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

by Gregg
Tags: confusion, i3>, notation, states
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Gregg
#1
Feb17-13, 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 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.
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Bill_K
#2
Feb17-13, 09:14 AM
Sci Advisor
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P: 4,160
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
#3
Feb17-13, 09:56 AM
P: 463
Quote Quote by Bill_K View Post
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.

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
#4
Feb17-13, 02:07 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
Notation confusion; |\pi N; I, I_3> states

Yep, that's correct.
Gregg
#5
Feb18-13, 11:51 AM
P: 463
Thanks for the help


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