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I How does the singlet arise

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  1. Sep 15, 2016 #1
    I'm currently reading "Quantum Mechanics, The Theoretical Minimum" and on page 166 the singlet state is introduced. However there is no explanation as to where this comes from and appears to be plucked from thin air. I had a look at a previous PF thread https://www.physicsforums.com/threads/how-do-i-get-the-singlet-state.67769/ however I'm a complete novice when it comes to QM and this is slightly beyond my knowledge. I would be very greatful is someone could possibly explain this to me either fully or partially.

    Many thanks :)
     
  2. jcsd
  3. Sep 15, 2016 #2
    A very general statement would be that a singlet (with regard to specific symmetry) is a state that is not affected by operations (that keep the symmetry)
    it is like a scaler.
    A good examples are:
    1) spin-0 particles are singlets with regards to rotations
    2) uncharged particles are singlets with regards to the em field (u(1) internal symmetry)
     
  4. Sep 15, 2016 #3
    Thank you for your reply :) what do you mean by "with regard to specific symmetry"? :)

    Many thanks
     
  5. Sep 15, 2016 #4
    in the example of spin-0 - rotations, the symmetry is rotation.
    singlet is relative abstract mathematical notion, so the answers will be usually abstruct.
     
  6. Sep 15, 2016 #5

    Zafa Pi

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    Google "spontaneous parametric down conversion" and select whatever option tickles your fancy.
     
  7. Sep 16, 2016 #6

    vanhees71

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    This is about the addition of angular momenta and not about "parametric downconversion", which is way more complicated. The mathematical question is that after the eigenvalue problem of the total-spin operator which is a sum of two independent spins. You have to get familiar with the tensor product of Hilbert spaces (in this case the somewhat simpler case of finite-dimensional Hilbert spaces) and how to solve the eigenvalue problem of angular-momentum operators (in this case simplifying to the eigenvalue problem of finite-dimensional hermitean matrices).

    As will turn out in your case, adding two spins 1/2, the sum contains vectors of total spin 0 (singlet) and total spin 1 (triplet), and the basis of eigenvectors is

    for total spin 1:
    $$|S=1,\Sigma_z=1 \rangle=|s_1=1/2,\sigma_z^{(1)}=1/2;s_2=1/2,\sigma_z^{(2)}=1/2 \rangle,$$
    $$|S=1,\Sigma_z=0 \rangle=\frac{1}{\sqrt{2}} (|s_1=1/2,\sigma_z^{(1)}=1/2;s_2=1/2,\sigma_z^{(2)}=-1/2 \rangle + |s_1=1/2,\sigma_z^{(1)}=-1/2;s_2=1/2,\sigma_z^{(2)}=1/2 \rangle),$$
    $$|S=1,\Sigma=-1 \rangle = |s_1=1/2,\sigma_z^{(1)}=-1/2;s_2=1/2,\sigma_z^{(2)=-1/2}.$$

    for total spin 0:
    $$|S=0,\Sigma_z=0 \rangle = \frac{1}{\sqrt{2}} (|s_1=1/2,\sigma_z^{(1)}=1/2;s_2=1/2,\sigma_z^{(2)}=-1/2 \rangle - |s_1=1/2,\sigma_z^{(1)}=-1/2;s_2=1/2,\sigma_z^{(2)}=1/2 \rangle).$$
     
  8. Sep 16, 2016 #7

    PeroK

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    Perhaps a simpler explanation. If you have two spin-1/2 particles, then together they form a two-particle system, where the spins can either add (which roughly means they are spinning in the same direction) or cancel (which roughly means they are spinning in opposite directions). So, the total spin of the system is either 0 or 1.

    In any case, if you measure the total spin of the two-particle system you will get either 0 or 1.

    If you get 0, then the particles are said to be in the singlet state. The reason it is called singlet is that there is only one combination of the two spin-1/2 particles that gives a total spin of 0.

    By contrast, the triplet state corresponds to a total spin of 1, and there are three linear combinations of the individual spins that give a total spin of 1.
     
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