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Homework Help: Eigenvalues and Eigenspinors

  1. Nov 30, 2007 #1
    [SOLVED] Eigenvalues and Eigenspinors

    1. The problem statement, all variables and given/known data
    (a) Find the eigenvalues and eigenspinors of [tex]S_{y}[/tex].

    2. Relevant equations
    [tex]\hat{Q}f(x) = \lambda f(x)[/tex]

    3. The attempt at a solution
    The above equation wasn't given specifically for this problem; but that's the one I'm trying to use.

    [tex]\hat{Q}f(x) = \lambda f(x)[/tex] --> [tex]\frac{\hbar}{2}\left(^0_i ^-i_0\right) f(x) = \lambda f(x)[/tex]

    But here is where I'm stuck. I'm not sure what I would be using for my function [tex]f(x)[/tex]. I guess the fact that it's a matrix kind of throws me off too.

    For the second part (finding the eigenspinors of [tex]S_{y}[/tex]), so far I have:

    [tex]X_{+}y = \frac{\hbar}{2}\left(^{0}_{i} ^{-i}_{0}\right)[/tex] --> [tex]\frac{\hbar}{2}\left(^{0}_{i}\right)[/tex]

    P.S. I'm not quite sure how to do matrices in this tex code. Does anyone know how?
  2. jcsd
  3. Nov 30, 2007 #2


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    f(x) is not a function here but a column vector. Just replace it by a column vector with entries "a" and "b" which are two constants to be determined.

    basically, this is a matrix algebra problem. You simply have to find the eigenvalues and eigenvectors of the matrix S_y. Just proceed as you did in linear algebra (find teh roots of the characteristic equation, etc)
  4. Nov 30, 2007 #3


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    Here is an example (click with yoru mouse on the tex output to see the code I typed in)
    1 & 1 & 1 \\
    1 & \omega & \omega^2 \\
    1 & \omega^2 & \omega
    \end{pmatrix} [/tex]
  5. Nov 30, 2007 #4
    nrqed - Thanks. So now I have done:

    [tex]\frac{\hbar}{2}\left( \begin{array}{cc}0 & -i\\i & 0\end{array}\right) \left(^{a}_{b}\right) = \lambda \left(^{a}_{b}\right)[/tex] --> [tex]\left(^{ai\frac{\hbar}{2}}_{-bi\frac{\hbar}{2}}\right) = \left(^{a\lambda}_{b\lambda}\right)[/tex]

    [tex]\lambda = \pm i\frac{\hbar}{2}[/tex].

    Is that correct? Also, for the second part, can I assume since you did not respond to that section, that it is also correct? Thanks again for the help.
  6. Nov 30, 2007 #5


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    Check this carefully.. Do you see the mistake in the above?
    final answer for the lambda is correct but assuming that it's not a typo in what you wrote, you could not have obtained those values of lambda from the equation you wrote. Hwo did you get your values of lambda?

    For the second part (finding the eigenvectors) we must first get the eigenvalues so let's clarify this first.

    Just a question: do you recall how to find eigenvalues of matrices? The best way is to find the characteristic equation which gives a polynomial in lambda and to find the eigenvalues.
  7. Nov 30, 2007 #6
    It is easier to simply rotate the eigenvectors of [tex]\sigma_z[/tex] around the x-axis.
  8. Nov 30, 2007 #7
    Well, for my values I just set:

    [tex]ai\frac{\hbar}{2} = a\lambda[/tex] ---> [tex]\lambda = i\frac{\hbar}{2}[/tex]


    [tex]-bi\frac{\hbar}{2} = b\lambda[/tex] ---> [tex]\lambda = -i\frac{\hbar}{2}[/tex]

    Oh, and yes, I see the mistake. My a and b are switched in the second equation with vectors; but it's still the same right?
    Last edited: Nov 30, 2007
  9. Nov 30, 2007 #8
    nrqed - I remember the characteristic equation being somthing like:

    [tex]y'' + 4y' + 4y = 0[/tex] ---> [tex]r^2 + 4r + 4[/tex]

    And then you solve for r to get the values you need. And as for the eigenvalues of matrices, I remember something like:

    [tex]\left(\begin{array}{cc}1-\lambda & 0\\0 & -1-\lambda \end{array}\right)[/tex]

    Count_Iblis - I'm not sure what you mean.
  10. Nov 30, 2007 #9
    i think it is not correct, the secular equation reads: Lambda^2-1=0 so lambda=+1 or-1. i've putted h=4pi. Now u have two eigenvalues in a space of dimension 2 so the diagonalized matrix is Sz=diag(1,-1) and the eigenvectors are |+> and |-> read (1,0) and (0,1) but in column. (Spin up or Spin down). This exercice suggest u that it is forbidden to simoultaneusly diagonalize Sy and Sz. To find a good basis for the angular momentum u should switch the axis at the occurence. the important u have to know is what follow ->>> [Sx,Sy]=Sz and cycles. read abou the SU(2) algebra... the pauli matrices.
  11. Nov 30, 2007 #10


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    But then you get lambda is equal to both i hbar/2 and -i hbar /2 which is impossible!

    yes you switched them which is why it does not work.
    If you do it correctly, you can't solve anything. You have two equations for two unknowns: a,b and lambda and you are stuck.
    What you are doing here is the second step of the calculation: finding the eigenvectors once you know the eigenvalues. You must first find the eigenvalues by solving the characteristic equation.
  12. Nov 30, 2007 #11


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    This is in a different context and is not what I am referring to.

    That's what I have in mind except you wrote the wrong expression You need to solve [tex] det(S_y - \lambda) =0 [/tex]. What you wrote above is not S_y - lambda (here I mean lambda times the identity matrix).
  13. Nov 30, 2007 #12
    nrqed - Right! Yeah, I didn't mean for my example of finding the eigenvalues as the problem that I was trying to solve. So...

    [tex]\begin{pmatrix}-\lambda -i\frac{\hbar}{2} \\i\frac{\hbar}{2} -\lambda \end{pmatrix} = 0[/tex]

    [tex]\lambda^{2} =\frac{\hbar^{2}}{4}[/tex] ---> [tex]\lambda = \pm \frac{\hbar}{2}[/tex]

    Is that right???
  14. Nov 30, 2007 #13


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    That's right. Now for each of those eigenvalues you go back to the eigenvalue equation and find the corresponding two eigenstates.
  15. Nov 30, 2007 #14

    Just consider cheating a bit. :smile: What you call the x, y and z direction is arbitrary, it cannot affect the physics. So, it immediately follows the eigenvalues of S_y must be the same as S_z as the eigenvalues are the possible values you can observe when measuring the spin and the particle doesn't care what direction you call x, y or z.

    Now, what we want to do is to express the eigenvectors of Sy in terms of the eigenvectors of Sz. But because Sy and Sz are the same up to a rotation, you can just as well rotate the eigenvectors of Sz. So, we need to know how spinors transform under rotations.

    Suppose you have some particle with spin. Suppose that you rotate your coordinate axis by [tex]\bf{\theta}[/tex] (magnitude = angle and direction is rotation axis, righthand rule convention). The particle itself is not affected in any way by the rotation (this is called a passive rotation). It can be shown that:

    [tex]|s'>=\exp\left( \frac{i}{\hbar} \bf{S}\cdot\bf{\theta}\right)|s>[/tex]

    where |s> is the spin of the particle and |s'> is the transformed spin. E.g. if you rotate by pi/2 around the y-axis, then the new z-direction will point in the direction of the old x-direction. A particle with spin polarized in the old z-direction will be polarized in the minus x-direction. If you take theta = pi/2, S = S_y and |s> |+> in the above formula, you get |s'> = the state corresponding to a particle polarized in the minus x-direction.

    To solve this problem we want to rotate the coordinate axis such that the new y-direction will point in the direction of the old z-axis. The eigenvectors of Sz will then transform into the eigenvectors of Sy. This means that you have to rotate around the x-axis by pi/2:

    [tex]|s'>=\exp\left( \frac{i}{\hbar} S_{x}\frac{\pi}{2}\right)|s>[/tex]

    If you work this out for spin 1/2 using properties of Pauli matrices you get

    [tex]\exp\left( \frac{i}{\hbar} S_{x}\frac{\pi}{2}\right) =\frac{\sqrt{2}}{2}\begin{pmatrix}1&i\\i&1\end{pmatrix}[/tex]
  16. Nov 30, 2007 #15


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    This is all good but I am not sure if it's the right level for the OP. It's always tricky to answer questions at the right level (people have to learn to walk before lerning to run :smile:). I felt that from the comments made, it was more appropriate to first learn to find the eigenavlues and eigenvectors explicitly. But if this is clear top the OP, that's a good way to do it!
  17. Nov 30, 2007 #16
    So, you see that |+> = (1,0) transforms into 1/sqrt(2) (|+> + i |-> ) = 1/sqrt(2) (1,i) so this is the eigenvector of Sy with eigenvalue h-bar/2 and |-> = (0,1) transforms into 1/sqrt(2) (i,-1) which is the eigenvector corresponding to the eigenvalue of -h-bar/2

    Of course, the moment you know one, the other is fixed as the eigenvectors are orthogonal.
  18. Nov 30, 2007 #17
    I agree that the student must master linear algebra first. But this means that the student must still learn about coordinate transformations, unitary transformations etc. And this problem is suitable to demonstrate coordinate transformations as well.
  19. Nov 30, 2007 #18


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    Very true.


  20. Nov 30, 2007 #19
    Count Iblis - I understand what you mean. And that makes perfect sense too, that's a great way of getting the different spins. So wikipedia shows the [tex]S_{y}[/tex] e-spinors as something different. They show:

    [tex]X^{y}_{+} = \frac{1}{\sqrt{2}} \left[ \begin{array}{cc}1 \\i \end{array}\right][/tex]
    [tex]X^{y}_{-} = \frac{1}{\sqrt{2}} \left[ \begin{array}{cc}1 \\-i \end{array}\right][/tex]

    I don't think our professor is looking for this kind of solution; but I'm sure he would accept it.

    Hmmm... in your next post you show what wikipedia has; but I don't really "see" how it transforms, so maybe nrqed is right. Maybe I'm not quite ready to grasp all of that yet.

    So use the eigenvalue equation and input those e-values for [tex]\lambda[/tex], and I get:

    [tex]\frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} \left( \begin{array}{cc}a \\b \end{array}\right) = \pm \frac{\hbar}{2}\left( \begin{array}{cc}a \\b \end{array}\right)[/tex] --->

    [tex]\left( \begin{array}{cc}-ib \\ia \end{array}\right) = \left( \begin{array}{cc}a \\b \end{array}\right)[/tex]

    So, [tex]a = \mp ib[/tex]. And then what? I don't see how I get from that point to what they show on wikipedia.
  21. Nov 30, 2007 #20


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    Now you have a relation between a and b.
    Then you must normalize your state and it will fix the eigenvectors completely (well, apart for an overall pure phase [itex] e^{i \theta}[/itex] which is irrelevant).
  22. Dec 4, 2007 #21
    Well, after a few days I'm still a bit confused with this problem. According to wikipedia the eigenspinors of [tex]S_{y}[/tex] are:

    [tex]X^{y}_{+} = \frac{1}{sqrt{2}}\left[ \begin{pmatrix}1 \\ i \end{pmatrix}\right][/tex]

    [tex]X^{y}_{-} = \frac{1}{sqrt{2}}\left[ \begin{pmatrix}1 \\ -i \end{pmatrix}\right][/tex]

    [tex]X^{y}_{+} = \frac{1}{sqrt{2}}\left[ \begin{pmatrix}1 \\ i \end{pmatrix}\right][/tex]

    [tex]X^{y}_{-} = \frac{1}{sqrt{2}}\left[ \begin{pmatrix}1 \\ -i \end{pmatrix}\right][/tex]
    Last edited: Dec 4, 2007
  23. Dec 4, 2007 #22
    The vector [1, -i] can also be written as -i[i, 1]. Since the overall factor -i is a phase factor (it has modulus 1 and can be written as exp(3/2 pi i) ) you can omit it and still have a normalized eigenvector. So, you can just as well take the eigenvector to be [i,1].

    The other eigenvector is [1, i] and you see that [1, -i] is orthogonal to this eigenvector: [1, i] dot [1, -i] =
    1*conjugate (1) + i*conjugate (-i) =1*1 + i*i = 0


    [1, i] dot [i, 1] = 1 Conjugate(i) + i Conjugate(1) = -i + i = 0
  24. Dec 4, 2007 #23
    I'm just not sure how to get from my relation: [tex]a = \mp ib[/tex] to:

    [tex]X^{y}_{+} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i \end{pmatrix}[/tex]

    [tex]X^{y}_{-} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i \end{pmatrix}[/tex]
  25. Dec 4, 2007 #24
    Take a = 1, then b = ±i. Now the norm of a vector [1,±i] is sqrt(2), so you have to divide the vector by sqrt(2) to get a vector with norm 1. Of course, this means amounts to taking a = 1/sqrt(2) from the start. However it is often easier to fix the normalization at the end because the norm is a nonlinear function of the coefficients.
  26. Dec 4, 2007 #25
    Oh, ok. Great, that's exactly what I needed. Thanks Count Iblis and nrqed and others.
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