Question

Find the eigenvalue of the total spin angular momentum operator, s^2.

2) Find the eigenvalue of the total spin angular momentum operator, $2.

Answer 1

Answer:

Remember that Sˆ is a vector, i.e. it is a triplet of operators. In Cartesian coordinates Sˆ = (Sˆ _x, Sˆ _y, Sˆ _z), and the commutation relations are:

[Sˆ_x, Sˆ_y] = iħJˆ_z , [Sˆ _y, Sˆ _z] =iħJˆ_x , [Sˆ _z, Sˆ_x] = iħJˆ_y.

The square of the angular momentum is represented by the operator Sˆ² ≡ Sˆ_x² + Sˆ_y ²+ Sˆ_z ²,

with the property that [Sˆ², Sˆ_x]=[Sˆ², Sˆy]=[Sˆ², Sˆz]=0.

The Compatibility Theorem tells us that, for example, the operators S^ˆ2 and Sˆ_z have simultaneous eigenstates. We denote these common eigenstates by |λ, m>. Looking back at the results obtained in the previous lectures, these are the kets associated with the spherical harmonics Y_l^m (θ, φ). We can write: Sˆ² |λ, m> = λħ² |λ, m>

Sˆ_z |λ, m> = λħ |λ, m>

so that the eigenvalues of Sˆ² are denoted by λħ² whilst those of Sˆ_z are denoted by mħ.