1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: The generic spinor as a linear combination

  1. Mar 28, 2008 #1
    I have trouble understanding the concept of spin (spin 1/2 in this case). In Introduction to Quantum Mechanics Griffiths states that "the generic spinor X can be expressed as a linear combination of [eigenvectors of the spin component Sx]
    [tex]\chi = \frac{a+b}{\sqrt{2}}\chi^{(x)}_+ + \frac{a-b}{\sqrt{2}}\chi^{(x)}_-[/tex]

    What I don't understand (among other things) is why Sy and Sz don't count. Because there are only to directions, up and down? (If so, how do I know which of Sx,Sy and Sz to choose?)

    Another thing is that I'm told that a spin in the direction (x,y,z) is defined as [tex]\textbf{S}=\frac{\hbar}{2}(x\sigma _x +y\sigma _y +z\sigma _z)[/tex]. What exactly is this and how does it relate to the spinor? It is a matrix but a spinor (that's also supposed to describe the spin state) is a vector.
  2. jcsd
  3. Mar 28, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    Well, Sy and Sz do count. Point is, [tex]\chi^{(x)}_+[/tex] and [tex]\chi^{(x)}_-[/tex] are eigenspinors of the [tex]S_x[/tex] operator. These two eigenspinors span the 1/2 spin states, so any general state can be written as a linear combination of the two.

    If you write the general spin state in terms of the eigenspinors of the [tex]S_z[/tex] operator,

    [tex] \chi_+ = \left( \begin{array}{c} 1 \\ 0 \end{array} \right) [/tex]

    [tex] \chi_-=\left( \begin{array}{c} 0 \\ 1 \end{array} \right) [/tex]


    [tex] \chi = a \chi_+ + b \chi_-[/tex], this general spin state can also be written in the basis of the eigenspinors of the [tex]S_x[/tex] operator as

    [tex]\chi = \frac{a+b}{\sqrt{2}}\chi^{(x)}_+ + \frac{a-b}{\sqrt{2}}\chi^{(x)}_-[/tex]

    S is not the spin state, but the operator to measure the spin in the required direction. In this notation, x,y,z are components of a unit vector in the required direction. [tex]\sigma _x, \sigma _y, \sigma _z[/tex] are the pauli matrices.
    Last edited: Mar 28, 2008
  4. Mar 28, 2008 #3
    Ok, many thanks. Though I'm still not sure if I understand this. I mean it's quite unclear to me how these are used.

    Edit: The problem is the following:
    We have vectors a=(a1,a2,a3) and b=(b1,b2,b3)
    Express the spin state |+>b as a linear combination of the normalized eigenvectors of the spin operator Sa and show that the probabilities to measure +/- hbar/2 are [tex]cos^2 (\theta/2)[/tex] and[tex]sin^2 (\theta/2)[/tex] where theta is the angle between a and b.

    I know the eigenvectors of Sa but can't get past that.

    Below is everything I have so far.

    How do I calculate the eigenvalue of S?

    [tex]\textbf{S}=\frac{\hbar}{2}(x \sigma _x + y \sigma _y + z \sigma _z)=
    \frac{\hbar}{2}\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)
    So the eigenvalues would be..?
    \left| \begin{array} {cc}\frac{\hbar}{2}z-\lambda & \frac{\hbar}{2}(x-iy) \\ \frac{\hbar}{2}(x+iy) & -\frac{\hbar}{2}z-\lambda \end{array}\right|
    \Rightarrow \lambda = ^+_- \frac{\hbar}{2}[/tex]
    because (x,y,z) is a unit vector.
    If I'm to find the eigenvector (of the positive eigenvalue):
    [tex]\textbf{Sv}=\frac{\hbar}{2}\textbf{v}\Rightarrow\frac{\hbar}{2}\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)\textbf{v}=\frac{\hbar}{2}\textbf{v}\Rightarrow\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)\textbf{v}=\textbf{v}\Rightarrow\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)\left( \begin{array}{c} u \\ v \end{array} \right)=\left( \begin{array}{c} u \\ v \end{array} \right)[/tex]
    I get two equations:
    [tex]zu+(x-iy)v=u \ \ ,\ \ (x+iy)u-zv=v[/tex]
    Rearranging these:
    [tex](z-1)u+(x-iy)v=0 \ \ = \ \ (x+iy)u-(z+1)v=0[/tex]
    And again:
    u=-x+iy-z-1 \ \ ,\ \ v=-x-iy+z-1
    So the eigenvector would be
    [tex]\left( \begin{array}{c} -x+iy-z-1 \\ -x-iy+z-1 \end{array}\right)[/tex]
    And for the negative eigenvalue
    [tex]\left( \begin{array}{c} -x+iy-z+1 \\ -x-iy+z+1 \end{array}\right)[/tex]
    These seem to work if I insert them into
    Normalizing the eigenvectors I get:
    [tex]\chi_+=\frac{1}{2\sqrt{x+1}}\left( \begin{array}{c} -x+iy-z-1 \\ -x-iy+z-1
    [tex]\chi_- =\frac{1}{2\sqrt{-x+1}}\left( \begin{array}{c} -x+iy-z+1 \\ -x-iy+z+1 \end{array}\right)[/tex]
    But I don't actually understand what I've just done. Is this physically reasonable?

    How do I actually use these to get some generic spinor?
    [tex]\chi = \frac{a+b}{2\sqrt{x+1}}\chi _+ + \frac{a-b}{2\sqrt{-x+1}}\chi _-\ \ ?[/tex]

    And to make things even more laborious, I'd have to know how to express the spin state of some other vector (i,j,k) as a linear combination of the previous eigenvectors. I know that when vectors m = (x,y,z) and n = (i,j,k) are unit vectors, then there's a relation [tex]m \cdot n =cos\ \theta[/tex]
    The problem is I don't understand how I can apply that relation to the spinors. If theta is the angle between m and n, how can I apply it to figure how the eigenvectors in the direction m relate to the eigenvectors in the direction n? If the eigenvectors X+ and X- are the eigenvectors of the spin operator S in the direction m, how can I use them to get the spinor in the direction n?
    Last edited: Mar 28, 2008
  5. Jun 7, 2008 #4
    I'd also like to see this topic's explanation developed, especially with an eye to the physical meaning of the operators and matrices... Anyone help us out?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook