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Momentum eigenstates of spin

  1. Aug 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Starting with [tex]\sigma_{y}[/tex], calculate the momentum eigenstates of spin in the y direction.
    [tex]\sigma_{y} = \left[\stackrel{0}{i} \stackrel{-i}{0}\right][/tex] (Pauli spin matrix in the y direction)
    [tex]S_{y} = \frac{\hbar}{2}\sigma_{y} [/tex] (spin angular momentum operator for the y direction)

    2. Relevant equations

    [tex]A\left|\psi\right\rangle = a\left|\psi\right\rangle [/tex] 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, [tex] det (A - \lambda I) = 0[/tex], where [tex] A \equiv S_{y} [/tex], [tex]\lambda[/tex]
    are the eigenvalues and I is the 2x2 identity matrix.

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

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

    Therefore, the momentum eigenstates I obtain are just

    [tex]S_{y}\left|\psi\right\rangle = \pm \frac{\hbar}{2} \left|\psi\right\rangle [/tex]

    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.


    p.s. Does anyone know how to write matrices in latex? Sorry, about my dodgy matrix up above for sigma y
  2. jcsd
  3. Aug 7, 2007 #2


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    Homework Helper

    "momentum eigenstates" doesn't make sense. I think what they want is just for you to find the eigenstates of [tex]S_y[/tex]. 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.
  4. Aug 8, 2007 #3
    For the eigenvalue [tex]\lambda = + \frac{\hbar}{2}[/tex],

    I get two simulatenous equations:
    [tex] -v_{1} + iv_{2} = 0[/tex] ... (1)
    [tex] iv_{1} - v_{2} = 0 [/tex] ... (2)

    Solving (1) for [tex] v_{1}[/tex] in terms of [tex]v_2[/tex]:
    [tex] -v_{1} = iv_{2}[/tex]
    [tex] v_{1} = -iv_{2}[/tex]

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

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

    Is this correct, for the positive eigenvalue?
  5. Aug 8, 2007 #4


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    Homework Helper

    Nope, you made a little mistake; if you look at the above [tex]\vec v[/tex]you will see that [tex]v_2 = -i[/tex], so that [tex]iv_2 = 1 = v_1[/tex] 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 [tex]S_y[/tex].
  6. Aug 9, 2007 #5
    Ah yes, you are correct. The eigenvector I did before was for [tex]-\frac{\hbar}{2}[/tex]. 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.
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