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Non-invariance under 2-Pi rotations?

  1. Oct 28, 2011 #1
    I have heard that quantum systems (and therefor all physical systems) are not truely invariant under 2∏ rotations. Something to do w/ the wave function changing sign. Is this true? Can someone point me to an on-line primer on this?

    Thanks!
     
  2. jcsd
  3. Oct 31, 2011 #2

    dextercioby

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    Well, for non-relativistic systems in which the spin is half integral (an electron from Pauli equation, for example), a rotation of angle 2[itex]\pi[/itex] indeed changes the sign of the electronic wavefunction. However, this is not really important, because in computing the probabilities, we always take a square, thus such a phase factor is eliminated.

    Anyways, Ballentine's book on QM (or any group theory book for physicists) explains how that (-1)^s comes from.
     
  4. Oct 31, 2011 #3

    Bill_K

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    There's a difference. Spinors are double-valued functions. When you write a spinor wavefunction there is always understood to be a ± sign in front. A spinor ψ is a single-valued function on the group space of SU(2), but when you go to SO(3) the mapping is twofold, and ψ becomes double-valued. Consequently for a spin one-half object, ψ and -ψ do not just "differ by a phase", they are literally the same.
     
  5. Oct 31, 2011 #4
    Here we define unit vector n in the polar coordinate.

    [tex]\vec{n} = ( \sin\theta \cos\varphi, \, \sin\theta \sin\varphi, \, \cos\theta )[/tex]
    Of course by 2pi rotation, this vector doesn't change,

    [tex]\varphi \to \varphi + 2\pi, \qquad \vec{n} \to \vec{n}[/tex]
    The n component of the spinor operator is

    [tex] \hat{S}_n = \vec{n}\cdot \vec{S} = \frac{\hbar}{2} \left[ (\sin\theta \cos\varphi) \sigma_1 + (\sin\theta \sin\varphi) \sigma_2 + (\cos\theta) \sigma_3 \right][/tex]
    where sigma is Pauli matrices. So,

    [tex] \hat{S}_n = \frac{\hbar}{2} \left( \begin{array}{cc} \cos\theta & e^{-i\varphi} \sin\theta \\ e^{i\varphi} \sin\theta & -\cos\theta \end{array} \right) [/tex]
    The eigenstate of this operator (which direction is "n" ) is

    [tex]\alpha_n = \left( \begin{array}{c} \cos \frac{\theta}{2} e^{-i\varphi /2} \\ \sin\frac{\theta}{2} e^{i\varphi /2} \end{array} \right) \quad \hat{S}_n \alpha_n = \frac{\hbar}{2} \alpha_n [/tex]
    If we rotate the direction of "n" by 2pi, the unit vector "n" doesn't change, as shown above.

    [tex]\vec{n} \to \vec{n} \quad (\varphi \to \varphi + 2\pi)[/tex]
    But its eigenstate change from +1 to -1 by 2pi rotation.

    [tex]\alpha_n \to - \alpha_n \quad (\varphi \to \varphi + 2\pi) [/tex]
     
  6. Oct 31, 2011 #5
    Another way to think of it is that if an object shares an elastic force or field interaction with other objects that have spatial extension, spinning the object [itex]2\pi[/itex] radians gives a different topological situation compared with [itex]4\pi[/itex] radians.

    Dirac is said to have invented this type of example to explain the concept:

    Spinor rotated twice

    http://www.youtube.com/watch?v=O7wvWJ3-t44&NR=1
     
    Last edited: Oct 31, 2011
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