• Problem 1, solution

  • Problem 2, solution

    The common eigenfunctions of S2 and Sz are

    |11>=|++>,
    |10>=½1/2(|+->+|-+>),
    |1-1>=|-->,
    |00>=½1/2(|+->-|-+>).

    From symmetry we therefore have that the common eigenfunctions of S2 and Sx are

    |11>x=|++>x,
    |10>x=½1/2(|+->x+|-+>x),
    |1-1>x=|-->x,
    |00>x=½1/2(|+->x+--+>x).

    We have

    |++>x=½1/2(|+>+|->)Ľ1/2(|+>+|->)=½(|++>+|+->+|-+>+|-->).
    |-->x=½1/2(|+>-|->)Ľ1/2(|+>-|->)=½(|++>-|+->-|-+>+|-->).
    |+->x=½1/2(|+>+|->)Ľ1/2(|+>-|->)=½(|++>-|+->+|-+>-|-->).
    |-+>x=½1/2(|+>-|->)Ľ1/2(|+>+|->)=½(|++>+|+->-|-+>-|-->).

    Therefore

    |11>x=½(|++>+|+->+|-+>+|-->),
    |10>x=½1/2(|++>-|-->),
    |1-1>x=½(|++>-|+->-|-+>+|-->),
    |00>x=½1/2(|+->-|-+>).

  • Problem 3, solution  

    (a)  The eigenvectors of Sy are |+>y=2-1/2(|+>+i|->) and |->y=2-1/2(|+>-i|->).

    Therefore: |+>=2-1/2(|+>y+|->y) and |->=(-i/21/2)(|+>y-|->y).

    |y1>=|+>y.

    |y2>=(-i/61/2)(|+>y+|->y) -(i/31/2)(|+>y-|->y)=(-i)(0.986|+>y+0.169|->y

    (b)  <y1|y2>=-i/61/2-i(2/6)1/2=-i0.986 in the Sz basis.

          <y1|y2>=-i0.986 in the Sy basis.

    (c)  |y1(t)>=2-1/2(exp(-iw0t/2)|+>+iexp(iw0t/2)|->) 

               =(h/4)i(-exp(-iw0t)+exp(iw0t))= -(h/2)sin(w0t).

    If <Sx>=±(h/2), then |y1(t)> is an eigenvector of Sx.  This happens when w0t=np/2, n=odd.

    Problem 4, solution