5.4 Angular Momentum

Classically, angular momentum is defined as

It is a useful quantity to study in classical mechanics because it is conserved for any particle under the influence of a central force such that the Coulomb force. We should therefore expect that angular momentum is conserved for the electron in hydrogen as well, so let’s investigate.

Quantum mechanically, we define angular momentum in the same way it is defined classically, except using (vector) operators instead of just vectors.

Recall that ${\widehat{r}}_i{\widehat{p}}_j$ commute as long as $i\ne j$, so the ordering does not matter when writing out the components. However, given that we cannot know all components of $\widehat{𝒓}$ and $\widehat{𝒑}$ simultaneously due to the uncertainty principle, it seems like we will not be able to know all components of angular momentum simultaneously either.

Let us test this out by finding some commutators. Note that as a general tip for calculating commutators, it is almost always useful to use standard commutator identities to simplify as much as possible before expanding the commutators themselves. For example, when finding the commutator of ${\widehat{L}}_x$ and ${\widehat{L}}_y$, we use the fact that

to write

The last two terms contain only things that all commute with each other, so they must be zero. Similarly, in the first two terms involving $y$ and $x$ commute with everything else in their commutators, so we can pull them out as constants:

Now we can insert the value of the commutator that we know, $[\widehat{z},{\widehat{p}}_z]=i\mathrm{\hslash }$, to get

Hence the $x$ and $y$ components of angular momentum do not commute and we cannot know them both simultaneously. Calculating the commutators between the other components, we find the same outcome. The values are

Notice that these commutators follow a cyclic pattern under the exchange of labels $x$, $y$, and $z$. This is just a consequence of the definition of $\widehat{𝑳}$ as a cross product.

What properties can be know about the electron’s angular momentum simultaneously? What about one of the components and the total magnitude? To try this, we calculate the commutator of one of the components with the operator ${\widehat{L}}^2$, which is defined as

So the commutator between ${\widehat{L}}_x$ and ${\widehat{L}}^2$ is

The first term will be zero, and using the identity

the second and third terms can be written as

Hence we can know the total magnitude of angular momentum and one of its components simultaneously, as one can show similarly that

It can also be shown that ${\widehat{L}}^2$ and the components of angular momentum also all commute with the Hamiltonian:

meaning that the eigenfunctions of $\widehat{H}$ are also simultaneously eigenfunctions of ${\widehat{L}}^2$ and one component of angular momentum, conventionally taken to be ${\widehat{L}}_z$. This is a maximal set of mutually commuting operators, a set of operators where each one commutes with all others.

Let us find the eigenvalues of the ${\widehat{L}}^2$ and ${\widehat{L}}_z$ operators. Given that angular momentum is a quantity relating to rotations, it is reasonable to assume that it will act on the angular part of the wavefunction. For this reason, we will write out the angular momentum operators explicitly in terms of angular position coordinates.

In the position basis, we can write

with similar results for ${\widehat{L}}_y$ and ${\widehat{L}}_z$. Using these and the chain rule, e.g.

we get the following results for angular momentum operators in the position basis:

So we were right to assume that no derivatives of $r$ would appear in these operators.

Notice that equation 5.99 looks very similar to the angular equation for the hydrogen atom electron (equation 5.11). In fact, if we take the angular equation and multiply it by $-{\mathrm{\hslash }}^2{Y}_{\mathrm{\ell }}^m(\theta ,\varphi )$, bearing in mind that $Y_{\mathrm{\ell }}^m(\theta ,\varphi )=\mathrm{\Theta }(\theta )\mathrm{\Phi }(\varphi )$ since the spherical harmonics were the solution to the angular equation, we get

where we have substituted ${\lambda }^2$ for $\mathrm{\ell }(\mathrm{\ell }+1)$. This looks like an eigenvalue equation, and in fact it looks exactly like what we would get if we acted ${\widehat{L}}^2$ on a wavefunction:

therefore the eigenfunctions of ${\widehat{L}}^2$ must be the spherical harmonics, which makes sense since they are the angular part of the eigenfunctions of the hydrogen Hamiltonian, and their eigenvalues are ${\mathrm{\hslash }}^2\mathrm{\ell }(\mathrm{\ell }+1)$.

This finally explains why $\mathrm{\ell }$ is called the orbital angular momentum quantum number, as it determines the magnitude of the angular momemtum of an electron in a certain orbital. An electron in a state with a given $\mathrm{\ell }$ will have orbital angular momentum with magnitude

This implies that the angular momentum of the electron is forced to be quantised. Depending on the total energy the electron has, which is determined by the principal quantum number $n$, the angular momentum can only have a magnitude given by the eigenvalue equation above for the allowed values of $\mathrm{\ell }$. For example, if an electron is sitting in the 2^nd excited state ($n=3$), then the magnitude of its angular momentum can either be $\mathrm{\hslash }$, $\sqrt{2}\mathrm{\hslash }$, or $\sqrt{6}\mathrm{\hslash }$.

The spherical harmonics are also eigenfunctions of ${\widehat{L}}_z$, which we can calculate explicitly by writing

So the allowed values of the $z$-component of angular momentum are

If we consider again the electron in the 2^nd excited state and suppose it has $\mathrm{\ell }=2$, the allowed values for the $z$-component of its angular momentum are $-2\mathrm{\hslash }$, $-\mathrm{\hslash }$, $0$, $\mathrm{\hslash }$, and $2\mathrm{\hslash }$. So the magnetic quantum number $m$ determines the alignment of the electron’s angular momentum with the $z$-axis, the larger the magnitude of $m$, the more aligned they are. Note that $L_z$ can never be as large as the total magnitude of angular momentum, i.e. the angular momentum can never point exactly along $z$. This is because if $L_z=L$, then we know that $L_x=L_y=0$, which is not allowed! However, as $\mathrm{\ell }$ gets larger, the maximum allowed value of $L_z$ gets closer and closer to $L$.

We can visualise the allowed values of angular momentum for a given state in the 3D space of angular momentum with $L_x$, $L_y$, and $L_z$ each on their own axis. The angular momentum vector has a fixed length determined by $\mathrm{\ell }$, and the different allowed values of $m$ sweep out cones around the $z$-axis since the values of $L_x$ and $L_y$ are unknown.

Note that this does not imply that $L_x$ and $L_y$ are allowed to take on a continuous set of values, they are also quantised. The cones are simply a visual aid which denotes that the values of $L_x$ and $L_y$ are unknown.

The angle between $𝑳$ and the $z$-axis in this diagram can be written

The minimal angle occurs for maximal $L_z$, which occurs when $|m|=\mathrm{\ell }$.

This implies that

as we have seen, however as $\mathrm{\ell }\to \mathrm{\infty }$, we get

which is what we expect from classical physics.