GATE 2026 PH – Question 40
Consider two particles with angular momenta $j_1=2\hbar$ and $j_2=\hbar/2$. If the expression $$|j=5/2,m=3/2\rangle=c_1|j_1=2,m_1=1\rangle|j_2=1/2,m_2=1/2\rangle+c_2|j_1=2,m_1=2\rangle|j_2=1/2,m_2=-1/2\rangle$$ gives an eigenstate of the total angular momentum of the two particles, using standard notation. Which of the following is true? (Hint: $\hat J_\pm|j,m\rangle=\sqrt{j(j+1)-m(m\pm1)}|j,m\pm1\rangle$)
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Correct answer: (A) $c_1=2/\sqrt5,\ c_2=1/\sqrt5$
Explanation
The total angular momentum $j=\tfrac52$ is the maximum value $j_1+j_2=2+\tfrac12$, so the state $|j=\tfrac52,m=\tfrac32\rangle$ is obtained from the top state $|\tfrac52,\tfrac52\rangle$ by applying the lowering operator $\hat J_-=\hat J_{1-}+\hat J_{2-}$.
**Top state:** $|\tfrac52,\tfrac52\rangle=|2,2\rangle|\tfrac12,\tfrac12\rangle$.
**Lowering** (using $\hat J_-|j,m\rangle=\sqrt{j(j+1)-m(m-1)}\,|j,m-1\rangle$):
- $\hat J_{1-}|2,2\rangle=\sqrt{6-2}\,|2,1\rangle=2|2,1\rangle$,
- $\hat J_{2-}|\tfrac12,\tfrac12\rangle=\sqrt{\tfrac34+\tfrac14}\,|\tfrac12,-\tfrac12\rangle=|\tfrac12,-\tfrac12\rangle$,
- $\hat J_-|\tfrac52,\tfrac52\rangle=\sqrt{\tfrac{35}4-\tfrac{15}4}\,|\tfrac52,\tfrac32\rangle=\sqrt5\,|\tfrac52,\tfrac32\rangle$.
So
$$\sqrt5\,|\tfrac52,\tfrac32\rangle=2\,|2,1\rangle|\tfrac12,\tfrac12\rangle+|2,2\rangle|\tfrac12,-\tfrac12\rangle .$$
$$c_1=\frac2{\sqrt5},\qquad c_2=\frac1{\sqrt5}\quad(\text{option A}).$$
(Check: $c_1^2+c_2^2=\tfrac45+\tfrac15=1$ ✓.)