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Addition of angular momentum

Combining angular momenta: allowed totals, the triplet and singlet, Clebsch–Gordan coefficients, spin–orbit fine structure and hyperfine lines.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to combine two angular momenta, build the triplet and singlet, use $\vec{J}_1 \cdot \vec{J}_2$ to find fine and hyperfine splittings, and read Clebsch–Gordan coefficients as probabilities.

2. What you already have

You know the eigenvalues of a single angular momentum, $j(j + 1)\hbar^2$ and $m\hbar$, the ladder operators, and spin-½. You know from hydrogen that each level holds several states. Real systems usually have more than one angular momentum — two electrons' spins, or an electron's orbit and its spin — and this lesson shows how they combine.

3. Words for this lesson

4. Adding quantum vectors

Two angular momenta $\hat{\vec{J}}_1$ and $\hat{\vec{J}}_2$ that act on different parts of a system commute with each other, and their sum $\hat{\vec{J}} = \hat{\vec{J}}_1 + \hat{\vec{J}}_2$ obeys the angular momentum algebra. So the total has eigenvalues $j(j + 1)\hbar^2$ and $m\hbar$, and the rules for combining are:

$$m = m_1 + m_2, \qquad j = |j_1 - j_2|, \ |j_1 - j_2| + 1, \ \ldots, \ j_1 + j_2.$$

Each $j$ contributes $2j + 1$ states, and together they account for all $(2j_1 + 1)(2j_2 + 1)$ product states. The states of definite $j$ — the coupled basis — are superpositions of product states with the same $m$, and the amplitudes are the Clebsch–Gordan coefficients.

The most important case is two spin-½ particles. Their four product states regroup into a triplet with $S = 1$,

$$|1, 1\rangle = \uparrow\uparrow, \quad |1, 0\rangle = \tfrac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow), \quad |1, -1\rangle = \downarrow\downarrow,$$

and a singlet with $S = 0$, $|0, 0\rangle = \tfrac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$. Because $S^2 = S_1^2 + S_2^2 + 2\vec{S}_1 \cdot \vec{S}_2$, the dot product is $\tfrac{1}{4}\hbar^2$ in the triplet and $-\tfrac{3}{4}\hbar^2$ in the singlet. Any energy proportional to $\vec{S}_1 \cdot \vec{S}_2$ — the hyperfine interaction, or the exchange interaction in magnets — splits the two by an amount set entirely by this algebra.

5. The method, step by step, and how to check it

1. Allowed totals. Subtract and add the two quantum numbers; step by one. Two half-integers give integers; an integer and a half-integer give half-integers.
2. Count states. The sum of $2j + 1$ over the allowed $j$ must equal $(2j_1 + 1)(2j_2 + 1)$.
3. Energies from dot products. Whenever an interaction is proportional to $\vec{J}_1 \cdot \vec{J}_2$, write it as $\tfrac{\hbar^2}{2}[j(j + 1) - j_1(j_1 + 1) - j_2(j_2 + 1)]$; the coupled states are then its eigenstates.
4. Probabilities. Expand a coupled state in product states; the squared coefficients give the probabilities of $m_1$ and $m_2$.

Checks. The largest $m$ must be $j_1 + j_2$, reached by only one product state, which must therefore be the top of the highest multiplet. Clebsch–Gordan probabilities for a given coupled state must add to one. Energy splittings from $\vec{L} \cdot \vec{S}$ must satisfy the interval rule: adjacent levels differ in proportion to the larger $j$.

6. Why $m$ adds but $j$ does not

The $z$ component of the total is the operator sum $\hat{J}_z = \hat{J}_{1z} + \hat{J}_{2z}$, and a product state is an eigenstate of both terms, so its $m$ is just $m_1 + m_2$. The magnitude is different: $\hat{J}^2 = \hat{J}_1^2 + \hat{J}_2^2 + 2\hat{\vec{J}}_1 \cdot \hat{\vec{J}}_2$ contains $\hat{J}_{1x}\hat{J}_{2x} + \hat{J}_{1y}\hat{J}_{2y}$, which do not have sharp values in a product state. They flip one angular momentum up while flipping the other down, mixing product states with the same $m$.

So product states generally do not have definite $j$, and coupled states do not have definite $m_1$ and $m_2$. Which basis to use depends on what the Hamiltonian contains. With a strong external magnetic field, which couples to each part separately, the product basis is natural; with spin–orbit coupling, which involves $\vec{L} \cdot \vec{S}$, the coupled basis is. When both matter, the levels are found by diagonalizing a small matrix, as in the two-state lesson.

7. Fine structure and the $j$ quantum number

An electron moving through the nucleus's electric field sees, in its own frame, a magnetic field proportional to its orbital angular momentum, and its spin's magnetic moment interacts with it. The energy is $A\,\vec{L} \cdot \vec{S}$, which commutes with $J^2$ but not with $L_z$ or $S_z$ separately, so the good quantum numbers become $l$, $s$, $j$ and $m_j$. For one electron, $j = l \pm \tfrac{1}{2}$, and

$$\vec{L} \cdot \vec{S} = \frac{\hbar^2}{2}\left[j(j + 1) - l(l + 1) - \tfrac{3}{4}\right] = \begin{cases} \tfrac{l}{2}\hbar^2 & j = l + \tfrac{1}{2} \\ -\tfrac{l + 1}{2}\hbar^2 & j = l - \tfrac{1}{2}\end{cases}$$

The two levels split by $A(2l + 1)/2$. Spectroscopists label them with term symbols such as $3p_{3/2}$ and $3p_{1/2}$. The splitting grows rapidly with the nuclear charge — about $2$ meV in sodium, $69$ meV in cesium — which is why heavy atoms have widely separated doublets and light ones nearly overlapping lines.

8. The singlet is special

The singlet $(\uparrow\downarrow - \downarrow\uparrow)/\sqrt{2}$ has total spin zero: it looks the same from every direction. Measure both spins along any axis and they always come out opposite. That makes it the prototype of an entangled state — neither particle has a definite spin of its own, yet their results are perfectly anticorrelated — and it is the state used in the Bell tests of the last lesson.

The singlet is also antisymmetric under exchange of the two particles, while the triplet states are symmetric. For two electrons, whose total state must be antisymmetric, a spin singlet pairs with a symmetric spatial state and a triplet with an antisymmetric one. This link between spin and spatial shape, developed in lesson 21, makes the singlet the bonding state of the hydrogen molecule and the triplet the basis of ferromagnetism.

9. Hyperfine structure

Nuclei have spin too, and the interaction of an electron's spin with the nucleus's is proportional to $\vec{S}_e \cdot \vec{I}$. In hydrogen's ground state, with $s = \tfrac{1}{2}$ and $I = \tfrac{1}{2}$, the total $F$ is $1$ or $0$, and the two levels are split by $5.874$ μeV. A spin flip from $F = 1$ to $F = 0$ emits the $21$ cm radio line at $1420.4$ MHz.

The hyperfine splitting of cesium-133 defines the second: exactly $9{,}192{,}631{,}770$ periods of the radiation between its two ground-state hyperfine levels, $F = 4$ and $F = 3$ from combining the electron's $\tfrac{1}{2}$ with the nucleus's $\tfrac{7}{2}$. The atomic clocks aboard GPS satellites use rubidium and cesium hyperfine transitions; a timing error of one nanosecond would move a GPS position by about $30$ cm, so those clocks must keep the hyperfine frequency to about a part in $10^{13}$.

10. Building the coupled states by lowering

To find Clebsch–Gordan coefficients without tables, start at the top. The state $|j_1, j_1\rangle|j_2, j_2\rangle$ is the only product state with $m = j_1 + j_2$, so it must be $|j_1 + j_2, j_1 + j_2\rangle$. Apply the total lowering operator $\hat{J}_- = \hat{J}_{1-} + \hat{J}_{2-}$ to both sides, using $\hat{J}_-|j, m\rangle = \hbar\sqrt{j(j + 1) - m(m - 1)}\,|j, m - 1\rangle$, and normalize; repeat down the ladder.

The state with $m = j_1 + j_2 - 1$ and the next lower total, $j_1 + j_2 - 1$, must be the combination orthogonal to the one just found, since it has the same $m$ but a different $j$. Lower it in turn, and continue. For two spins-½, one lowering of $\uparrow\uparrow$ gives $(\uparrow\downarrow + \downarrow\uparrow)/\sqrt{2}$, and the orthogonal combination is the singlet. The method works for any pair and is how the tables are built.

11. In the world: atomic clocks and the definition of the second

Since 1967 the second has been defined by the hyperfine splitting of cesium-133: exactly $9{,}192{,}631{,}770$ oscillations of the radiation that flips the atom between its two ground-state hyperfine levels. Those levels, $F = 4$ and $F = 3$, come from adding the valence electron's spin $\tfrac{1}{2}$ to the nucleus's spin $\tfrac{7}{2}$, exactly as in this lesson.

The primary frequency standard of the United States, the cesium fountain clock NIST-F2 in Boulder, Colorado, launches a ball of laser-cooled cesium atoms up through a microwave cavity and lets them fall back, giving each atom about a second to interact with the microwaves. It keeps time to about one second in $300$ million years. GPS satellites carry rubidium and cesium clocks built on the same hyperfine physics, and every GPS fix on a phone depends on them.

12. In the world: the 21 cm line and the structure of galaxies

Hydrogen's hyperfine line at $1420.4$ MHz passes easily through the dust that hides much of the Milky Way from optical telescopes. Its detection in 1951 at Harvard by Harold Ewen and Edward Purcell let astronomers map the galaxy's spiral arms for the first time, measuring each cloud's speed from the Doppler shift of the line.

Measured this way, the outer parts of spiral galaxies rotate far faster than their visible stars and gas can explain: the rotation curves stay flat well beyond the visible disk. Radio observations of the 21 cm line, alongside Vera Rubin's optical measurements, became central evidence for dark matter. The same line is now the target of the Square Kilometer Array and of experiments such as HERA in South Africa, which aim to detect the faint 21 cm glow of hydrogen from the universe's first billion years — all from two spins-½ combining into $F = 1$ and $F = 0$.

13. Angular momenta do not add like numbers

Two spins of $\tfrac{1}{2}$ do not simply make a spin of $1$. They can make $1$ or $0$, and each of the four product states either belongs to one of those totals or is a mixture of them. The same goes for orbit and spin: an electron with $l = 1$ can have $j = \tfrac{3}{2}$ or $\tfrac{1}{2}$, not simply $\tfrac{3}{2}$.

Nor do the magnitudes add. With $s_1 = s_2 = \tfrac{1}{2}$, each $|\vec{S}_i|$ is $\tfrac{\sqrt{3}}{2}\hbar$, yet the triplet's magnitude is $\sqrt{2}\,\hbar$, less than their sum $\sqrt{3}\,\hbar$: even the triplet's spins are not exactly parallel. Only the $z$ components add as numbers, $m = m_1 + m_2$, because $\hat{J}_z$ is a plain sum of commuting operators.

14. The triplet and singlet of two spins

1. Start from the top state.

   $|1, 1\rangle = |\uparrow\uparrow\rangle$

   The only product state with $M = 1$.

2. Lower the left side.

   $\hat{S}_-|1, 1\rangle = \hbar\sqrt{2}\,|1, 0\rangle$

   $\sqrt{s(s + 1) - m(m - 1)} = \sqrt{2}$ for $s = m = 1$.

3. Lower the right side one spin at a time.

   $(\hat{S}_{1-} + \hat{S}_{2-})|\uparrow\uparrow\rangle = \hbar(|\downarrow\uparrow\rangle + |\uparrow\downarrow\rangle)$

   Each spin lowers from up to down with coefficient $\hbar$.

4. Equate the two sides.

   $|1, 0\rangle = \tfrac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)$

   The middle member of the triplet.

5. Find the orthogonal combination.

   $|0, 0\rangle = \tfrac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$

   Same $M = 0$, orthogonal, so it must be the singlet.

15. Spin–orbit splitting of a $p$ level

1. List the possible $j$ for $l = 1$.

   $j = 1 + \tfrac{1}{2} = \tfrac{3}{2}, \qquad j = 1 - \tfrac{1}{2} = \tfrac{1}{2}$

   One electron's spin with its orbit.

2. Write the dot product in terms of totals.

   $\vec{L} \cdot \vec{S} = \tfrac{1}{2}(J^2 - L^2 - S^2)$

   From $J^2 = (\vec{L} + \vec{S})^2$.

3. Evaluate it for $j = \tfrac{3}{2}$.

   $\dfrac{\vec{L} \cdot \vec{S}}{\hbar^2} = \tfrac{1}{2}\left(\tfrac{15}{4} - 2 - \tfrac{3}{4}\right) = \tfrac{1}{2}$

   $j(j + 1) = \tfrac{15}{4}$.

4. Evaluate it for $j = \tfrac{1}{2}$.

   $\dfrac{\vec{L} \cdot \vec{S}}{\hbar^2} = \tfrac{1}{2}\left(\tfrac{3}{4} - 2 - \tfrac{3}{4}\right) = -1$

   $j(j + 1) = \tfrac{3}{4}$.

5. Find the splitting in terms of $A$.

   $\Delta E = A\left(\tfrac{1}{2} - (-1)\right) = \tfrac{3}{2}A$

   $A(2l + 1)/2$ with $l = 1$.

6. Use sodium's measured splitting.

   $\tfrac{3}{2}A = 2.13\ \text{meV} \quad\Rightarrow\quad A = 1.42\ \text{meV}$

   From the D-line wavelengths.

7. Check the weighted average.

   $4 \times \tfrac{1}{2}A + 2 \times (-A) = 0$

   The $j = \tfrac{3}{2}$ level has four states and $j = \tfrac{1}{2}$ two: the center of gravity is unshifted.

16. Probabilities in a coupled state

1. Take a $p$ electron in $|j = \tfrac{3}{2}, m_j = \tfrac{1}{2}\rangle$ and list the product states with that $m_j$.

   $|m_l = 0\rangle|\uparrow\rangle, \qquad |m_l = 1\rangle|\downarrow\rangle$

   $m_l + m_s = \tfrac{1}{2}$.

2. Start from the top of the quartet.

   $|\tfrac{3}{2}, \tfrac{3}{2}\rangle = |1\rangle|\uparrow\rangle$

   The only product state with $m_j = \tfrac{3}{2}$.

3. Lower the left side.

   $\hat{J}_-|\tfrac{3}{2}, \tfrac{3}{2}\rangle = \hbar\sqrt{3}\,|\tfrac{3}{2}, \tfrac{1}{2}\rangle$

   $\sqrt{\tfrac{15}{4} - \tfrac{3}{4}} = \sqrt{3}$.

4. Lower the right side.

   $(\hat{L}_- + \hat{S}_-)|1\rangle|\uparrow\rangle = \hbar\sqrt{2}\,|0\rangle|\uparrow\rangle + \hbar|1\rangle|\downarrow\rangle$

   $\hat{L}_-$ gives $\sqrt{2}$ for $l = m = 1$; $\hat{S}_-$ gives one.

5. Equate and normalize.

   $|\tfrac{3}{2}, \tfrac{1}{2}\rangle = \sqrt{\tfrac{2}{3}}\,|0\rangle|\uparrow\rangle + \sqrt{\tfrac{1}{3}}\,|1\rangle|\downarrow\rangle$

   Divide by $\sqrt{3}$.

6. Find the probability of spin up.

   $P_\uparrow = \tfrac{2}{3}$

   Agrees with $(l + m_j + \tfrac{1}{2})/(2l + 1) = 2/3$.

7. Find the average $L_z$.

   $\langle L_z\rangle = \tfrac{2}{3} \times 0 + \tfrac{1}{3} \times \hbar = \tfrac{1}{3}\hbar$

   Weighted by the probabilities.

8. Check against $m_j$.

   $\langle L_z\rangle + \langle S_z\rangle = \tfrac{1}{3}\hbar + \left(\tfrac{2}{3} - \tfrac{1}{3}\right)\tfrac{\hbar}{2} = \tfrac{1}{2}\hbar$

   The components add to $m_j\hbar$, as they must.

17. Your turn: what values of $j$ can an electron with $l = 3$ have, and how many states are there in all?

1. List the totals.

   $j = 3 - \tfrac{1}{2}, \ 3 + \tfrac{1}{2} = \tfrac{5}{2}, \ \tfrac{7}{2}$

   From $|l - s|$ to $l + s$.

2. Count the states of each.

   $2j + 1 = 6, \ 8$

   One per $m_j$.

3. Your turn: work this step out. Its working is at the end of the packet.

   Add the two counts.

18. Guided practice

Two electrons' spins, each $s = \tfrac{1}{2}$, are combined. What values can the total spin quantum number $S$ take, and how many states does each have?

• $S = 1$ with three states and $S = 0$ with one
• only $S = 1$, since $\tfrac{1}{2} + \tfrac{1}{2} = 1$
• $S = 1$ with two states and $S = 0$ with two
• $S = \tfrac{1}{2}$ with four states

19. Guided practice

Complete the worked solution: an electron with $l = 2$ has spin–orbit energy $E = A\,\vec{L} \cdot \vec{S}/\hbar^2$ with $A = 6$ meV. Find the energies of the $j = l + \tfrac{1}{2}$ and $j = l - \tfrac{1}{2}$ levels and their splitting.

1. Evaluate the upper level's energy.

   $E_{j = l + 1/2} = A \cdot \dfrac{l}{2} =$ _u meV

   $\vec{L} \cdot \vec{S}/\hbar^2 = l/2$ when the spin is aligned.

2. Evaluate the lower level's energy.

   $E_{j = l - 1/2} = -A \cdot \dfrac{l + 1}{2} =$ _d meV

   $\vec{L} \cdot \vec{S}/\hbar^2 = -(l + 1)/2$ when it is opposed.

3. Subtract to find the splitting.

   $\Delta E = A \cdot \dfrac{2l + 1}{2} =$ _s meV

   The fine-structure splitting.

20. Guided practice

Two angular momenta $j_1$ and $j_2$ combine into a total $j$. Match each rule to its expression.

21. Practice

An angular momentum $j_1 = 4$ combines with $j_2 = 1$. Fill in the number of states for each allowed total $j$, and the total.

22. Practice

The potassium atom's first excited $p$ level is split by spin–orbit coupling into $j = \tfrac{3}{2}$ and $j = \tfrac{1}{2}$, giving two lines at $766.490$ nm and $769.896$ nm. What is the splitting, in meV? Use $hc = 1239.84$ eV nm.

Answer: meV

23. Practice

An electron in a $d$ orbital ($l = 2$) is in the state with total angular momentum $j = \tfrac{5}{2}$ and $m_j = \tfrac{-1}{2}$. What is the probability that a measurement of $S_z$ gives $+\hbar/2$?

Answer:

24. Somewhere new

In the ground state of muonium, the electron's spin and the nucleus's (or muon's) spin combine into two levels $18.459$ μeV apart. Radio astronomers and precision labs detect the photon emitted when the spins flip between them. What is its frequency, in MHz?

Answer: MHz

25. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

26. Test question

An angular momentum $j_1 = 4$ combines with $j_2 = 1$. Fill in the number of states for each allowed total $j$, and the total.

27. What you can do now

You can combine angular momenta. Explain to someone why two spin-½ particles can have total spin zero.

Working for the steps left to you

17. Your turn: what values of $j$ can an electron with $l = 3$ have, and how many states are there in all?, step 3