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Spin-X particles

  1. Mar 16, 2009 #1
    Sorry for this being a really general question but I am a bit confused what it means when it is stated that a particle or a bound state of several particles has a spin-X. For instance a deuteron is said to have spin-1.

    Does this number refer to the total spin angular momentum, the total angular momentum, or something else that I am missing?
     
  2. jcsd
  3. Mar 16, 2009 #2
    Spin is a little bit tricky, and a course in QM usually involves a lecture or two about the addition of spin. A deuteron is made up of two spin-1/2 particles, so it can have a spin of 0 or 1, where a spin-0 state is when the spin number and the magnetic number are both 0 for both particles. The spin-1 state is actually a superposition of three states. In all three, both particles have s=1/2, and then the total m is given by summing the two different m's together. Remember that for s=1/2, m=-1/2 or 1/2, so if we take all the combinations of those three, we get all of the spin-1 states. Note that we can get M=0 either by adding 1/2 and -1/2, or by adding -1/2 and 1/2 (i.e. these are not distinguishable particles), so this state is also a superposition.

    Check out http://itl.chem.ufl.edu/4412_u97/angular_mom/node10.html [Broken] if you want some more details.
     
    Last edited by a moderator: May 4, 2017
  4. Mar 17, 2009 #3

    malawi_glenn

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    when one talks of spin for a composite system, one can also mean the total angular momentum: sum of orbital and spin angular momentum.

    But the deuteron has S = 1, and it has also J = 1 (J=S+L). So both spin S and 'nuclear spin" J is 1.

    So it can have L = 0 or/and 2 (not L = 1 since then you would not have an antisymmetric wave-function)

    http://en.wikipedia.org/wiki/Deuterium
     
  5. Mar 17, 2009 #4

    Meir Achuz

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    In simple terms, the "spin" of particle is the total angular momentum it has in its rest system. Details of how it is calculated get more complicated.
     
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