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)
<br /> (S_y - \frac{\hbar}{2}I) \vec v = 0<br />
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
<br /> (S_y + \frac{\hbar}{2}I) \vec u =0<br />
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 vyou 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.