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Eigenvalues and Eigenspinors 
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#1
Nov3007, 09:47 AM

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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
Nov3007, 09:56 AM

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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) 


#3
Nov3007, 09:59 AM

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Here is an example (click with yoru mouse on the tex output to see the code I typed in) [tex] \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{pmatrix} [/tex] 


#4
Nov3007, 10:25 AM

P: 223

Eigenvalues and Eigenspinors
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. 


#5
Nov3007, 10:46 AM

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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. 


#6
Nov3007, 10:53 AM

P: 2,157

It is easier to simply rotate the eigenvectors of [tex]\sigma_z[/tex] around the xaxis.



#7
Nov3007, 11:05 AM

P: 223

Well, for my values I just set:
[tex]ai\frac{\hbar}{2} = a\lambda[/tex] > [tex]\lambda = i\frac{\hbar}{2}[/tex] and [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? 


#8
Nov3007, 11:10 AM

P: 223

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. 


#9
Nov3007, 11:16 AM

P: 173

i think it is not correct, the secular equation reads: Lambda^21=0 so lambda=+1 or1. 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.



#10
Nov3007, 11:25 AM

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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. 


#11
Nov3007, 11:29 AM

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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). 


#12
Nov3007, 11:58 AM

P: 223

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??? 


#13
Nov3007, 12:07 PM

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#14
Nov3007, 12:39 PM

P: 2,157

Rahmuss,
Just consider cheating a bit. 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 yaxis, then the new zdirection will point in the direction of the old xdirection. A particle with spin polarized in the old zdirection will be polarized in the minus xdirection. 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 xdirection. To solve this problem we want to rotate the coordinate axis such that the new ydirection will point in the direction of the old zaxis. The eigenvectors of Sz will then transform into the eigenvectors of Sy. This means that you have to rotate around the xaxis 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] 


#15
Nov3007, 12:43 PM

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#16
Nov3007, 12:52 PM

P: 2,157

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 hbar/2 and > = (0,1) transforms into 1/sqrt(2) (i,1) which is the eigenvector corresponding to the eigenvalue of hbar/2
Of course, the moment you know one, the other is fixed as the eigenvectors are orthogonal. 


#17
Nov3007, 12:57 PM

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#18
Nov3007, 01:34 PM

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Regards Patrick 


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