Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Understanding spin, spinors, and rotations

  1. Feb 14, 2012 #1
    There's something I don't think I quite understand about spin and how it acts a generator of rotations. I'll start with quickly going over what I do understand. Suppose you want to do an infinitesimal rotation around the z-axis on some state:

    [itex]
    \def\ket#1{\left | #1 \right \rangle}
    \ket{\psi '} = (1 - \frac{i\epsilon _z J_z}{\hbar})\ket{\psi}
    [/itex]

    Now, if the wavefunction is just a scalar, then projecting on the coordinate basis gives us:

    [itex]
    \psi '(x,y,z) = (1 - \frac{i\epsilon _z J_z}{\hbar})\psi (x,y,z)
    [/itex]

    and so, as usual, we can do a finite rotation of [itex]\phi[/itex] around the z-axis by breaking it down into infinitesimal rotations and composing them, giving the relation:

    [itex]
    \psi '(x,y,z) = e^{-i \phi L_z / \hbar}\psi (x,y,z)
    [/itex]

    Now, because we know what this operator does, we don't actually have to expand the exponential. We can just (passively) rotate the coordinates:

    [tex]
    \begin{pmatrix}
    x' \\
    y'
    \end{pmatrix} =
    \begin{pmatrix}
    \cos\phi & +\sin\phi\\
    -\sin\phi & \cos\phi
    \end{pmatrix}
    \begin{pmatrix}
    x \\
    y
    \end{pmatrix} = R^{-1}(\phi)
    \begin{pmatrix}
    x \\
    y
    \end{pmatrix}
    [/tex]
    Giving:
    [tex]
    \psi '(x,y,z) = \psi (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z)
    [/tex]

    So far, so good. Now, if ##\ket{\psi}## actually requires a spinor to describe the wavefunction, [itex]\ket{\psi} \rightarrow \begin{pmatrix}
    \psi _+(x,y,z) \\
    \psi _-(x,y,z)
    \end{pmatrix}[/itex] then ##J_z = L_z + S_z## and the act of rotating the ket has two effects: a passive rotation of the coordinate system and a shuffling together of the two wavefunction components. So, we repeat the exercise we went through for ##L_z##, derive the Pauli spin matrices from the commutators, etc., eventually finding that:
    $$
    e^{-i \phi S_z / \hbar} =
    \begin{pmatrix}
    \cos(\phi /2) - i \sin(\phi /2) & 0\\
    0 & \cos(\phi /2) + i \sin(\phi /2)
    \end{pmatrix}
    $$

    What I don't get is how this makes any sense. I mean, I can follow all the steps in the derivation, I just don't get the final result. Why isn't the overall rotation due to spin just given by the matrix ##R(\phi)## above? In other words, taking into account both ##L_z## and ##S_z##, why isn't the rotated spinor given by:

    $$
    \begin{align}
    \begin{pmatrix}
    \psi _+ '(\vec{r}) \\
    \psi _- '(\vec{r})
    \end{pmatrix} &= R(\phi)
    \begin{pmatrix}
    \psi _+ (R^{-1}(\phi)(\vec{r})) \\
    \psi _- (R^{-1}(\phi)(\vec{r}))
    \end{pmatrix} \\ &=
    \begin{pmatrix}
    (\cos\phi) \psi _+ (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) - (\sin\phi) \psi _- (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) \\
    (\sin\phi) \psi _+ (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z) + (\cos\phi) \psi _- (x\cos\phi + y\sin\phi,-x\sin\phi + y\cos\phi,z)
    \end{pmatrix}
    \end{align}
    $$

    That's what rotating something does, so why do spinors behave differently?
     
    Last edited: Feb 14, 2012
  2. jcsd
  3. Feb 15, 2012 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Forget about spinors for a moment, and think about vector-valued wavefunctions. These turn up for spin 1 particles, such as the photon.

    A vector-valued function is an object that assigns a little arrow to each point in space. How does the rotation group act on such an object? Keeping with active rotations, the rotations should act by moving the whole field of arrows around. But a rotation does not just move the points to which the arrows are attached; it also has to rotate the arrows, so that the entire assembly moves rigidly. So one would write

    [tex]\tilde{A}^i (\vec x) = R^i{}_j A^j (R^{-1} \vec x),[/tex]
    where R is some rotation matrix.

    Now back to spinors. When you rotate a spinor field, you also have to move everything rigidly. The missing piece of information, which is not always clear from the algebraic definitions, is that a spinor is a kind of object that indicates a direction in space. Specifically, given a 2-component spinor

    [tex]\begin{pmatrix} \alpha \\ \beta \end{pmatrix},[/tex]
    one forms the complex number [itex]z \equiv \alpha / \beta[/itex], which is then related to a spatial direction by stereographic projection onto the Riemann sphere. It isn't hard to show that the eigenspinors of the Sx, Sy, and Sz operators correspond, respectively, to directions along the x, y, and z axes.

    So this is the reason that the spinor components must also mix. They must change to represent a new direction in space, obtained by rigidly rotating the old direction, just as you would do with a vector field. The catch is that the direction is represented by stereographic projection (hence why this is called a projective representation of the rotation group), so the exact manner in which the components mix may seem strange at first.
     
    Last edited: Feb 15, 2012
  4. Feb 15, 2012 #3
    Thanks for your response! So then it's just the fact that spinor elements are complex that makes them mix differently than vector functions?

    Edit: or is it because a spinor has two components instead of three?
     
  5. Feb 15, 2012 #4
    Basically this. This is properly understood using representation theory, in whose language we would say that two-component spinors live in a different representation of the rotation group than 3-vectors. Spin 1/2 objects are in the 2-dimensional representation, while vectors are in the 3-dimensional representation. You can define rotation matrices for both of them, but they are different rotation matrices (not least because spinors use 2x2 rotation matrices and vectors use 3x3 rotation matrices).
     
  6. Feb 15, 2012 #5
    Do you know of any good book on representation theory and Lie groups which explains all this stuff?
     
  7. Feb 15, 2012 #6
    Spin isn't really rotation in the classical sense, it's more of the pattern of how a wave oscillates, and angular momentum effects waves differently than spheres. You could try shaking a glass of water to see what happens. I myself have found that merely changing the direction in which energy is put into oscillating a fluid effects the shapes that somehow almost perfectly coincides with different orbital models.
     
  8. Feb 15, 2012 #7
    It's true that spin has some differences with classical rotation, but they're much more similar than you seem to think. For instance, you need to rotate an electron 720 degrees to get it back to where it started, whereas you only need to rotate classical objects by 360 degrees. But notice for both kinds of rotations there exist angles which bring you back to square one. And that's just the start of their similarities.

    In more formal terms, the rotation group of ordinary groups is SO(3), whereas if you want to have spin you need the closely related group SU(2). These two groups are locally isomorphic, which means they're infinitesimal elements can be identified. To put it another way, during an infinitesimal time dt both electron spin and the rotation of a top are doing exactly the same thing. It's only after a finite time t that you notice that the two behaviors diverge from each other.
     
  9. Feb 16, 2012 #8
    I second this!
    A book that is perhaps more geared towards showing it's applications to problems like this :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Understanding spin, spinors, and rotations
  1. Spin Rotation (Replies: 1)

Loading...