What's Wrong with My Eigenvector Calculation?

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athrun200
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Homework Statement


Printed homework problem on spin-1/2 particles: Sx, Sy and Sz given as 2x2 matrices with hbar/2, and a request to find eigenvalues and eigenfunctions of Sxy = Sx + Sy, then measurement probabilities of Sz for state beta, 30 marks.



Homework Equations





The Attempt at a Solution


I don't know what's wrong with my work. I can't obtain the eigenvector provided in the model answer.

My work
Handwritten derivation of eigenvalues and normalized eigenvectors of a 2x2 matrix operator, with eigenvalues plus and minus 1/sqrt(2) and normalization giving |beta| = 1/sqrt(2).


Model Answer
Typed text giving eigenvalues hbar/sqrt(2) and -hbar/sqrt(2), eigenstates |beta> = (e^(-i pi/8), e^(i pi/8)) and |alpha> = (e^(-i pi/8), -e^(i pi/8)) as column vectors, and that measuring Sz in state |beta> gives hbar/2 and -hbar/2 each with probability 0.5.
 
Attachments
  • Handwritten notes computing eigenvalues and eigenvectors of a 2x2 matrix S_xy, with eigenvalues of plus and minus 1/sqrt(2) and normalization of the eigenvector components.
    IQM.jpg
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  • Printed exam question on a spin-1/2 particle: Sx, Sy and Sz given as 2x2 matrices with hbar/2 prefactors, asking for eigenvalues and eigenfunctions of Sxy = Sx + Sy and measurement probabilities of Sz for the maximal-eigenvalue state (30 marks).
    IQM1.jpg
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  • Typed text stating eigenvalues hbar/sqrt(2) and -hbar/sqrt(2), two column-vector eigenstates |beta> and |alpha> with entries involving e^(-i pi/8), and that measuring S_z in state |beta> gives hbar/2 and -hbar/2 each with probability 0.5.
    IQM2.jpg
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Your work is fine. Remember that an eigenvector is only unique up to a multiplicative constant. The eigenvector you found can be written
$$\left\lvert \frac{\hbar}{\sqrt{2}} \right\rangle = \begin{bmatrix} \frac{1-i}{2} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix}\frac{1-i}{\sqrt{2}} \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}e^{-i\pi/4} \\ 1 \end{bmatrix}.$$ If you multiply that by ##e^{i\pi/8}## and ignore the normalization factor of ##1/\sqrt{2}##, you'll get the answer in the solution.
 
Oh! Thank you very much.
But why do we bother to have such a complicated eigenvector?
The one with (1-i)/2 and 1/(sqrt2) is much easier to find. Why do we need to change it to the form of exp?
 
You don't need to, but you have to admit there's certain symmetry there. And physicists like symmetry. :wink: