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# 4.4: Physical Meaning of Angular Quantum Numbers

As was stated in the previous section, separation of variables in spherical coordinates is particularly useful for us, as so many physical systems are based on central forces. We found three quantum numbers ($$n$$, $$l$$, and $$m_l$$), associated with separable stationary-state solutions to the Schrödinger equation with a potential that depends only upon the radial component.  In the 1-dimensional cases the only quantum number was associated with a physical property – the energy level. For the cartesian cases (3D infinite square well and 3D harmonic oscillator), we saw that the three quantum numbers combined together to give the energy level, but didn't really have any special interpretation for each individual quantum number. This is mainly because there aren't any fundamental difference between the $$x$$, $$y$$, and $$z$$ directions. Rotate our coordinate system by $$90^o$$, and just the labels change. But this is not the case for spherical coordinates, as we will see.

#### Orbital Angular Momentum

If we want to interpret quantum-mechanical quantities, it sometimes helps to review what we know from classical mechanics. The one problem we solved in classical mechanics involving a central potential was that of an orbit due to a gravitational attraction, known as the Kepler problem. [You can review some of the details of the solution of this problem here.] A useful approach to this problem is to break the velocity vector of the orbiting body into tangential and radial parts, giving a total energy that breaks up into the kinetic energy due to radial motion, the kinetic energy due to tangential motion, and potential energy. The tangential part oc the kinetic energy can be written in terms of the angular momentum of the orbiting body, giving:

$E_{tot} = KE_{radial} + KE_{tangential} + U = \frac{1}{2}mv_{radial}^2 + \dfrac{L^2}{2mr^2} - \dfrac{GMm}{r}$

Note that only the third term above is specific to gravitation, so if we substitute a generic radial potential and compare this with the hamiltonian acting on the radial wave function in Equation 4.3.15, we find a link between the angular momentum and the quantum number $$l$$:

$L^2=l\left(l+1\right)\hbar^2,\;\;\;\;\;\;\;\; l=0,\;1,\;2,\;\dots$

Assuming this association is correct, it means that besides energies, three-dimensional bound states in quantum mechanics also quantize the magnitude of total angular momentum. We stated in Equation 4.3.14 that the $$l$$ and $$m_l$$ quantum numbers were tied to each other – specifically, the maximum (absolute) value of $$m_l$$ for a given state is limited by whatever the value of $$l$$ happens to be. We don't yet have an interpretation of this, but we can ask if the value of $$l$$ will be similarly limits, and if so, by what?

Clearly the energy levels will be quantized, though we don't know how they will depend upon the three quantum numbers (this will depend upon the specific potential energy function). Returning to our Kepler example, we can show that the total energy of the system puts a limit on the angular momentum it can possess. Suppose we look at the orbiting body at an instant of time when it has a speed $$v$$ and is a distance $$r$$ from the gravitating body. These two quantities totally define the energy of the system – the speed gives us it total kinetic energy, and the separation the potential energy. The kinetic energy is the sum of the radial contribution and the term involving the angular momentum, so if we want to max-out the angular momentum for a given energy, we simply need to choose the state with identically zero radial velocity – circular orbits.

The minimum angular momentum obvious comes from the case of only radial motion (along a straight line through the origin), though one could hardly call such motion an "orbit." Still, if the gravitating body didn't get in the way, one could imagine the orbiting body oscillating back-and-forth through the origin to maximum distances determined by the total energy. It is not difficult to show that the maximum distance from the origin for this case is twice the distance from the origin for the circular orbit case.

Now of course for the Kepler problem, given a fixed total energy, an infinite number of orbits are possible, ranging from circular to linear, with the elliptical orbit having every eccentricity in between. The relationship between $$l$$ and the angular momentum proposed above indicates that there will not be an infinite number of different angular momentum states for a given energy, but like the case of $$m_l$$ with $$l$$, the value of $$l$$ will also be limited. The number $$l$$ is referred to as the orbital angular momentum quantum number, or simply the orbital quantum number.

#### Components of Angular Momentum

It is interesting that we have a quantum number that characterizes the magnitude of the angular momentum, but angular momentum is a vector, so what about its components? Rather than trying to make a classical argument as we did above, let's use what we know from quantum mechanics. We are working in spherical coordinates, which means we have given the $$z$$-axis particular stature (it is the axis from which the polar angle is measured and around which the azimuthal angle is measured), so we'll look at the $$z$$-component of angular momentum. We recall from classical mechanics the definition of this quantity:

$\overrightarrow L \equiv \overrightarrow r \times \overrightarrow p \;\;\; \Rightarrow \;\;\; L_z=x\;p_y - y\;p_x$

A component of angular momentum is an observable quantity, so it has associated with it a self-adjoint operator. We know how to construct such operators from the operators we already know. In the position basis, the operator for $$L_z$$ must be:

$\widehat L_z = \widehat x\;\widehat p_y - \widehat y\;\widehat p_x = -i\hbar\left[ x\dfrac{\partial}{\partial y} - y\dfrac{\partial}{\partial x} \right]$

This doesn't do us much good in cartesian coordinates, since we are currently working in spherical coordinates, so we need to change this operator using a coordinate transformation from $$\left(x,y,z\right)$$ to $$\left(r,\theta,\phi\right)$$. A brief examination of Figure 4.3.1 should be enough to convince the reader that the length of the projection onto the $$\left(x,y\right)$$ plane is $$r\sin\theta$$, and then the components of that in the $$x$$ and $$y$$ directions introduce another factor $$\cos\phi$$ and $$\sin\phi$$, respectively:

$x = r\sin\theta\cos\phi,\;\;\;\;\;\;\;\; y = r\sin\theta\sin\phi$

We also need the derivatives. Doing this part gets a bit tedious, so we'll jump to the pleasingly compact answer:

$\widehat L_z = -i\hbar\;\dfrac{\partial}{\partial \phi}$

Let's see what happens when we operate on the wave function with this operator:

$\widehat L_z \psi_{nlm_l}\left(r,\theta,\phi\right) = -i\hbar\;\dfrac{\partial}{\partial \phi}\;\left[R_{nl}\left(r\right)\Theta_{lm_l}\left(\theta\right)\Phi_{m_L}\left(\phi\right)\right] = -i\hbar\;\dfrac{\partial}{\partial \phi}\;\left[Ae^{im_l\phi}\right]R_{nl}\left(r\right)\Theta_{lm_l}\left(\theta\right) = \left(m_l\hbar\right)\psi_{nlm_l}\left(r,\theta,\phi\right)$

We see that our wave function actually represents an eigenstate of $$\widehat L_z$$, and all we have assumed is that the potential is radial! Well, that is almost all we assumed. We also assumed the wave function was separable in the three spherical components with our selected choice for the $$z$$-axis. This provided a bias, which made what we chose as the $$z$$-direction the direction in which the angular momentum component is an eigenvalue. If we construct the angular momentum component operators $$L_x$$ or $$L_y$$, and operate them on this wave function, we'd find that it is not an eigenstate of these observables. It turns out that there is no way to simultaneously observe any more details about the angular momentum beyond its magnitude (operator $$\widehat {L^2}$$ with eigenvalues $$l\left(l+1\right)\hbar^2$$) and one component that we define as being along the $$z$$-axis (operator $$\widehat L_z$$ with eigenvalues $$m_l\hbar$$).

If we can't know the values of $$L_x$$ or $$L_y$$, that means that neither can be equal to zero, which means that the angular momentum vector can never be entirely along the $$z$$-axis. This is consistent with the eigenvalues, as $$m_l$$ can have a maximum value of only $$l$$, which makes the largest possible $$z$$-component $$l\hbar$$.  Meanwhile, the magnitude of the angular momentum vector is $$\sqrt{l\left(l+1\right)}\hbar$$, which is always greater than $$l\hbar$$. So there is always some (indeterminate) $$x$$ and $$y$$ components of the angular momentum. This may sound a bit strange, but it is also consistent with the uncertainty principle – if all of the angular momentum vector pointed along the $$z$$ direction, then that means that the particle has no component of momentum perpendicular to the $$x$$-$$y$$ plane, and orbits the origin in that plane. That would simultaneously determine the particle's $$z$$-components of momentum and position. [Actually, there is one way that all the components of the angular momentum are known simultaneously – when the total angular momentum is zero.]

We can depict this quantization of angular momentum with the possible angular momentum vectors for a given value of $$l$$ and the associated values of $$m_l$$.  Let's let $$l=1$$. If the angular momentum vector were to lie in the $$y$$-$$z$$ plane, then the diagram below would represent the three possible angular momentum vectors.

Figure 4.4.1 – Angular Momentum Quantization

But the angular momentum cannot be defined so precisely, so the diagrams above are not the best representation. Instead, we need to somehow leave the $$x$$ and $$y$$ components indeterminate. We can do this by saying that the angular momentum vector must lie within the surface of a cone centered around the axis of quantization (i.e. the $$z$$-axis). Of course, for $$m=0$$ the "cone" is a disk...

Figure 4.4.2 – Uncertainty in Components of Angular Momentum

#### Angular Probability Distribution

We might be inclined to think that with a radially-symmetric potential, the probability of finding the particle would also only depend upon the distance from the origin. This is actually true, if the form of the potential is all we know. But with three quantum numbers available, we can know the potential is radially-symmetric, and at the same time know that the particle is in a specific $$l,m_l$$ state.  There is still nothing that we can do to distinguish probability densities at different azimuthal angles, as the probability density for that part of the wave function is just equal to unity:

$\Phi_{m_l}^*\left(\phi\right)\Phi_{m_l}\left(\phi\right)=e^{-im_l\phi}e^{im_l\phi}=1$

But the part of the wave function dependent upon the polar angle is a different matter. This depends on both angular quantum numbers, and even though we can't pinpoint the $$x$$ and $$y$$ components of angular momentum, the fraction of the total angular momentum that is in the $$z$$ direction affects the probability of finding the particle at various polar angles. Perhaps a classical picture – though obviously not quite right – will help visualize this: Suppose the particle is following a circular orbit around the origin, and the direction of the angular momentum of the particle can point in any direction. This vector has a component along the $$z$$ direction. The bigger this component is, the closer the orbit of the particle is to the $$x-y$$ plane.  Also, if there is any component of angular momentum along the $$z$$-axis, then its motion will never intersect the $$z$$-axis. These are properties of the wave functions as well.