Question

System A consists of two spin-1/2 particles, and has a four-dimensional Hilbert space. 1. Write down a basis for the Hilbert space of two spin-1/2 particles. 2. Calculate the matrix of the angular momentum operator, Sfot = (ŜA, ŠA, ŜA) for system A, in the basis of question 4A.1, and express them in this basis. 3. Calculate the square of the total angular momentum of system A , Spotl?, and express this operator in the basis of question 4A.1. 4. Show that Is) = (1z, Iz) – ITz,tz))/V2 and T.) = (1+z, 1z) + ]Tz,tz))/v2 are eigenstates of the square total angular momentum. Find the other two eigenstates, and categorise all four total angular momentum eigenstates by their eigenvalues, ħstot (stot + 1), where stot is the total angular momentum quantum number, and by the z-component of their total angular momentum. 5. The three-dimensional triplet subspace has total spin quantum number stot = 1. In the triplet/singlet basis of question 4A.4, write out the matrix of the total angular momentum operator.

Answer 1

1 1. There are four states in the Hilbert space : 111), 111), 11), 11) 2. For a single spin the angular momentum operators are the Pauli matrices times half given by: σα (1) 1 - (5 ( ; ) 0:- ( 6 ) Oy = (2) (3) The angular momentum of spins in the complete 4 dimensional hilbert space are obtained by taking the tensor ptoduct with Identity. The total angular momentum operators are obtained by adding the individual angular momentum operators. 1 1 S: -40®I+10,6 (4) 1 1 1 0 0 1 0 0 0 1 + 1 0 (5) 2 1 2 1 0 1 1 1 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 + 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 (6) 2 2 2 Similarly, other operators are: Sy 1 1/2 Oy ØI+I Oy (7) 0 0 - i -i 0 0 i i (8) 0 i 0 i -i 0 0 Sz 2 OI+10-02 2 (9) 2 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -2 (10) 3. $2 is obtained just by squaring the above matrices and summing them up: S2 = S2 + S3, +52 (11 ) -2 1 $2 2 0 0 0 2 2 0 0 2 2 0 2 0 0 2 + 2 0 0 0 2 2 0 2 2 -2 0 0 0 0 + 4 0 0 0 0 0 0 0 0 0 0 4 (12) 4 4 0 2

2 2 0 0 0 0 1 1 0 0 1 1 0 0 0 0 2 (13) 4. The given states in the choosen basis are: 1 1 1S) = (14) {0,- va 250) {o za to 1 TO) (15) These are eigenvectors of S2 as shown below: 0 1 1 $2|S) = 2 0 0 0 0 1 1 0 0 1 1 0 0 0 0 2 =0 (16) V2 0 Eigenvalue is 0 1 1 SPT) 2 0 0 0 0 1 0 0 0 0 0 0 2 V2 (17) 1 1 2 Eigenvalue is 2. The other two eigenstates are just: 0 Ti) (18) Eigenvalue is 2 |T-1) (19) Eigenvalue is 2 The stae (S) has stot = 0) while the states \T;) have stot = 1. THe S, values for the states |S), To),Ti) |T-1) are 0,0,1, -1 respectively Note: the factors of ħ have been ignored here.