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Quantum Entanglement - electron spins

  1. Mar 4, 2010 #1
    I'm currently writing a report on quantum entanglement however getting a little confused with some of the concepts.

    If anyone could help with the following it would be greatly appreciated.

    I know that an electron has a spin; which means that is has both an electric field and a magnetic field. In otherwords, it has an electric field (an arrow) that is also rotating (with the arrow the axis of rotation). This means electron spin is described by a complex vector - a vector because its an arrow, a complex vector because its also rotating and therefore has a phase. But I know its not possible to add the spins of two electrons together, and instead you need to take a tensor product, but I'm sure sure as to why you can't add the two spins together.

    Can anyone help?
  2. jcsd
  3. Mar 5, 2010 #2
    When you think about the electron spin 1/2 in the quantum entanglement, do you understand the basic property of the electron spin as follows?

    Probably you imagine the spin 0 state like Helium atom (spin up + spin down = spin 0).
    But the electron spin magnetic moment is stronger and the distance between the two electrons of helium is much longer than that of the protons or neutrons. So, actually, (If we suppose that the real electron spin exists), the magnetic field generated by the two electrons of helium doesn't become zero unless the two electrons stick to each other.

    And the spin can be expressed by the "spinor" not by the vector.
    If we rotate the +z spin to the direction which has the angle [tex]\theta[/tex] between the z-axis and the angle [tex]\phi[/tex] between the x-axis, the spinor becomes,

    [tex]\begin{pmatrix}1\\0 \end{pmatrix} \quad \to \quad \begin{pmatrix}\cos\frac{\theta}{2}e^{-i\phi/2} \\ \sin\frac{\theta}{2}e^{i\phi/2}\end{pmatrix}[/tex]

    This result shows the spinor doesn't go back to the original by [tex]2\pi[/tex] rotation (in both the cases [tex]\theta, \phi[/tex]). By [tex]4\pi[/tex] rotation, it return as shown in this relation.

    And if the electron spin is caused by the real spinning of the electron, the electron rotation speed becomes much faster than the speed of light. So the spin has the strong "mathematical" property.
    So can we really use this "mathematical" spinor in the real entanglement experiment?

    And if we use "some external magnetic field" in this experiment, the experimental result becomes different from the original spin state? (Because the spin direction may change by the magnetic field.)
    Last edited: Mar 5, 2010
  4. Mar 6, 2010 #3
    So the measured angle of the spin is [tex]\theta=2\pi[/tex](the original direction), the measured amplitude becomes [tex]\cos (2\pi/2)[/tex] =-1 !
    For the photon case ([tex]\theta \to 2\theta[/tex]), the vector (not the spinor) can be used.

    (I have one more question about the entanglement.)
    For example, suppose one photon is a real stick pointing its polarization direction.
    When the angle between this "stick" (photon) and the polarizing filter is [tex]\theta[/tex], the probability([tex]f(\theta)[/tex]) that the stick passes through the filter can be clearly determined in each [tex]\theta[/tex], because this "real" stick has a definite figure and character.

    Suppose the two photons(sticks)(A, B) having the opposite polarization axes are flying in the opposite directions.
    And the photon A bumps into the two kinds of the filters A1 and A2 (pass(+) or not(-) in each filter).
    The photon B also bumps into the two filters B1 and B2.
    There are [tex]2^4=16[/tex] patterns in which these two photons pass these filters.
    The sum of all these probabilities becomes 1, when we repeat this experiment.

    [tex]\sum_{i=1}^{16} P_i = 1[/tex]

    If this relation is satisfied, the Bell inequallity(CHSH) is satisfied, which means the "local" theory. Right?

    But can we clearly say the photon is a real stick (or particle), even if it can be counted by the counter?
    Because if the photon is a real stick, it can't interfere with itself.

    If we suppose the photon is an electromagnetic waves,
    each time the photon A (its electronic field E) passes the filter A1, the probability that the photon B (of the opposite polarization axis) passes the filter B1(which has the angle [tex]\theta[/tex] between A1) becomes [tex](\cos \theta)^2[/tex]. (due to [tex](E\cos\theta)^2[/tex])

    Because the wave can be divided like([tex] E\times\cos\theta_{1}\times\cos\theta_{2}\times\cos\theta_{3}....[/tex])
    This case (of the electromagnetic wave) violates the Bell inequality in some angle([tex]\theta[/tex]). Right?

    So I'm afraid the interpretaion of the Bell inequality violation is related to wheter the photon is considered as a real particle or wave. (If the photon is a wave, we need not consider the nonlocal theory such as the entanglement?)

    Of course we don't know what the photon really is, so we don't know whether the entanglement really occur or not?
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