Solution

j = 0

state exists only when

j_{1} = j_{2} =: j

. Since its total magnetic quantum number is zero, every uncoupled basis term must have

m_{1} + m_{2} = 0

, so

∣0, 0 ⟩ = \sum_{m = - j}^{j} α_{m} ∣ j, m ⟩_{1} ∣ j, - m ⟩_{2} .

Apply the total raising operator

J_{+} = J_{1 +} + J_{2 +}

. The coefficient of

∣ j, m + 1 ⟩_{1} ∣ j, - m ⟩_{2}

(j - m) (j + m + 1) (α_{m} + α_{m + 1}) .

J_{+}

, hence

α_{m + 1} = - α_{m}, - j \leq m < j .

All coefficients consequently have the same modulus and alternating signs. Normalization fixes that modulus to

(2 j + 1)^{- 1/2}

, and an arbitrary overall phase may be chosen so that

α_{m} = \frac{( - 1 ) ^{j - m}}{2 j + 1} .

Thus

∣0, 0 ⟩ = \frac{1}{2 j + 1} \sum_{m = - j}^{j} (- 1)^{j - m} ∣ j, m ⟩_{1} ∣ j, - m ⟩_{2} .

Solved by gpt-5.6-sol high.

 Ancestors (11)