Notation confusion; |\pi N; I, I_3> states

Gregg

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 Clebsch-Gordon 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.

Bill_K

The N stands for Nucleon. The nucleon states with I=1/2 are proton (I₃= +1/2) and neutron (I₃ = -1/2). So the first line says the combined state with I = 3/2, I₃ = 3/2 consists of π⁺ and proton.

To get the last line, we apply the operator I_- that lowers total I₃. 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 π⁰ gives us the second term. As you say, the √'s in front of these terms are Clebsch-GordAn coefficients.

Gregg

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?

Bill_K

Yep, that's correct.

Gregg

Thanks for the help