Momentum eigenstates of spin

1. Aug 7, 2007

PsiPhi

1. The problem statement, all variables and given/known data
Starting with $$\sigma_{y}$$, calculate the momentum eigenstates of spin in the y direction.
$$\sigma_{y} = \left[\stackrel{0}{i} \stackrel{-i}{0}\right]$$ (Pauli spin matrix in the y direction)
$$S_{y} = \frac{\hbar}{2}\sigma_{y}$$ (spin angular momentum operator for the y direction)

2. Relevant equations

$$A\left|\psi\right\rangle = a\left|\psi\right\rangle$$ where A is some linear operator and a is the corresponding eigenvalue

3. The attempt at a solution

The solution I tried was determining the eigenvalues for the matrix, $$det (A - \lambda I) = 0$$, where $$A \equiv S_{y}$$, $$\lambda$$
are the eigenvalues and I is the 2x2 identity matrix.

After working through the determinant expression, I obtain eigenvalues of $$\lambda = \pm \frac{\hbar}{2}$$

Then for momentum eigenstates, since the eigenstates aren't given I just used an arbitrary eigenstate, defined as $$\left|\psi\right\rangle$$

Therefore, the momentum eigenstates I obtain are just

$$S_{y}\left|\psi\right\rangle = \pm \frac{\hbar}{2} \left|\psi\right\rangle$$

I'm just wondering if my logic is correct as I step through my calculations. First I tried operator the spin angular momentum (y-direction) operator in the known matrices for spin-up, spin-down states. But, I realised that these were states in the z-direction. So, for momentum eigenstates in the y-direction the only way I could think of was the eigenvalue equation method.

Thanks.

p.s. Does anyone know how to write matrices in latex? Sorry, about my dodgy matrix up above for sigma y

2. Aug 7, 2007

olgranpappy

"momentum eigenstates" doesn't make sense. I think what they want is just for you to find the eigenstates of $$S_y$$. Solve the following matrix equation (matrices are a pain in tex, so I didn't write the matrices explicitly--I used I for the 2x2 unit matrix)
$$(S_y - \frac{\hbar}{2}I) \vec v = 0$$
for v_1 in terms of v_2 (you only get one independent equation from the above matrix equation) and then also use the fact that v should be normalized. This gives you the eigenstate of S_y with eigenvalue +hbar/2.

Then solve
$$(S_y + \frac{\hbar}{2}I) \vec u =0$$
for u_1 in terms of u_2 and normalize to get the other eigenstate.

3. Aug 8, 2007

PsiPhi

For the eigenvalue $$\lambda = + \frac{\hbar}{2}$$,

I get two simulatenous equations:
$$-v_{1} + iv_{2} = 0$$ ... (1)
$$iv_{1} - v_{2} = 0$$ ... (2)

Solving (1) for $$v_{1}$$ in terms of $$v_2$$:
$$-v_{1} = iv_{2}$$
$$v_{1} = -iv_{2}$$

Therefore, looking at the comparison of $$v_{1}$$ and $$v_2$$, the eigenvector for $$\lambda = + \frac{\hbar}{2}$$ is $$\left[\stackrel{1}{-i}\right]$$

And for the negative eigenvalue it should follow the same logic, haven't determined it yet though.

Is this correct, for the positive eigenvalue?

4. Aug 8, 2007

olgranpappy

Nope, you made a little mistake; if you look at the above $$\vec v$$you will see that $$v_2 = -i$$, so that $$iv_2 = 1 = v_1$$ which is not what your equations say.

But don't fear, the above vector is actually still an eigenvector, it's the eigenvector with eigenvalue -\hbar/2 as you can easily check by acting on it with the matrix $$S_y$$.

5. Aug 9, 2007

PsiPhi

Ah yes, you are correct. The eigenvector I did before was for $$-\frac{\hbar}{2}$$. But a weird thing happens, if i solve v_2 in terms of v_1 you will get a different eigenvector. However, I finally realised they differ by a multiplicative constant of i.

Thanks for the help, olgranpappy.