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Finding eigenvectors and eigenspinors of S_y

  1. Dec 7, 2005 #1
    I have a homework problem whose first part asks for the eigenvectors and eigenspinors of [tex] S_y [/tex]. My problem is following the text to figure out what (mathematically) is [tex] S_y [/tex].
    Also, The book first derives S^2, using the following method using [tex] 3/4\hbar^2 [/tex] as the eigenvalue. I was orignally going to follow the books example for S^2 and use the same strategy, however since they didnt explain themselves, I am a little confused. Any help would be appreciated. Thanks.
     
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  3. Dec 7, 2005 #2

    Physics Monkey

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    Presumably you are trying to calculate the eigenvectors of [tex] S_y [/tex] in terms of the eigenvectors of [tex] S_z [/tex]. The electron is spin half so the spin is represented by the Pauli matrices i.e. [tex] S_y = \frac{\hbar}{2} \sigma_y [/tex]. Since you know the matrix representation of [tex] S_y [/tex] in the [tex] S_z [/tex] basis, what should you do to find the eigenvectors?
     
  4. Dec 8, 2005 #3
    Ooops I ment to say eigenvalues, not eigenvectors. Sorry about that. I do not know of S_y expressed as a matrix in the S_z basis though. I know that

    [tex] S_y= (\frac{1}{1i})(S_+ - S_-) [/tex] if I substitue [tex] S_y = \frac{\hbar}{2} \sigma_y [/tex] in to that first equation, I get [tex] S_y = \frac {h}{2i}[/tex]times the 2x2 matrix: first row (0,1) second row (-1,0).

    so is [tex] \frac{\hbar}{2i} and -\frac{\hbar}{2i}[/tex] my eignevalues?

    sry not sure how to get matricies done with latex.
     
    Last edited: Dec 8, 2005
  5. Dec 8, 2005 #4

    Galileo

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    Since [itex]S_y[/itex] is an observable (Hermitian) its eigenvalues are ofcourse real, so it can't be the right answer.
    BTW: I think the relation should be [itex]S_y=\frac{1}{2i}(S_+-S_-)[/itex].

    What do you know about [itex]S_+[/itex] and [itex]S_-[/itex]? Do you know how they act on the eigenstates of [itex]S_z[/itex]?
     
  6. Dec 8, 2005 #5
    ah yes [tex] 1/2i [/tex] sry bout that.


    Well I know S_+ and S_-, and knowing that [tex] S_y= (\frac{1}{2i})(S_+-S_-) [/tex] we can simplify and get

    [tex] S_z = \frac{\hbar}{2} \sigma_z [/tex]


    but im not sure what to do with that, or what it means to the eigenvalues of [tex] S_y [/tex]
     
  7. Dec 8, 2005 #6

    Galileo

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    So what happens if you let [itex]S_y[/itex] act on [itex]|\frac{1}{2} \frac{1}{2}\rangle[/itex] and [itex]|\frac{1}{2} -\frac{1}{2}\rangle[/itex]?
     
  8. Dec 9, 2005 #7

    dextercioby

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    It's not cheating but i think your problem's solved in Schaum's outline on Quantum Mechanics (solved problems book).

    Daniel.
     
  9. Dec 10, 2005 #8
    ah thx guys for the hints, I think I got it. However, I did get trying to find the probibility of finding [tex] S_y [/tex] in the general state of [tex] \chi [/tex]. My general state [tex] \chi = a(\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}}) + b(\frac{1}{\sqrt{2}},\frac{-i}{\sqrt{2}}) [/tex] where those row vectors are actually collum vectors. So obviously a,b are gonna be my coefficents, but when I try to normalize this, I do not get 1/2 and 1/2 for my answer. I am not sure what I'm doing wrong! thx in advance.
     
  10. Dec 11, 2005 #9

    Galileo

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    Ok, so you've found the right eigenstates of [itex]S_y[/itex] and you've written some general state [itex]\chi[/itex] as a linear combination of these eigenstates. Did you also find the corresponding eigenvalues? (That should be really easy now)

    What do the postulates say about the probability of measuring, say [itex]\frac{\hbar}{2}[/itex], if you measure [itex]S_y[/itex] for the state [itex]\chi[/itex]?

    BTW: You wrote: 'the probability of finding [itex]S_y[/itex]. A hope that was a typo. You MEASURE an observable. You FIND an eigenvalue with some probability. The probability ofcourse depends on the state, so it's not always 1/2 and 1/2.
     
    Last edited: Dec 11, 2005
  11. Dec 11, 2005 #10
    Yes thats what i ment, i guess i was trying to be a little too pithy. I think I know of the posulate you are thinking of,
    [tex] \chi = (\frac{a + b}{\sqrt{2}})\chi_+ + (\frac{a - b}{\sqrt{2}})\chi_- [/tex]
    where [tex] \chi_+ [/tex] coressponds to the eigenvaule [tex] \frac {\hbar}{2} [/tex] and [tex] \chi_- [/tex] coressponds to [tex]\frac {-\hbar}{2} [/tex]



    Oh wait, I forgot that when I find the probabilities, its [tex] |a+b|^2 and |a-b|^2 [/tex]

    which in both cases means [tex] (a+b)(a-b^*) [/tex]
     
    Last edited: Dec 11, 2005
  12. Dec 12, 2005 #11

    Galileo

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    Wishbone. I would advice you to read your book and go through each of the postulates carefully. I find it hard to understand what you mean exactly with most of the questions you post. This question is the simplest possible nontrivial exercise in applications of the QM postulates in the sense that the state space is simply 2-dimensional. But it seems as though the relationship between and importance of observable operators, eigenvectors and eigenvalues is not clear. Please read your book and come with questions if something is not clear. If you make exercises before understanding the fundamentals you will not learn much from them.
     
    Last edited: Dec 12, 2005
  13. Dec 12, 2005 #12
    I am not exactly sure what you mean, this specific question was to find the probiblities. I read the book, I do not understand all of it, obviously, which is why I came here. What about my question is unclear?
     
  14. Dec 12, 2005 #13

    Galileo

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    This post is vague. I guess, but am not sure that [itex]\chi_+[/itex] and [itex]\chi_-[/itex] are supposed to be the eigenstates of [itex]S_x[/itex]. Then, if the general state in terms of the eigenfunctions of [itex]S_z[/itex] is [tex]\chi=a\chi_{z_+}+b\chi_{z_-}[/tex], then the expansion [itex]\chi[/itex] as a linear combination of the eigenstates of [itex]S_x[/itex] is [tex] \chi = (\frac{a + b}{\sqrt{2}})\chi_+ + (\frac{a - b}{\sqrt{2}})\chi_- [/tex]
    as you wrote down.
    Anyway I shouldn't have to guess this and it also doesn't immediately apply to your problem since you must have the eigenstates of [itex]S_y[/itex], not those of [itex]S_x[/itex].

    Ok, here is a question to see if you got the postulates down right. Suppose [itex]\{|\varphi_n\rangle\}_{n=1}^{N}[/itex] is a complete set of eigenstates of the observable Q with [itex]|\varphi_i\rangle[/itex] corresponding to the eigenvalue [itex]a_i[/itex] Assume the spectrum is nondegenerate. Let [itex]|\psi\rangle[/itex] be an arbitrary ket and expand it in the given eigenbasis:

    [tex]|\psi \rangle =\sum_{i=1}^N c_n |\varphi_n\rangle[/tex]

    Now, for a system in the state [itex]|\psi\rangle[/itex], what is the probability of measuring [itex]a_i[/itex] when a measurement of Q is made?

    If you can answer this, you can also answer the question on your problem.
     
    Last edited: Dec 12, 2005
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