Addition of Angular Momenta | Quantum Science Philippines

Sandra Manulat and Michael Jabines

Consider two spin 3/2 particles at the ground state. Suppose that their total spin is 2 with its z-component equal to zero. If you measure [eq]J_z[/eq] on the second particle, what values would you get and the probabilities of measuring such?

In order to answer the problem, we first need to learn the addition of angular momenta.

Let us define the total angular momentum operator:

[eq]\vec{J} =\vec{ J}_1 + \vec{J}_2 [/eq]

where [eq]\vec{J}_1[/eq] can be the orbital angular momentum operator and [eq]\vec{ J}_2[/eq] is the spin angular momentum operator.

Moreover, [eq]\vec{J_1}[/eq] and [eq]\vec{J_2}[/eq] can also denote the angular momentum operators of two different particles in a multi-particle system just like the problem above.

By definition, angular momentum operators are Hermitian, hence [eq]J[/eq], which is a just the sum of two Hermitian operators is also Hermitian satisfying the commutation relation

[eq]\vec{ J} \times \vec{J} = i\hbar\vec{ J} [/eq]

Thus, [eq]\vec{J}[/eq] possesses the expected properties of an angular momentum operator:

[eq] J^2 = j(j + 1)\hbar^2 [/eq]

[eq] J_z = m\hbar^2 [/eq]

where [eq]m = j, j-1, \cdots, -j+1, -j[/eq]

Now, we find a relationship between the quantum number of the total angular momentum, [eq]j[/eq] and [eq]m[/eq] with the individual quantum numbers [eq]j_1[/eq], [eq]j_2[/eq], [eq]m_1[/eq], and [eq]m_2[/eq].

Recall that,

[eq][J^2, J_1] = [J^2, J_2] = 0[/eq]

this implies that we can solve for the quantum numbers [eq]j[/eq], [eq]j_1[/eq], and [eq]j_2[/eq] simultaneously.  On the other hand,

[eq][J^2, J_{z1}] \neq 0[/eq]

[eq][J^2, J_{z2}] \neq 0[/eq]

which tells us that the quantum numbers [eq]m_1[/eq] and [eq]m_2[/eq] cannot be solved simultaneously with the quantum number [eq]j[/eq].

What to do?

Now we need to form two alternate groups of mutually commuting operators.

1st group: the eigenkets of [eq]{J_1}^2, {J_2}^2, J_{z1}[/eq] and [eq]J_{z2}[/eq] are [eq]|j_1, j_2; m_1, m_2>[/eq]

2nd group: the eigenkets of [eq]{J_1}^2, {J_2}^2, J, J_z[/eq] are [eq]|j_1, j_2; j, m>[/eq]

Each set of eigenkets are complete, mutually orthogonal and have unit norms, then we can write a conventional completeness relation,

[eq]\sum_{m_1}\sum_{m_2} |j_1, j_2; m_1, m_2><j_1, j_2; m_1, m_2 | = 1[/eq]

[eq]\sum_{j}\sum_{m} |j_1, j_2; j, m><j_1, j_2; j, m | = 1[/eq]

The summation is over all allowed values of [eq]m_1, m_2, j[/eq] and [eq]m[/eq].

These relation tells us that if we know [eq]j_1, j_2, j, m[/eq] we will be able to solve for a range of possible values of  [eq]m_1[/eq] and [eq]m_2[/eq] .

Now we ask: Given a state [eq]|j_1, j_2 ; m_1, m_2>[/eq], how do we express it in terms of [eq]|j_1, j_2; j, m>[/eq]?

It’s simple, we can use the completeness relation above to write,

[eq]|j_1, j_2; j, m> = \sum_{m_1}\sum_{m_2}<j_1, j_2; m_1, m_2|j_1, j_2; j, m>|j_1, j_2; m_1, m_2>[/eq]

Writing it more simply we have,

[eq]|j,m> = \sum_{m_1,m_2} C_{j_1, j_2; m_1, m_2}^{j, m} |j_1, m_1>|j_2, m_2>[/eq]

where [eq]C_{j_1, j_2; m_1, m_2}^{j, m}\;\;\;\; =\;\;\;\; \langle j_1, j_2; m_1, m_2 | j_1, j_2; j, m\rangle [/eq] is called the Clebsch-Gordan coefficients.

At this point, we are ready to answer our problem. We don’t need to worry solving for the Clebsch-Gordan coefficients because a table is made ready for our use.

Given: two spin 3/2 particles:

[eq]j_1= \frac{3}{2},\;\;\;\; j_2 =\frac{3}{2}[/eq]

[eq]j \;\;\;\; = \;\;\;\; 2, \;\;\;\; m\;\;\;\;=\;\;\;\; 0[/eq]

[eq]m = m_1 + m_2 = 0[/eq]

Now [eq]m_1 [/eq] and [eq]m_2[/eq] can have values in the range [eq]j_1, j_1 – 1, \cdots, -j_1 + 1, -j_1[/eq], and [eq]j_2, j_2 – 1, \cdots, -j_2 + 1, -j_2[/eq] respectively.

Thus,

[eq]|2\;\;\;\;0>\;\;\;\; = \;\;\;\;C_1 |2\;\;\;\;\frac{3}{2}>|0\;\;\;\;-\frac{3}{2}>\;\; +\;\; C_2 |2\;\;\;\;\frac{1}{2}>|0\;\;\;\;-\frac{1}{2}>[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + C_3 |2\;\;\;\;-\frac{1}{2}>|0\;\;\;\;\frac{1}{2}>\;\; +\;\; C_4 |2\;\;\;\;-\frac{3}{2}>|0\;\;\;\;\frac{3}{2}>[/eq]

For the values of [eq]C_1, C_2, C_3, C_4[/eq] we’ll use the table of Clebsch-Gordan coefficients, specifically the 3/2 by 3/2 table since we have two spin 3/2 particles:

Clebsch-Gordan coefficients for spin 3/2 two particle system. The shaded region corresponds to the probability of the corresponding spin state.

Now we have,

[eq]|2\;\;\;\;0>\;\;\;\; = \;\;\;\;\frac{1}{\sqrt{4}} |2\;\;\;\;\frac{3}{2}>|0\;\;\;\;-\frac{3}{2}>\;\; +\;\; \frac{1}{\sqrt{4}} |2\;\;\;\;\frac{1}{2}>|0\;\;\;\;-\frac{1}{2}>[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -\frac{1}{\sqrt{4}} |2\;\;\;\;-\frac{1}{2}>|0\;\;\;\;\frac{1}{2}>\;\; \;\;-\frac{1}{\sqrt{4}} |2\;\;\;\;-\frac{3}{2}>|0\;\;\;\;\frac{3}{2}>[/eq]

Finally, we are told to measure the z-component of the spin on the second particle: [eq]J_{z2} = m_2\hbar[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;J_{z2} \;\;\;\;\;\;\;\;\;\;Probability[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{3}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

[eq]\;\;\;\;\;\;\;\;-\frac{3}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

[eq]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]

[eq]\;\;\;\;\;\;\;\;-\frac{1}{2}\hbar\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{4}[/eq]