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# Angular Momentum in Quantum Mechanics: Commutators and Eigenvalues

The angular momentum (more precisely: orbit angular momentum) $$\class{green}{\boldsymbol{L}}$$ of a classical particle is given by the cross product between the distance $$\boldsymbol{r}$$ of the particle from the axis of rotation and the particle momentum $$\boldsymbol{p}$$:

Thus, the angular momentum is perpendicular to $$\boldsymbol{r}$$ and $$\boldsymbol{p}$$ because of the cross product and has the following three components, each in one spatial direction:

1. Angular momentum component in $$x$$-direction:

2. Angular momentum component in $$y$$ direction:

3. Angular momentum component in $$z$$ direction:

## Angular momentum commutators

Using a commutator $$[A,\, B]$$, we can directly determine whether the quantities (observables) $$A$$ and $$B$$ in quantum mechanics are simultaneously measurable, in principle, with arbitrary accuracy. If the commutator of two observables does not vanish, the uncertainty relation applies to them.

In classical mechanics places $$x,y,z$$ as well as momentum $$p_{\text x}, p_{\text y}, p_{\text z}$$ but also angular momentum components $$L_{\text x}, L_{\text y}, L_{\text z}$$ are scalar quantities, i.e. 'ordinary' numbers, which do not make any difference if you multiply this way $$z \, p_{\text z}$$ or that way $$p_{\text z} \, z$$. The commutator of two quantities gives the deviation when two quantities are multiplied in reversed order. It is obvious that ordinary numbers are commutative. Therefore, all commutators with classical locations, momenta and angular momentum vanish. Physically it means that we could measure the classical angular momentum components in principle arbitrarily exactly in an experiment. The uncertainty relation does not apply to them.

In quantum mechanics, on the other hand, things are quite different, because here we are interested in operators with the help of which (and with the help of the wave function) we can predict the measurement results. Let us therefore consider the angular momentum components 2, 3, and 4 as operators.

To do this, we replace the momentum occurring in angular momentum components 2, 3, and 4 with the following axiomatic assignments:

• Momentum component $$p_{\text x}$$ becomes operator: $$\hat{p}_{\text x} ~=~ -\mathrm{i} \, \hbar \partial_x$$.
• Momentum component $$p_{\text y}$$ becomes operator: $$\hat{p}_{\text y} ~=~ -\mathrm{i} \, \hbar \partial_y$$.
• Momentum component $$p_{\text z}$$ becomes operator: $$\hat{p}_{\text z} ~=~ -\mathrm{i} \, \hbar \partial_z$$.

Here $$\mathrm{i}$$ is the imaginary unit, $$\hbar$$ is reduced Planck constant and $$\partial_x$$ is the derivative operator which, applied to a function, gives the derivative of this function with respect to $$x$$. Of course, a stand-alone derivative makes no sense. Therefore an operator unfolds its effect only if it is applied to a function. The result is a new function modified by the operator.

What about the $$x$$, $$y$$ and $$z$$ positions? Don't they also have to be replaced somehow? Of course, the positions are now also operators $$\hat{x}$$, $$\hat y$$ and $$\hat z$$. But since we work here in the so-called position space, the positions remain unmodified. The operator $$\hat{x}$$ applied to a function $$f$$ simply yields a multiple of the function: $$\hat{x}\,f ~=~ x \, f$$. By the way, in momentum space it would be the other way around. Here the positions must be replaced, while the momenta remain unchanged.

Since we now understand positions, momenta and angular momenta as operators, you have to be careful now. You cannot handle operators as carelessly as numbers. Therefore it is sometimes useful to put hats on the momenta $$p_{\text x}$$, $$p_{\text y}$$ and $$p_{\text z}$$ to mark them as operators: $$\hat{p}_{\text x}$$, $$\hat{p}_{\text y}$$ und $$\hat{p}_{\text z}$$. Analogously, you would also have to provide positions and angular momenta with little hats. Since they make the equations look a bit more cluttered, I omit the hats. We have agreed that we have now switched to the operators.

A useful property of the commutator is that it is distributive:

We can exploit this property to write down the commutator of two components of angular momentum in a useful way for the further procedure:

• Commutator of $$L_{\text x}$$ and $$L_{\text y}$$:

• Commutator of $$L_{\text y}$$ and $$L_{\text z}$$:

• Commutator of $$L_{\text y}$$ and $$L_{\text z}$$:

Swapping the components, for example $$[ L_{\text x}, \, L_{\text y} ]$$ to $$[ L_{\text y}, \, L_{\text x} ]$$ just gives a minus sign. This is the antisymmetric property of the commutator.

Let's break down the commutators a little further. For this we use another property of the commutator, namely the product rule:

Let us first look at the first angular momentum commutator 6. Let us apply the product rule 9 twice to each summand. Let's start with the first summand of 6:

The first term in 10 vanishes because $$[ y, \, z ] = 0$$. The third and fourth terms also vanish since $$[ p_{\text z}, \, p_{\text x} ] = 0$$ and $$[ y, \, p_{\text x} ] = 0$$. You can easily check this, e.g. for the term $$[ p_{\text z}, \, p_{\text x} ] ~= -\hbar^2 \partial_{\text z} \partial_{\text x} + \hbar^2 \partial_{\text x} \partial_{\text z} = 0$$, because it doesn't matter whether you first partially differentiate a function with respect to $$x$$ or with respect to $$z$$. The partial derivatives can be interchanged so that the two terms cancel exactly.

The term that does not vanish in 10 contains the commutator of the position $$z$$ and momentum $$p_{\text z}$$. Hopefully you already know this commutator. If not, you can easily calculate it by applying it to some test function $$f$$:

So the commutator must be $$[ p_{\text z}, \, z ] = -\mathrm{i} \, \hbar$$. Of course, the commutator does not vanish for other positions $$x$$ and $$y$$ if the momentum points in the same direction: $$[ p_{\text x}, \, x ] = -\mathrm{i} \, \hbar$$ und $$[ p_{\text y}, \, y ] = -\mathrm{i} \, \hbar$$.

Let's move on to the next summand in 6. For the second summand we get:

You can say that right away without having to do much math, since there is no $$z$$ in this commutator. Indeed, we know that only momenta and positions result in a non-vanishing commutator if the momentum points along this direction. We can also apply this directly to the third summand in 6. It is also zero, because in this commutator the momenta do not point along $$z$$:

The last summand in 6, on the other hand, does not vanish, since noncommutative $$z$$ and $$p_{\text z}$$ appear there:

Overall, the angular momentum commutator 6, together with the non-vanishing terms 10 and 14 become:

The expression in the parenthesis is exactly the $$L_{\text z}$$ component of the angular momentum, as seen in 4. Thus we have:

Similarly you proceed to find out the other two commutators 7 and 8. As a result you get:

For example, if you remember the commutator 16, you can easily get the other two commutators by cyclically permuting the indices.

If you like, you can compactly combine the three angular momentum commutators 16, 17 and 18. To do this, we replace the position labels $$x,y,z$$ with three indices $$i, j, k$$, which can take the values 1, 2, and 3. For any two angular momentum components $$L_i$$ and $$L_j$$ the commutator is then:

Here $$\varepsilon_{ijk}$$ is the Levi-Civita Symbol, which is always $$\varepsilon_{ijk} = 1$$ when indices $$ijk$$ are rotated (all indices are permuted). And the symbol is $$\varepsilon_{ijk} = -1$$ exactly if only two indices are swapped. The symbol is $$\varepsilon_{ijk} = 0$$ if at least two indices are equal. For example, this results in the following: $$[ L_1, \, L_2 ] ~=~ \mathrm{i}\hbar \, L_3$$ or $$[ L_2, \, L_1 ] ~=~ -\mathrm{i}\hbar \, L_3$$ or $$[ L_1, \, L_1 ] ~=~ 0$$ and so on.

With the Levi-Civita symbol, we can represent all possible (16 in total) angular momentum commutators and do it in a single equation! Isn't that handy?

Using the angular momentum commutators, we can also say that there are no functions $$Y$$ that are simultaneous eigenfunctions of $$L_{\text z}$$ and $$L_{\text y}$$. So it is not possible that the following is valid: $$L_{\text z} \, Y = \lambda \, Y$$ AND simultaneously $$L_{\text y} \, Y = \kappa \, Y$$, for all eigenfunctions $$Y$$ of $$L_{\text z}$$. Two operators which do not commute have no common eigenfunctions.

Common eigenfunctions exist, for example, for $$L_{\text z}$$ and the square of angular momentum operator $$L^2 = {L_{\text x}}^2 + {L_{\text y}}^2 + {L_{\text z}}^2$$. Thus, the commutator of $$L^2$$ and $$L_{\text z}$$ vanishes:

Apply product rule 9. Thereby commutator $$[ {L_{\text z}}^2, \, L_{\text z} ]$$ with equal components is eliminated:

Now insert angular momentum commutators 16 and 17:

By the way, the commutator with $$L^2$$ vanishes for all angular momentum components:

Since the commutator of $$L^2$$ and an angular momentum component $$L_{\text z}$$ vanishes, eigenfunctions $$Y$$ exist which are simultaneously eigenfunctions of $$L_{\text z}$$ and $$L^2$$:

Here $$\mu$$ is the eigenvalue of $$L^2$$ and $$\lambda$$ is the eigenvalue of $$L_{\text z}$$. Why do we need $$L^2$$ at all? Well, this operator will help us to find out the eigenvalues, but also the eigenfunctions $$Y$$ concretely.

Ladder operator method is a powerful algebraic method that allows you to determine the eigenvalues $$\lambda$$ of $$L_{\text z}$$ WITHOUT having to know eigenfunctions $$Y$$. We define the ladder operators $$L_+$$ and $$L_-$$ as follows:

First, let's calculate four useful commutators, which we will exploit immediately afterwards:

• Commutator of $$L_{\text z}$$ and $$L_+$$:

• Commutator of $$L_{\text z}$$ and $$L_-$$:

• Commutator of $$L^2$$ and $$L_+$$:

All commutators containing the same components are dropped:

Apply product rule:

Insert angular momentum commutators:

• Commutator of $$L^2$$ and $$L_-$$:
You can work it out analogously to 28.

So far so good. We know that $$Y$$ is an eigenfunction of both $$L^2$$ and $$L_{\text z}$$. If we apply our defined ladder operator $$L_+$$ to $$Y$$, we get a new modified function: $$(L_+ \, Y)$$. Next, we can ask whether $$(L_+ \, Y)$$ is also an eigenfunction of $$L^2$$ and $$L_{\text z}$$. Let's check this:

After the first equal sign in Eq. 33 we have exchanged $$L_+$$ and $$L^2$$. We are allowed to do so, because in 31 we have tediously calculated that their commutator vanishes. Is the commutator of two operators equal to zero? Then we may swap both operators as much as we want. With 33 we can confirm that $$(L_+ \, Y)$$ is an eigenfunction of $$L^2$$. We get the same function $$(L_+ \, Y)$$, scaled with the eigenvalue $$\mu$$.

Also $$(L_- \, Y)$$ is an eigenfunction of $$L^2$$, as we can easily verify:

Using the previously calculated commutators between $$L_{\text z}$$ and $$L_{\pm}$$, we can easily prove that $$L_+\,Y$$ and $$L_-\,Y$$ are also eigenfunctions of $$L_{\text z}$$. But beware! Here the commutators, as we calculated in 26 and 27, are not zero. So, for example, if we swap $$L_{\text z}$$ and $$L_+$$, then we need to add the associated commutator 26 to compensate for the difference in swapping. With this in mind, we get:

Have you noticed how useful commutators are when working with operators?

With Eq. 35 we have shown that $$(L_+ \, Y)$$ is an eigenfunction of $$L_{\text z}$$, with an increased eigenvalue $$\lambda + \hbar$$.

Also, reapplying $$L_+$$ to $$(L_+ \, Y)$$ increases the eigenvalue $$\lambda + \hbar$$ by another $$\hbar$$:

And so on... Since $$L_+$$ increases the eigenvalue of $$L_{\text z}$$ by $$+\hbar$$ for each further application to $$Y$$, we call $$L_+$$ an raising operator. Its "opposite" is the lowering operator $$L_-$$.

We can quickly check this with the help of their commutator 27:

You will encounter the raising and lowering operators not only in the case of quantum mechanical angular momentum, but also, for example, in the case of the quantum mechanical harmonic oscillator or in quantum field theory.

## Eigenvalues of Lz and L²

If we apply the raising operator $$L_+$$ to $$Y$$ again and again, we increase the eigenvalue $$\lambda$$ by $$+\hbar$$ each time. Can the new eigenvalue become arbitrarily large if we apply $$L_+$$ incessantly?

In quantum mechanics, the eigenvalue represents, physically speaking, a possible measured value of the angular momentum component in the $$z$$ direction. A closed system, for example being an H atom, has an finite angular momentum $$\langle L \rangle$$. This total angular momentum is preserved. The eigenvalue $$\lambda$$ representing its $$L_{\text z}$$ component cannot become larger than the total angular momentum. So there must be a topmost' eigenfunction $$\overline{Y\,} := L_+...L_+ \, Y$$ of $$L_{\text z}$$ which is not an eigenfunction of $$L_+$$:

The eigenvalue $$\lambda$$ of $$L_{\text z}$$ is maximal when applied to the topmost' eigenfunction $$\overline{Y\,}$$. We write this maximum eigenvalue as a multiple of $$\hbar$$, namely as $$\lambda = l\, \hbar$$:

We do not know the quantity $$l$$. We still have to find out how it behaves exactly.

And, if we apply $$L^2$$ to $$\overline{Y\,}$$, we still get the eigenvalue $$\mu$$ because we saw in Eq. 33 that $$L_+$$ does not change the eigenvalue of $$L^2$$ at all:

Let us express the eigenvalue $$\mu$$ of $$L^2$$ with the maximum eigenvalue $$l \, \hbar$$ of $$L_{\text z}$$. For this purpose we first determine $$L_-\,L_+$$:

Rearrange Eq. 41 for $$L^2$$:

Let us apply the transformed operator $$L^2$$ to the topmost' eigenfunction $$\overline{Y\,}$$:

The term $$L_- \, (L_+ \, \overline{Y\,}) = 0$$ vanishes according to Eq. 38, since $$\overline{Y\,}$$ is already the topmost' eigenfunction. A further application of $$L_+$$ eliminates the term. And, what yields $$L_{\text z} \, \overline{Y\,}$$ is in Eq. 39. After substituting the results of Eq. 39 into Eq. 43 we get:

The eigenvalue $$\mu$$ of $$L^2$$ expressed with $$l$$ is: $$\mu ~=~ l \, (l ~+~ 1) \, \hbar^2$$.

Now what happens if we excessively apply the lowering operator $$L_-$$ to $$Y$$: $$L_- ... L_-\,Y$$? At some point we land or cross zero as a possible eigenvalue and the eigenvalue becomes negative. Can the angular momentum eigenvalue $$\lambda$$ become negative arbitrarily large? It cannot become negative arbitrarily large for the same reason as it cannot become positive arbitrarily large. The eigenvalue of the $$L_{\text z}$$ component, which is maximum in magnitude but negative, cannot exceed the total angular momentum.

Let us denote the 'lowest' eigenfunction of $$L_{\text z}$$ as $$\underline{Y}$$ and the smallest eigenvalue as $$\underline{l} \, \hbar$$. The lowering operator $$L_-$$ is not an eigenfunction of $$\underline{Y}$$:

And the eigenvalue equations for $$L_{\text z}$$ and $$L^2$$ are:

Let us first calculate $$L_+ \, L_-$$, analogously to Eq. 41. This time, however, the ladder operators are reversed:

Let us use Eq. 47, rearranged for $$L^2$$, to express the eigenvalue $$\mu$$ of $$L^2$$ with the lowest eigenvalue $$\underline{l} \, \hbar$$ of $$L_{\text z}$$:

Thus the eigenvalue ist $$\mu ~=~ \underline{l} \, (\underline{l} ~-~ 1) \, \hbar^2$$. This eigenvalue is not changed by ladder operators, that is, the $$\underline{l} \, \hbar$$ eigenvalue expressed with lowest eigenvalue $$\underline{l} \, \hbar$$ is still the same eigenvalue if we express it with the highest eigenvalue $$l \, \hbar$$ of $$L_{\text z}$$. When we expressed $$\mu$$ with the highest eigenvalue in 44, we had gotten $$\mu ~=~ l \, (l ~+~ 1) \, \hbar^2$$. Both expressions must be the same:

Here are two possibilities for $$\underline{l}$$. Either $$\underline{l} ~=~ l ~+~ 1$$ or $$\underline{l} ~=~ -l$$. The first solution would mean that the lowest eigenvalue of $$L_{\text z}$$ is larger than the highest eigenvalue with $$l$$, which makes little sense. On the other hand, the second solution $$\underline{l} ~=~ -l$$ is plausible and therefore we continue to use this solution.

Let's keep our results together:

1. The highest eigenvalue of $$L_{\text z}$$ is: $$l \, \hbar$$.
2. The lowest eigenvalue of $$L_{\text z}$$ is: $$-l \, \hbar$$.
3. We get from one eigenvalue to another by increasing or decreasing the eigenvalue by $$\hbar$$, staying between the highest and lowest eigenvalues.
4. The eigenvalue of $$L^2$$ is expressed with the highest $$L_{\text z}$$ eigenvalue: $$\mu = l \,(l ~+~ 1) \, \hbar^2$$.

The $$L_{\text z}$$ angular momentum component has eigenvalues that are multiples of $$\hbar$$: $$\lambda = \class{violet}{m} \, \hbar$$. The value of $$\class{violet}{m}$$ ranges from $$-l$$ to $$l$$:

Here $$\class{violet}{m}$$ runs in integer steps (two neighboring eigenvalues differ by $$\hbar$$). If there are total $$N$$ values between $$-l$$ and $$l$$, then it follows from $$l ~=~ -l ~+~ N$$ that: $$l = N/2$$. For example, for $$N=5$$ values, $$l = 2.5$$ would be half-integer and for $$N = 10$$, $$l = 5$$ would be integer. The half-integer values of $$l$$ describe the spin states of quantum particles and integer values of $$l$$ describe orbital angular momentum of quantum particles:

For each value of $$\class{violet}{m}$$ there is an eigenfunction $$Y_{l}^{\class{violet}{m}}$$. With the help of the ladder operators we have found out eigenvalues of $$L^2$$ and $$L_{\text z}$$. In the next lesson we will determine the eigenfunctions $$Y_{l}^{\class{violet}{m}}$$. As it will turn out: They are spherical harmonics!