phi| \left(1 - {i\over\hbar}\hat{L}_z d\phi\right) |\Psi}.\]

Equating the first order terms, we see that

\[\braket{r, \theta, \phi| \hat{L}_z |\Psi} = {\hbar\over i} {\partial\over\partial \phi} \braket{r, \theta, \phi|\Psi},\]

i.e. we can write the position space representation of \(\hat{L}_z\) as (using the notation of Townsend)

\[\hat{L}_z\rightarrow {\hbar\over i}{\partial\over\partial\phi}.\]

This makes sense since we know that the position space representation of linear momentum is \(\hat{\mathbf{p}}\rightarrow (\hbar/i)\mathbf{\nabla}\). The \(z\)-component of angular momentum is \(rp_\phi\), so we could therefore write \(\hat{L}_z \rightarrow (i/\hbar)(\partial/\partial \phi)\).

Now let this act on an eigenstate

\[\braket{r, \theta, \phi| \hat{L}_z |\ell, m} = {\hbar\over i} {\partial\over\partial \phi} \braket{r, \theta, \phi|\Psi} = m \hbar \braket{r, \theta, \phi|\Psi}\]
\[\Rightarrow {\partial\over\partial \phi} \braket{r, \theta, \phi|\Psi} = i m\braket{r, \theta, \phi|\Psi}\Rightarrow \braket{r, \theta, \phi|\Psi}\propto e^{im\phi}.\]

It makes sense that the wavefunction of the angular momentum eigenstate of \(L_z\) is a plane wave in the \(\phi\) direction, representing motion around the \(z\)-axis.

The requirement that \(\psi(r,\theta,\phi)\) be single-valued leads to integer \(m\), since we need to have

\[e^{im(\phi + 2\pi)} = e^{im\phi}.\]

Since \(m\) ranges from \(-\ell\) to \(+\ell\), it follows that \(\ell\) is also integer. The half-integral values are not allowed because we are dealing with orbital angular momentum which involves real space rotations.

Eigenstates of orbital angular momentum in position space#

To get the full picture of the eigenstates of orbital angular momentum, we also need the variation in the \(\theta\) direction. There a couple of ways to get this. One is to write out the components of \(\hat{\mathbf{L}} = \hat{\mathbf{r}}\times\hat{\mathbf{p}}\) in spherical coordinates and compute \(\hat{L}^2\). This is covered in Townsend 9.8 (see eqs. 9.125-9.129). The other thing we can do is go back to the two-body Hamiltonian and recognize that the non-radial kinetic energy must be \(L^2/2\mu r^2\), ie. \((1/2)L^2/I = (1/2)I\Omega^2\) (we’ll see this more formally next time). Since \(\hat{p}^2\rightarrow -(\hbar^2/2\mu)\nabla^2\), we can make the identification

\[\hat{L}^2\rightarrow -\hbar^2 r^2 \nabla^2_\perp\]

where \(\nabla^2_\perp\) is the non-radial part of the Laplacian. This gives

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

(this is equation 9.129 of Townsend). You probably already know that the eigenfunctions of the angular part of the Laplacian are the spherical harmonics \(Y_{\ell m}(\theta,\phi)\) which satisfy

\[r^2\nabla^2 Y_{\ell m}(\theta,\phi) = -\ell(\ell+1) Y_{\ell m}(\theta,\phi),\]

which is exactly what we need for the angular momentum eigenstates since the total angular momentum is \(\ell(\ell+1)\hbar^2\). In addition, the \(\phi\)-dependence is \(Y_{\ell m}\propto e^{im\phi}\) as expected. The angular momentum eigenstates can therefore be written as

\[\braket{\theta,\phi | \ell,m} = Y_{\ell m}(\theta,\phi).\]

Next time we will have a look at these functions! See below for some code to plot them if you want to have a look.

Plotting the spherical harmonics

Here is some Python code to plot the spherical harmonics that we can take a look at in class.

Code
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from scipy.special import sph_harm

def sph2cart(r, phi, theta):
   x = r* np.sin(theta)* np.cos(phi)
   y = r* np.sin(theta)* np.sin(phi)
   z = r* np.cos(theta)
   return x, y, z

phi = np.linspace(0, 2*np.pi, 1000)
theta = np.linspace(0, np.pi, 1000)
phi, theta = np.meshgrid(phi, theta)

l, m = 3,0

Y = sph_harm(m, l, phi, theta)

r = np.abs(Y.real)
#r = np.abs(Y)**2     # this is what is plotted in Townsend

frac = (Y.real - np.min(Y.real)+1e-10)/ (np.max(Y.real)-np.min(Y.real)+1e-10)

x, y, z = sph2cart(r, phi, theta)

fig = plt.figure(figsize=(10,8))
ax = fig.add_subplot(111, projection='3d')
ax.set_aspect('equal')
lim = (-0.5,0.5)
ax.set_xlim(lim)
ax.set_ylim(lim)
ax.set_zlim(lim)
ax.plot_surface(x, y, z, facecolors=cm.cool(frac))

Further reading#

  • Townsend 9.6-9.9.