Question

Consider two particles with angular momenta \(j_1 = 2\hbar\) and \(j_2 = \hbar/2\). If the expression
\[ |j=\tfrac52, m=\tfrac32\rangle = c_1|j_1=2,m_1=1\rangle|j_2=\tfrac12,m_2=\tfrac12\rangle + c_2|j_1=2,m_1=2\rangle|j_2=\tfrac12,m_2=-\tfrac12\rangle \]gives an eigenstate of the total angular momentum of the two particles, using standard notation, which of the following is true?
(Hint: \(\hat J_{\pm}|j,m\rangle = \sqrt{j(j+1)-m(m\pm1)}\,|j,m\pm1\rangle\))

• A. \(c_1 = \dfrac{2}{\sqrt5},\quad c_2 = \dfrac{1}{\sqrt5}\)
• B. \(c_1 = \dfrac{1}{\sqrt5},\quad c_2 = \dfrac{2}{\sqrt5}\)
• C. \(c_1 = \dfrac{1}{\sqrt2},\quad c_2 = \dfrac{1}{\sqrt2}\)
• D. \(c_1 = 0,\quad c_2 = 1\)

Hint:

Start from the top state \(|j=5/2,m=5/2\rangle\), which is just the simple product state, then apply the lowering operator to both sides and match coefficients.

Answer 1

The Correct Option is A

Step 1: Start from the fully stretched top state.
For \(j_1=2\) and \(j_2=\tfrac12\), the largest possible total is \(j=j_1+j_2=\tfrac52\), and its top state is simply the product of both particles at their own maximum \(m\):
\[ \left|\tfrac52,\tfrac52\right\rangle = |j_1=2,m_1=2\rangle\,|j_2=\tfrac12,m_2=\tfrac12\rangle \]
This is true because there is only one way to combine \(m_1\) and \(m_2\) to reach the maximum \(m=\tfrac52\), so no mixing coefficient is needed here.

Step 2: Lower the total state by one step using the hint formula.
Apply the total lowering operator \(\hat J_- = \hat J_{1-}+\hat J_{2-}\) to \(\left|\tfrac52,\tfrac52\right\rangle\). On the left side, using the hint with \(j=\tfrac52,m=\tfrac52\):
\[ \hat J_-\left|\tfrac52,\tfrac52\right\rangle = \sqrt{\tfrac52\cdot\tfrac72 - \tfrac52\cdot\tfrac32}\,\left|\tfrac52,\tfrac32\right\rangle = \sqrt{\tfrac{35}{4}-\tfrac{15}{4}}\,\left|\tfrac52,\tfrac32\right\rangle = \sqrt5\,\left|\tfrac52,\tfrac32\right\rangle \]

Step 3: Apply the same lowering operator to the product state.
On the right side, \(\hat J_{1-}\) acts only on the \(j_1=2\) part and \(\hat J_{2-}\) only on the \(j_2=\tfrac12\) part. Using the hint for each:
\[ \hat J_{1-}|2,2\rangle = \sqrt{2\cdot3-2\cdot1}\,|2,1\rangle = \sqrt{6-2}\,|2,1\rangle = 2|2,1\rangle \]
\[ \hat J_{2-}\left|\tfrac12,\tfrac12\right\rangle = \sqrt{\tfrac12\cdot\tfrac32-\tfrac12\cdot(-\tfrac12)}\left|\tfrac12,-\tfrac12\right\rangle = \sqrt{\tfrac34+\tfrac14}\left|\tfrac12,-\tfrac12\right\rangle = 1\cdot\left|\tfrac12,-\tfrac12\right\rangle \]
So
\[ \hat J_-\Big(|2,2\rangle\left|\tfrac12,\tfrac12\right\rangle\Big) = 2\,|2,1\rangle\left|\tfrac12,\tfrac12\right\rangle + 1\cdot|2,2\rangle\left|\tfrac12,-\tfrac12\right\rangle \]

Step 4: Match the two sides and normalise.
Both sides describe the same physical state, so
\[ \sqrt5\left|\tfrac52,\tfrac32\right\rangle = 2\,|2,1\rangle\left|\tfrac12,\tfrac12\right\rangle + |2,2\rangle\left|\tfrac12,-\tfrac12\right\rangle \]
Dividing through by \(\sqrt5\):
\[ \left|\tfrac52,\tfrac32\right\rangle = \frac{2}{\sqrt5}\,|2,1\rangle\left|\tfrac12,\tfrac12\right\rangle + \frac{1}{\sqrt5}\,|2,2\rangle\left|\tfrac12,-\tfrac12\right\rangle \]

Step 5: Read off the coefficients.
Comparing term by term with the expression given in the question, \(c_1 = \dfrac{2}{\sqrt5}\) (the coefficient of \(|2,1\rangle|\tfrac12,\tfrac12\rangle\)) and \(c_2 = \dfrac{1}{\sqrt5}\) (the coefficient of \(|2,2\rangle|\tfrac12,-\tfrac12\rangle\)).

Step 6: Why the other options are wrong.
Option (B) simply swaps \(c_1\) and \(c_2\), which would be the answer only if the labeling of the two basis kets were reversed. Option (C) treats the two terms as equally weighted, ignoring that lowering \(m_1\) from \(2\) to \(1\) and lowering \(m_2\) from \(\tfrac12\) to \(-\tfrac12\) do not carry equal weight here. Option (D) would mean the state is just the unmixed product state, which is only true for the top state \(m=\tfrac52\), not for \(m=\tfrac32\).

Final Answer:
\(c_1=\dfrac{2}{\sqrt5}\) and \(c_2=\dfrac{1}{\sqrt5}\), option (A).\[ \boxed{c_1=\tfrac{2}{\sqrt5},\ c_2=\tfrac{1}{\sqrt5}} \]