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I Quarks and isospin ladder operators

  1. Jun 22, 2016 #1
    Hi, guys.

    This is actually a question about quantum mechanics, but since the context in which it appeared is particle physics, I'll post it here.
    On Thompson's book (page 227, equation (9.32)), we have
    $$T_+ |d\bar{u}\rangle = |u\bar{u}\rangle - |d\bar{d}\rangle$$

    But I thought ##T_+=T_+(1)\otimes T_+(2)##, and in that case
    $$T_+ |d\bar{u}\rangle = -|u\bar{d}\rangle$$

    It seems like he uses ##T_+=T_+(1) + T_+(2)##, and I don't know why he does that.
     
  2. jcsd
  3. Jun 22, 2016 #2
    I wasn't thinking correctly: I now know the answer. I don't know if I should delete this thread (or even if I can delete it) or if I should answer myself, for others who may have the same question to see...
     
  4. Jun 22, 2016 #3

    mfb

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    Staff: Mentor

    That's the best option, you are certainly not the first or last one with that question.
     
  5. Jun 22, 2016 #4
    You're right.
    Well, since ##T\equiv T(1) + T(2)## (the total angular momentum operator is the sum of the angular momentum operators of each particle), we have
    $$T_+\equiv T_1+iT_2=(T_1(1)+T_1(2))+i(T_2(1)+T_2(2))=T_+(1)+T_+(2)$$
     
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