# Quarks and isospin ladder operators

• I
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.

Related High Energy, Nuclear, Particle Physics News on Phys.org
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...

mfb
Mentor
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
$$T_+\equiv T_1+iT_2=(T_1(1)+T_1(2))+i(T_2(1)+T_2(2))=T_+(1)+T_+(2)$$

vanhees71 and mfb