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Notation confusion; \pi N; I, I_3> states 
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#1
Feb1713, 08:04 AM

P: 463

In my book it says for the ##\pi N ## state:
##\pi N; \frac{3}{2},\frac{3}{2}\rangle =\pi ;1,1\rangle  N; \frac{1}{2},\frac{1}{2}\rangle## firstly, does this mean: ##\pi N; \frac{3}{2},\frac{3}{2}\rangle =\pi ;1,1\rangle \otimes  N; \frac{1}{2},\frac{1}{2}\rangle## ? Not that it really matters, but next it says that you can use quantum mechanical shift ladder to get ##\pi N; \frac{3}{2},\frac{1}{2}\rangle =\sqrt{\frac{1}{3}}\pi ^+n\rangle +\sqrt{\frac{2}{3}} \pi ^0 p\rangle## I'm really not too sure what this means, and the notation is not too clear to me, could someone explain it to me? In the title I and I_3 refers to isospin. I don't know what the N indicates, Baryon number? Is the tensor product a key to get the coefficients, are they similar to ClebschGordon coefficients or what is the ladder operation for spin in terms of these vectors? I really don't have much understanding of the notation or implications at the moment. 


#2
Feb1713, 09:14 AM

Sci Advisor
Thanks
P: 4,160

The N stands for Nucleon. The nucleon states with I=1/2 are proton (I_{3}= +1/2) and neutron (I_{3} = 1/2). So the first line says the combined state with I = 3/2, I_{3} = 3/2 consists of π^{+} and proton.
To get the last line, we apply the operator I_{} that lowers total I_{3}. It acts on both the pion and the proton. Lowering the proton to a neutron gives us the first term, while lowering the π^{+} to a π^{0} gives us the second term. As you say, the √'s in front of these terms are ClebschGordAn coefficients. 


#3
Feb1713, 09:56 AM

P: 463

OK, so would a way to look at that operation be: ##I_ \pi N; \frac{3}{2},\frac{3}{2}\rangle = I_ \pi ;1,1\rangle \otimes  N; \frac{1}{2},\frac{1}{2}\rangle + \pi ;1,1\rangle \otimes I_ N; \frac{1}{2},\frac{1}{2}\rangle## ? Which will give me ## k_1 \pi^0 p \rangle + k_2  \pi^+ n\rangle ## ? where the coefficients are CG coefficients? 


#4
Feb1713, 02:07 PM

Sci Advisor
Thanks
P: 4,160

Notation confusion; \pi N; I, I_3> states
Yep, that's correct.



#5
Feb1813, 11:51 AM

P: 463

Thanks for the help



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