Matrix Elements of Angular Momentum

The concept of angular momentum is perhaps the hardest concept to grasp in the first semester of quantum mechanics. At least for me, putting a hat on the symbol of angular momentum does not help me much to understand it. For those interested, I would refer you to J.J. Sakurai’s chapter on rotation in quantum mechanics to see angular momentum from a different perspective. Regardless of understanding the true physical meaning of angular momentum during the first semester, it is very often that you will be asked on manipulating their commutators and matrix elements. Here I will go over some important aspects of angular momentum.

The Ladder Operators

We define the ladder operators in angular momentum as

$$J_\pm \equiv J_x \pm iJ_y$$

And specifically, the have the following commutator relations:

$$\begin{gathered} [J_+, J_-]=2\hbar J_z \\ [J_z, J_\pm] = \pm \hbar J_\pm \\ [J^2, J_\pm] = 0 \end{gathered}$$

What do these ladder operators do exactly? Let’s examine their effect on a ket. But before we do that, we need to specify a basis that we will be using. Recall the following relations

$$J^2 \equiv J_xJ_x + J_yJ_y + J_zJ_z$$

and $J^2$ commutes with all $J_i$‘s; i.e.

$$[J^2, J_i] = 0$$

(To show this, use the cyclic relations of angular momentum.) Since $J^2$ and $J_z$ commute, by definition they share an eigenbasis, and hence we define a basis called $\ket{a, b}$, where $a$ is the eigenvalue of $J^2$, and $b$ is the eigenvalue of $J_z$. Now we can think about what happens when we apply $J_z$ on state $J_\pm \ket{a,b}$.

$$\begin{gathered} J_z(J_\pm \ket{a,b}) = ([J_z, J_\pm] + J_\pm J_z)\ket{a,b} \\ = (\pm \hbar J_\pm + J_\pm J_z)\ket{a,b} \\ = (\pm \hbar J_\pm)\ket{a,b} + b J_\pm \ket{a,b} \\ = (b \pm \hbar)J_\pm \ket{a,b} \end{gathered}$$

Notice that $(b\pm \hbar)$ is a constant, this means that $J_\pm \ket{a,b}$ is still an eigenstate of $J_z$! But now with eigenvalue shifted by $\pm \hbar$. However, $J_\pm$ does not have the “power” to shift the eigenvalues of $J^2$. i.e.

$$\begin{gathered} J^2 J_\pm \ket{a,b} = J\pm J^2\ket{a,b} \\ = a J_\pm \ket{a,b} \end{gathered}$$

To summarize:

$$J_\pm\ket{a,b} = c_\pm \ket{a, b\pm \hbar} \tag{1}$$

Eigenvalues of $J^2$ and $J_z$

To obtain the eigenvalues of $J^2$ and $J_z$, that is, the values for $a$ and $b$, we can play with the above ladder operators. Many online notes show a clear procedure; I would recommend Griffths, or MIT course notes. For the purpose of context, we will quote the results

$$a = \hbar^2 j(j+1)$$ $$b=m\hbar$$

where we call $m$ the magnetic quantum number sometimes. Similarly, $j$ is the angular momentum quantum number. To put it in a context of chemistry, $j$ is similar to the orbital number $l$, e.g. for $p$ orbital $l=1$. It is worth noting that many sources interchange the use of $l$ and $j$, and sometimes it can get confusing. In my understanding, $l$ is the pure orbital quantum number, and $j$ is used when the spin degrees of freedom arise from the inclusion of electrons. In my notes, I use $j$ throughout to be consistent, but do keep this point in mind in the future. Another result is that if $j$ is an integer, then $m$ must also be integers, and if $j$ is a half-integer, then $m$ must also be half-integers. Specifically, $m\in\{-j, -j+1, -j+2, ..., j-1, j\}$. Since we have learned that $j$ and $m$ can explicitly specify $J^2$ and $J_z$‘s eigenvalues, respectively, then we will now rewrite our basis as $\ket{j, m}$, where we have

$$J^2\ket{j,m} = j(j+1)\hbar^2\ket{j,m}$$ $$J_z\ket{j,m} = m\hbar\ket{j,m}$$

Matrix elements

Assuming $\ket{j,m}$ is normalized, we have

$$\braket{j'm'|J^2|j,m} = j(j+1)\hbar^2 \delta_{j'j} \delta_{m'm}$$ $$\braket{j'm'|J_z|j,m} = m\hbar \delta_{j'j} \delta_{m'm}$$

What about the matrix elements for $J_\pm$? First note that, $J^2 - J_z^2 = 1/2(J_+ J_- + J_-J_+)$. Then we can write

$$\begin{gathered} J_-J_+ = 2J^2 -2J_z^2 - J_+J_- \\ = 2J_x^2 +2J_y^2 -J_+J_- \\ = 2J_x^2+2J_y^2-2\hbar J_z - J_-J_+ \\ \implies J_-J_+ = J_x^2 + J_y^2 - 2\hbar J_z \\ = J^2 - J_z^2 -\hbar J_z \end{gathered}$$

(I used the relation $[J_+, J_-] = 2\hbar J_z$ from the second line to third)
So now let’s consider

$$\begin{gathered} \braket{j, m|J_+^\dagger J_+|j,m} = \braket{j,m|J_-J_+|j,m} \\ = \braket{j,m|J^2-J_z^2-\hbar J_z|j,m} \\ = \hbar^2 (j(j+1)-m^2-m) \braket{j,m|j,m} \\ = \hbar^2 (j(j+1)-m^2-m) \end{gathered}$$

Furthermore, we also know that $J_+\ket{j,m} = c_+\ket{j, m+1}$ must be normalized, and its bra is $\bra{j,m}J_+^\dagger$, so we have

$$\begin{gathered} \braket{j, m|J_+^\dagger J_+|j,m} = |c_+|^2 \\ =\hbar^2(j(j+1)-m^2-m) \\ =\hbar^2(j(j+1) - m(m+1)) \\ =\hbar^2(j-m)(j+m+1) \end{gathered}$$

We could just take the square root of $|c_+|^2$ to get $c_+$; usually, we get a phase factor of $\exp{-i\theta}$, but it does not affect physical significance, so we just let $\theta=0$, and we arrive at:

$$c_+ = \sqrt{(j-m)(j+m+1)}\hbar \tag{2}$$

Similarly,

$$c_- = \sqrt{(j+m)(j-m+1)}\hbar \tag{3}$$

Altogether,

$$\braket{j'm'|J_\pm|j,m} = \sqrt{(j\mp m)(j_\pm m+1)}\hbar \delta_{j'j} \delta{m'm+1} \tag{4}$$

A Simple Example of Adding Angular Momentum

Let’s consider 2 electrons in the 1s orbital, so we would not need to consider orbitals degree of freedom here. Here we will also substitute the previous notation of $J$ to $S$ for explicitly showing that we are only talking about spin. In this system, we have a total spin operator: $S = s_1 + s_2$, which is simply the sum of the individual spin operators of the electrons. The total spin operator satisfies the same rules as the individual spin operators, such as

$$[S_x, S_y] = i\hbar S_z$$

The relevant eigenvalues are:

$$S^2 \implies S(S+1)\hbar^2$$ $$S_z \implies m\hbar$$ $$s_{1z} \implies m_1\hbar$$ $$s_{2z} \implies m_2\hbar$$

The question is, how do we represent our basis so these eigenvalues have corresponding eigenvectors? Previously, we write \ket{j,m}, but that really only applies to one particle. To write a new basis, think of which operators commute? We have two options, $S^2$ and $S_z$, or $s_{1z}$ and $s_{2z}$. In fact, both representation works, but often the former representation conveys more information of a total system, whereas the second representation is more of simply combining the pieces of the individual subsystems.

Two representations

i) $\{m_1, m_2\}$ representation that is based on the eigenkets of $s_{1z}$ and $s_{2z}$. In case of electrons: $m_1 = \pm 1/2$, $m_2 = \pm 1/2$. Representing $+$ as $+1/2$ and $-$ as $-1/2$, we have 4 possible combination: $\ket{+,-}$, $\ket{+,+}$, $\ket{-,+}$, $\ket{-,-}$

ii) $\{S,m\}$ representation that is based on the eigenkets of $S^2$ and $S_z$. In our case here, we have possible states: $\ket{S=1, m=\pm1, 0}$, $\ket{S=0, m=0}$. (Recall that $m=-S, -S+1,...,S-1, S$)

To connect the two representations, $\ket{S=1,m=1} = \ket{+,+}$, but what is $\ket{S=1, m=0}$? To explicitly obtain the transformation, we would need to apply the ladder operator. Recall that the ladder operator in a combined system is $S_- \equiv S_{1-} + S_{2-}$, and we have

$$\begin{gathered} S_- \ket{+,+} \\ = (S_{1-} + S_{2-})\ket{+,+} \\ = (a\ket{-,+} + b\ket{+,-}) \end{gathered}$$

(note that $S_{1-}$ would only apply onto the first system, etc.)

The coefficients $a$ and $b$ are the same coefficients you would obtain when you apply the ladder operators on eigenstates of the angular momentum… i.e. $c_- = \sqrt{(j+m)(j-m+1)}\hbar$.

Observe the effect of applying the ladder operator on both sides:

$$\sqrt{(1+1)(1-1+1)}\ket{S=1,m=0} = \sqrt{(1/2+1/2)(1/2-1/2+1)}\ket{-,+} + \sqrt{(1/2+1/2)(1/2-1/2+1)}\ket{+,-}$$ $$\sqrt{2} \ket{S=1,m=0} = \ket{-,+} + \ket{+,-}$$ $$\ket{S=1,m=0} = \frac{1}{\sqrt{2}}( \ket{-,+} + \ket{+,-})$$

We would then repeat the same procedure to obtain $\ket{S=1,m=-1} = \ket{-,-}$
Obviously, as the system gets more complicated, this procedure could get tedious. But the coefficients that connect the two representations above are computed already, and they are called the Clebsh-Gordan coefficients! Essentially, when asked to add angular momentums, it is to represent the $\{S, m\}$ representation in terms of the $\{m_1, m_2,...\}$ representations.