Eigenvectors of spin operators of a spin 1/2 system

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You can solve this problem simply by diagonalizing S_y, which is given by:

S_y = hbar/2 [[0,-i],

i, 0]]

Subtract lambda from the diagonal and equate the determinant to zero:

lambda^2 - hbar^2/4 = 0 ------>

lambda = ± hbar/2

The eigenvector corresponding to hbar/2 is thus the solution of the equation:

S_y |v> = hbar/2 |v>

Write |v> = a |up> + b|down>, where |up> is the column vector (1,0) and |down> is the column vector (0,1), so v = (a,b) written as a column vector. Then we have two equations:

-i b = a

i a = b

The second equation is depend on the first, multiplying the first equation by i yields the second equation. Of course, the equations have to be dependent as the determinant is equal to zero; this is the necessary condition to get a nontrivial solution for the eigenvector. We can then chosse one component in any way we like, e.g. we can put a = 1 and then we get b = i. The state is thus:

|v> = |up> + i |down> = (1, i)

But this is not normalized, the norm of this state is sqrt(2). So, we must divide by sqrt(2) to get the correctly normalized state for the spin polarizid in te plus y-direction:

1/sqrt(2) [|up> + i |down>] = 1/sqrt(2) (1,i)

We can find the other eigenvector by using that it is orthogonal to the above eigenvector. If this is |s> = (c,d), then the inner product with the above state is:

<v|s> = c - i d = 0

If we put c = 1, then we find that d = -i. Now the state |s> = |up> - i |down> = (1,-i) has norm sqrt(2), so the correctly normalized state for the spin polarized in the minus y direction is:

1/sqrt(2) [|up> - i |down>] = 1/sqrt(2) (1, -i)

You can also solve this problem from first principles without invoking S_y simply by using rotational symmetry. I'll give a rough sketch of a derivation. First consider the case of ordinary vectors. An ordinary ...