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Spinors (quantum mechanics)

  1. Aug 7, 2010 #1
    This is not an assignment problem, but I am studying for my quantum mechanics final exam and came across a derivation in the book which I can't seem to get my head around :(

    The example in the book is solving for the probabilities of getting +h(bar)/2 and -h(bar)/2 if we are to measure the spin angular momentum Sx.

    I was able to follow the derivation up to the point where they obtained the eigenspinors:

    X+ = [1/sqrt2 1/sqrt2]' and X- = [1/sqrt2 -1/sqrt2]'

    But I don't get how they go from those to formulating the spinor:

    X = [(a+b)/sqrt2]X+ + [(a-b)/sqrt2]X-

    Any guidance would be much appreciated - thanks in advance.
     
  2. jcsd
  3. Aug 8, 2010 #2

    diazona

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    Homework Helper

    I'm assuming a and b are the coefficients in
    [tex]\chi = \begin{pmatrix}a \\ b\end{pmatrix}[/tex]
    right?

    Think of it this way: all possible spinors form a vector space. The vectors (1,0) and (0,1) form the standard basis for that space - in other words, when you have a state [itex]\chi[/itex] defined by two coefficients a and b, that's actually saying
    [tex]\chi = a\begin{pmatrix}1 \\ 0\end{pmatrix} + b\begin{pmatrix}0 \\ 1\end{pmatrix}[/tex]
    But you can express the same state in terms of any other basis. For example, [itex]\chi_+[/itex] and [itex]\chi_-[/itex] form a basis (just like (1/√2,1/√2) and (1/√2,-1/√2) form a basis for the xy plane), so you can write the state [itex]\chi[/itex] as
    [tex]\chi = c\chi_+ + d\chi_- = c\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix} + d\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{pmatrix}[/tex]
    The coefficients c and d are the components of [itex]\chi[/itex] in the +- basis. Can you find them?

    (Hint: if you're familiar with the vector projection formula, that's probably the quickest - though certainly not the only - way to do it)
     
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