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QM - Eigenspinors

  1. Sep 15, 2007 #1
    Please read the following posts. I know what is contained in this original is wrong, and I have since approached it differently.

    1. The problem statement, all variables and given/known data
    Consider a measurement of spin in the x-y plane, with the operator [tex]\hat{Q} = \frac{3\hat{S_x}+4\hat{S_y}}{5} [/tex]
    a) What are the eigenvalues and eigenstates (eigenspinors) of this operator?
    b) Given a spinor [tex]X = \frac{1}{\sqrt{5}} \binom{2}{1} [/tex], what is the probability that the above measurement yields the result of [tex]\frac{-\hbar}{2}[/tex]?

    2. Relevant equations

    3. The attempt at a solution
    Ok, for part a, im not sure where to start.
    Can I "say" that +-h/2 (from now on h=hbar) are the eigenvalues for each of the Sx,Sy operators in the main operator, and then just sum them up over all possibilities?
    3/5 * h/2 + 4/5 * h/2 = 7h/10
    3/5 * h/2 - 4/5 * h/2 = -h/10
    -3/5 * h/2 + 4/5 * h/2 = h/10
    -3/5 * h/2 - 4/5 * h/2 = -7h/10

    Or do I have to say that,
    [tex]QX^+ = (3/5)S_x X^+_x + (4/5)S_ yX^+_y [/tex]

    And thus when [tex]Q X^+ [/tex] is considered I use seperate vectors for each of the operators?

    [tex]X^+_x = \frac{1}{\sqrt{2}} \binom{1}{1}[/tex]
    [tex]X^-_x = \frac{1}{\sqrt{2}} \binom{1}{-1}[/tex]
    [tex]X^+_y = \frac{1}{\sqrt{2}} \binom{1}{i}[/tex]
    [tex]X^-_y = \frac{1}{\sqrt{2}} \binom{1}{-i}[/tex]

    Doing it that way, using the Pauli spin matrices for Sx,Sy, I get the possible eigenvalues to be:
    [tex]\frac{7\hbar}{10\sqrt{2}}[/tex] for [tex]X^+[/tex]

    [tex]\frac{-7\hbar}{10\sqrt{2}}[/tex] for [tex]X^-[/tex]

    Which of these methods are correct? Or am I doing this all wrong?
    The reason I ask, is part (b):

    To measure a probability of the [tex]\frac{-\hbar}{2}[/tex], I would assume that means that [tex]\hat{Q}X^? = \frac{-\hbar}{2}[/tex]

    and I believe from there you would do something like
    [tex]C_? = X^? X = X^? \frac{1}{\sqrt{5}} \binom{2}{1}[/tex]

    and then [tex]|C_?|^2 [/tex] would be the probability.

    Any hints or suggestions please?

    I feel like I understand what they're asking, but my concept of the eigenspinor may not be that clear.
    Wouldn't the eigenspinors of some operator Q be whatever eigenvectors you can use with it? But what are the constraints? Do I have to normalize the results that I got? Or are they correct?
    If so how would I ever measure -h/2 for part b, if my eigenspinors don't contain that value? So I assume they should. But how?

    Thank you!

    -Kristopher Healey
    Last edited: Sep 16, 2007
  2. jcsd
  3. Sep 16, 2007 #2


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    I don't understand what you're trying to do, but it doesn't look right. Just write the matrix form of the operator Q (from the matrix forms of Sx and Sy) and compute its eigenvalues and eigenvectors.
  4. Sep 16, 2007 #3
    Hmm, like this?
    [tex]\hat{Q} = \frac{3}{5}\frac{\hbar}{2}\left[\begin{array}{ c c }0 & 1 \\ 1 & 0\end{array} \right]+\frac{4}{5}\frac{\hbar}{2}\left[\begin{array}{ c c }0 & -i \\ i & 0\end{array} \right][/tex]

    \hat{Q} = \frac{\hbar}{10}\left[\begin{array}{ c c }0 & 3-4i \\ 3+4i & 0\end{array} \right]


    Well doing the determinate I get the eigenvalues to be [tex]\pm\frac{\hbar}{2}[/tex]

    then if i try to find the eigenspinors:

    \hat{Q}\binom{a}{b} = \left[\begin{array}{ c c }0 & \frac{3\hbar}{10}-\frac{4\hbar}{10}i \\ \frac{3\hbar}{10}+\frac{4\hbar}{10}i & 0\end{array} \right]-\frac{\hbar}{2}\left[\begin{array}{ c c }1 & 0 \\ 0 & 1\end{array} \right]=0

    I solve for the first one, just positive hbar/2, and I get:

    How do I solve for (a b)? I seem to just go in loops, but how do I find the basis?

    And thank you StatusX for the help!
    Last edited: Sep 16, 2007
  5. Sep 16, 2007 #4
    So far so good... in fact, it makes good intuitive sense: measurements of spin for a spin 1/2 system *always* gives either +1/2 or -1/2, no matter which direction you measure in. What's the eigenstates though?
  6. Sep 16, 2007 #5


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    You have to build the matrix for Q, as you did in your other post. As you found, it has eigenvalues [tex] \pm \hbar/2 [/tex]. Now you will have to find teh corresponding eigenstates and you will have to project the state they give you on the two eiegnstates of Q to find the amplitudes of getting [tex] \pm \hbar/2 [/tex].
  7. Sep 16, 2007 #6
    Please see my above post.
    I'm not sure how to get them, its been a while since I've had quantum and this is 2nd week homework in my applications class.

    Also, the problem did not specify it was a spin 1/2 system, but I just assume it is (though we did do the general case as well).

    Are the eigenstates just such that Q(a b) = +-hbar/2 (a b) ?
    or is there an easier way to find eigenstates?
  8. Sep 16, 2007 #7
    thanks nrged, yeah I realized that was wrong, should I change my OP?

    How do I find eigenvectors/spinors from this?
  9. Sep 16, 2007 #8
    Just find the eigenvectors for the matrix... if you can't remember, look back in your old notes. It's important for you to learn that stuff solidly (*really* solidly -- you'll need it lots). Suffice to say that, no, it's not what you've written so far. Can you remember the definition of an eigenvector and its geometric significance? If so, you'll see that what you've written *can't* be correct.
  10. Sep 16, 2007 #9


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    No need to change the OP, no.

    Yes, you must indeed solve Q(a b) = hbar/2 (a b) (where 9a b) means a two components column vector and a, b are complex numbers. and then you must solve Q (c d) = - hbar/2 (c d).

    At this point, it is just a linear algebra problem: finding the eigenvectors of a matrix. You will also need to normalize them, i.e. impose |a|^2 + |b|^2 = 1 etc.

    If you haven't seen an example, you should really look up an example worked out in details in a book.
  11. Sep 16, 2007 #10
    I can do linear algebra, thats not the problem (its been about 5 years since I took the class though).
    I looked at some examples and they're all real, not complex. The example we did in class was just for figuring out Sz, Sx, Sy, and S^2, which I can do. But for some reason, I can't seem to get an eigenvector out of this.

    And Genneth, when you say what ive written is wrong so far, which post did you mean? My first is obviously wrong, and I've agreed.

    I know that the eigenvector will be such that the vector does not change under the operator. And the examples that I've seen I can do, but for some reason I'm getting caught up in this.
    Has anyone solved it? I don't care about the answer, just that you got something that makes sense. I feel like I made a mistake earlier and its making things worse now.
    Last edited: Sep 16, 2007
  12. Sep 16, 2007 #11


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    The calculation for a complex matrix is not really different from real marices. If you follow the examples of Sx and Sy 9which are not diagonal) then you should be able to do Q. Just post your calculation and we can point out any mistake.
  13. Sep 16, 2007 #12
    You're now on the right track -- stick with it. If you really want, we can just put the answer up, but it wouldn't be as helpful.
  14. Sep 16, 2007 #13
    \hat{Q}\binom{a}{b} = \left[\begin{array}{ c c }0 & \frac{3\hbar}{10}-\frac{4\hbar}{10}i \\ \frac{3\hbar}{10}+\frac{4\hbar}{10}i & 0\end{array} \right]\binom{a}{b}=\frac{\hbar}{2}\binom{a}{b}
    \hat{Q}\binom{c}{d} = \left[\begin{array}{ c c }0 & \frac{3\hbar}{10}-\frac{4\hbar}{10}i \\ \frac{3\hbar}{10}+\frac{4\hbar}{10}i & 0\end{array} \right]\binom{c}{d}=\frac{-\hbar}{2}\binom{c}{d}

    Ok so start with the (a b) first:

    [tex]b(\frac{3\hbar}{10}-\frac{4\hbar}{10}i) = \frac{\hbar}{2}a[/tex]

    [tex]a(\frac{3\hbar}{10}+\frac{4\hbar}{10}i) = \frac{\hbar}{2}b[/tex]

    multiply through by 10/hbar:

    [tex]b(3-4i) = 5a[/tex]

    [tex]a(3+4i) = 5b[/tex]

    is a system of linear dependent equations right?
    How do I solve that? If I just solve for a and plug it in it just is true for any "a" or "b".
    Also the (0 0) case.

    Its already true a*a + b*b = 1 if I solve for either a or b. But that doesn't really help me, does it?
    Last edited: Sep 16, 2007
  15. Sep 16, 2007 #14
    The answer would be very helpful. I'm not going to take it for granted. I don't care about my homework score or anything, I really want to learn this. Maybe with the answer I could see where I went wrong. I know how to do the part(b) of this problem, but without the eigenspinors I can't really do it.
    I hope its obvious I'm actually working on figuring it out, and not just looking for a solution to put on paper. I want to understand it, and if you have a solution please let me know, or at least guide me a little more.

  16. Sep 16, 2007 #15
    Remember that an eigenvector is only defined up to a direction -- its length could be anything: Abx = abx, for any b. So your solution should be a line, through the origin. Mathematically, this means that the two equations are dependent, so you might as well just ignore one of them. Thus, a = 3-4i and b = 5 is a valid solution. Now, for it to make physical sense, you have to normalise the vector, but that should be straightforward for you.
  17. Sep 16, 2007 #16


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    If you isolate a from the first equation and plug that into the second equation
    you get a trivial result b=b. This is always like that in those equations: the equations are not linearly independent.

    So now impose that |a|^2 + |b|^2 = 1. Using either of your two equations to isolate b (let's say) and using this normalization condition will allow you to solve for "a" up to a phase. Assume that the phase (which is physically irrelevant) is zero, in other words assume that "a" is real. Once you have "a", plug it back in one of the two equations to find b.
  18. Sep 16, 2007 #17
    ahh i see thanks!

    so [tex]\frac{1}{5\sqrt{2}}\binom{3-4i}{5}[/tex]

    should work right?
    Now does that span all possible eigenspinors for the +h/2 case?

    I'll finish the 2nd case now...

    same thing, but
    Last edited: Sep 16, 2007
  19. Sep 16, 2007 #18


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    yes, it looks good.

    Note that you can always double check your answer by plugging this back in the eigenvalue equation to see if you get +hbar/2 as an eigenvalue.
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