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Half integer spin

  1. Jan 25, 2010 #1
    Lets asume that electron is in state:
    [tex]
    \left[
    \begin{array}{c}
    \psi(\vec{r})\\
    \phi(\vec{r})
    \end{array}
    \right]
    [/tex]
    It's a vector because electron has two spin components (up and down). If we rotate our labolatory by the angle [tex] 360^0 [/tex] we got:
    [tex]
    \left[
    \begin{array}{c}
    -\psi(\vec{r})\\
    -\phi(\vec{r})
    \end{array}
    \right]
    [/tex]
    How one can explain why this discontinuity doesn't affect physics.
    Is it possible to prove that even in some interference experiments
    we can't measure this sign difference (that is how to prove that
    we can't discern the system from the system rotated by 360).
     
  2. jcsd
  3. Jul 5, 2011 #2
    My best guess is, everything else rotates phase as well, so the relative phases of everything is unchanged. I'm not sure what happens if you walk around one particle and not the other, though. Sadly my QM is not as strong as I would like.
     
  4. Jul 5, 2011 #3

    Bill_K

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    paweld, For half-integer spin, the wavefunction is not single-valued ψ(x), it is double-valued ±ψ(x). There literally is no difference at all between +ψ and -ψ. We're used to saying that ψ and -ψ just differ by a phase, but for spin one-half they don't even differ by a phase, they are the same (double-valued) function.
     
    Last edited: Jul 5, 2011
  5. Jul 5, 2011 #4

    vanhees71

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    This is indeed a very important point for the understanding of the foundations of quantum mechanics, which is often not well explained in standard textbooks.

    In quantum mechanics not the normalized Hilbert-space vectors themselves are representing the (pure) states, but the rays in Hilbert space, i.e., if

    [tex]|\psi' \rangle=\exp(\mathrm{ii} \varphi) |\psi \rangle, \quad \varphi \in \mathbb{R}[/tex]

    both vectors, [itex] |\psi' \rangle[/itex] and [itex]|\psi \rangle[/itex] represent the same state.

    Thus, symmetries are represented a priori not by unitary representations of the corresponding groups but by ray representations, i.e., by unitary representations up to arbitrary phases.

    Now one can show that any unitary ray representation of a continuous symmetry Lie group can be lifted to a unitary representation on Hilbert space of the corresponding central extensions of the covering group of the original symmetry group.

    Here, we have the Galilei group (nonrelativistic particles) or the Poincare group (relativistic particles) in view. In the former case the physically relevant representations lead to a nontrivial central extension with the mass as central charges and to the substitution of the rotation group, SO(3), by its covering group, SU(2). The fundamental representation of the latter corresponds to spin 1/2.

    A full rotation (i.e. the rotation around an arbitrary axis with rotation angle, [itex]2 \pi[/tex]) leads to a phase factor of (-1) of the spinor valued wave functions. This doesn't affect any physically observable predictions (related to probabilities calculated by the absolute squares of the wave function!). Thus the full rotation is as good as doing nothing to the states as far as observable prediction of QT are concerned. Thus, everything is fine with this transformation behavior of spinors for half-integer spin particles.
     
  6. Jul 5, 2011 #5

    alxm

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    What discontinuity?? The change of phase is continuous with the angle of rotation.

    The overall phase of a wave function does not correspond to any physical observable, but relative phases can be measured. As it were, this particular property of electrons is the reason for the Pauli exclusion principle, so it's not only experimentally verified, but crucial to the existence of chemical bonds and the existence of matter as we know it.
     
  7. Jul 5, 2011 #6

    dextercioby

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    To add on Hendrik's post above, the reason we're dealing with Dirac spinors each time we speak of massive spin 1/2 particles is that the only way the irreducibility contained in the symmetry condition/axiom is achived for the full Poincare (Lorentz) group is by considering a 4-dimensional space, instead of the normal 2-dimensional space of the Weyl (left or right) spinors which would have been enough, if the discrete transformations were excluded. But we can't exclude those.

    As far as I'm aware the only serious (perhaps too serious in terms of algebra at the expense of functional analysis) treatment of this important point is in J. Cornwell's group theory treatise in the second volume.

    Bottom line: Without the discrete symmetries, there's no existence of the Dirac spinor.
     
    Last edited: Jul 5, 2011
  8. Jul 5, 2011 #7
    The probability density (not amplitude) of the spin 1/2 particle is,

    [tex] \left[ \begin{array}{cc} \psi (r) & \phi (r) \end{array} \right] \left[ \begin{array}{c} \psi (r) \\ \phi (r) \end{array} \right] = \left[ \begin{array}{cc} -\psi (r) & -\phi (r) \end{array} \right] \left[ \begin{array}{c} -\psi (r) \\ -\phi (r) \end{array} \right][/tex]

    This means 360 degrees rotation is meaningless ?


    But if we suppose that 360 degrees rotation doesn't affect the spin 1/2 particle's amplitude at all, the next equation must be satisfied.

    [tex] \left[ -\begin{array}{cc} \psi (r) & -\phi (r) \end{array} \right] \left[ \begin{array}{c} \psi (r) \\ \phi (r) \end{array} \right] = \left[ \begin{array}{cc} \psi (r) & \phi (r) \end{array} \right] \left[ \begin{array}{c} \psi (r) \\ \phi (r) \end{array} \right][/tex]

    But this equation is not satisfied, this means by 360 degrees rotation, electron becomes different from the original.

    I think in the case of calculating the (precession) frequency of (Ex. Larmor), we don't discriminate the minus sign by 360 degrees.
    (If we discriminate this, Thomas precession factor 2 is an "excessive" value, I think.)
    But in the case of the interference by neutron (precession) we discriminates this.
     
  9. Jul 6, 2011 #8

    alxm

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    That's just wrong. The amplitude of a wave function is [tex]\psi^*\psi[/tex] and won't change with multiplication by -1 or any phase-factor [tex]e^{i\theta}[/tex] in general.

    Which is _not_ the same thing as saying you'll get the same result if you arbitrarily change the sign of the wave function's complex conjugate but not the wave function itself.
     
  10. Jul 6, 2011 #9
    You say if 1 x 1 =1 and -1 x -1 = 1, it means 1 = -1 ?
    This is strange.

    The important point is whether wavefunction (amplitude) or wave phase exist or not.
    (What you are saying is the same as Tomonaga in Story of Spin or Weisskopf, which statement are older than 1975 neutron expetiment )

    Here we suppose they (amplitude or phase) exist.
    If by 360 degrees rotation (of object itself or our observers) they don't change (this seems very natural), we can shuffle them freely. (Because they are completely the same things.)
    In this case the strange thing as I said above would occur. So the 360 degrees rotation changes the wavefunction itself.

    Here let's recall when we started to use the "wavefunction" (or wave "phase") for the point-like electron.
    It originates from wave-particle duality of the electron.

    Where did the wave property of the electron come from ?
    It originates from the interference of the elecron. (ex. Davisson-Germer.)
    This means we can discriminate between the probability density and wave phase in this real world, and we started to use "wave phase".
    (Wavefunction or wave phase exist "independently" .)

    If actual "interference" experiment didn't exist, we would not have used the wave phase.

    So the only experiment we can confirm the wave change by 360 rotation is one taking interference.

    As you know, the experiments using the neutron interference showed the probability density (= neutron number) of

    [tex]|e^{ikL}|^2 + |e^{ikL'} e^{i\phi/2}|^2 \neq |e^{ikL} + e^{ikL'} e^{i\phi/2}|^2[/tex]

    And this actual experiments showed just 720 rotation returned the neutron wavefunction (wave phase)
    (Not 360 rotation.)
     
  11. Jul 6, 2011 #10

    alxm

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    No, i'm saying the amplitude of Ψ and -Ψ are the same. You said they didn't because that would imply that Ψ(-Ψ) = Ψ^2. Which means you don't know what the amplitude or absolute value of a complex number is.
     
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