Calculating Momentum Eigenstates of Spin in the Y Direction

[PsiPhi]

Homework Statement

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)

Homework Equations

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

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 realized 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

 

[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.

 

[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?

 

[olgranpappy]

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
$$\left[\stackrel{1}{-i}\right]$$

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 $$S_y$$.

 

[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 realized they differ by a multiplicative constant of i.

Thanks for the help, olgranpappy.