Undergrad Quarks and isospin ladder operators

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SUMMARY

The discussion centers on the interpretation of the isospin ladder operator \( T_+ \) in quantum mechanics, specifically in the context of particle physics as presented in Thompson's book. The equation \( T_+ |d\bar{u}\rangle = |u\bar{u}\rangle - |d\bar{d}\rangle \) is clarified as a result of using the total angular momentum operator \( T \equiv T(1) + T(2) \), rather than treating \( T_+ \) as a product of operators. The correct interpretation confirms that \( T_+ \) operates as a sum of individual particle operators, leading to the conclusion that \( T_+ |d\bar{u}\rangle = -|u\bar{d}\rangle \).

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  • Understanding of quantum mechanics principles
  • Familiarity with particle physics concepts
  • Knowledge of angular momentum operators in quantum systems
  • Proficiency in interpreting mathematical expressions in physics
NEXT STEPS
  • Study the properties of angular momentum operators in quantum mechanics
  • Explore the implications of isospin in particle physics
  • Review Thompson's book, focusing on the relevant sections regarding isospin and angular momentum
  • Learn about the mathematical framework of quantum operators and their applications
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Students and researchers in quantum mechanics and particle physics, particularly those looking to deepen their understanding of isospin and angular momentum operators.

Xico Sim
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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.
 
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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...
 
Xico Sim said:
or if I should answer myself, for others who may have the same question to see...
That's the best option, you are certainly not the first or last one with that question.
 
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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