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Probability that Measurement of Spinor Yields Spin-Down State

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the spinor [tex]\frac{1}{\sqrt{5}}\left(\begin{array}{cc}2\\1\end{array}\right) [/tex]. What is the probability that a measurement of [tex]\frac{3S_{x}+4S_{y}}{5}[/tex] yields the value [tex]-\frac{\hbar}{2}[/tex]?

    2. Relevant equations

    [tex]S_{x}=\frac{1}{2}\hbar\left(\begin{array}{cc}0&1\\1&0\end{array}\right)[/tex]
    [tex]S_{y}=\frac{1}{2}\hbar\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)[/tex]

    3. The attempt at a solution

    I'm looking at two different ways to solve this and figuring how to choose between.

    My first instinct was to be real nice, clean, and simple by expanding the spinor in the basis
    [tex]\frac{1}{\sqrt{5}}\left(\begin{array}{cc}2\\1\end{array}\right)=\frac{2}{\sqrt{5}}\left(\begin{array}{cc}1\\0\end{array}\right)+\frac{1}{\sqrt{5}}\left(\begin{array}{cc}0\\1\end{array}\right)[/tex]
    And then going directly to squaring the value next to the down state so that P = 1/5

    But that seems too easy to trust.

    So I dig in my book and it says that I can expand a state in the eigenstates of the operator S_xcos(phi)+S_ysin(phi) (which are complex) which would end up looking like
    [tex]\frac{1}{\sqrt{5}}\left(\begin{array}{cc}2\\1\end{array}\right)=\alpha_{+}\frac{1}{\sqrt{2}}\left(\begin{array}{cc}e^{-i\phi/2}\\e^{i\phi/2}\end{array}\right)+\alpha_{-}\frac{1}{\sqrt{2}}\left(\begin{array}{cc}e^{-i\phi/2}\\-e^{i\phi/2}\end{array}\right)[/tex]

    I solve for alpha-minus in an <alpha|S sort of way...
    [tex]\alpha_{-}=\frac{1}{\sqrt{10}}\left(\begin{array}{cc}e^{i\phi/2}&-e^{-i\phi/2}\end{array}\right)\left(\begin{array}{cc}2\\1\end{array}\right) \Rightarrow \frac{1}{\sqrt{10}}(cos(\phi/2)+3isin(\phi/2))[/tex]

    And square it to get probability so that
    [tex]|\alpha_{-}|^2 = \frac{1}{10}(cos^2(\phi/2)+9sin^2(\phi/2))[/tex]
    Where phi happens to be the arctan(4/3) (because when you look at how the operator is put together it has the complex number 3-4i bouncing around which corresponds to a 3-4-5 Pythagorean triple where phi would be the angle between x=3 and r=sqrt(x^2+y^2)=5) and this results in P = .259

    Two different paths to two different solutions, the first just too simple and the second pretty complex... I'd like to know, which, if either, is right, and also why.

    Thank You So Much!
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    The first method is incorrect. You expanded in the z basis. But your operators are Sx and Sy. So the basis vectors in the z basis are not eigenvectors of the operator you're looking at.

    Your second method looks more or less correct. I'm not going to check the algebra, though.

    Another way to tackle this is to look at the expectation value

    [tex]\langle \chi | \left( \frac35 S_x + \frac45 S_y \right) | \chi \rangle[/tex]

    which can be evaluated by simple matrix multiplication. Since there are only two possible spin states ([itex]+\hbar/2[/itex] and [itex]-\hbar/2[/itex]), the expectation value will tell you exactly how their probabilities must be weighted. For example, if the expectation value is 0, then the spin states must be weighted equally, etc.

    Note, however, that this trick only works for spin 1/2. For spin 1 and higher, in general, there are more than two possible spin states, so merely calculating the expectation value of the spin is not enough information to determine all the weights. In general, you must expand the state spinor in eigenvectors of the spin operator to find the probabilities for each value of spin angular momentum.
     
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