- #1
Gregg
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Notation confusion; ## |\pi N; I, I_3 \rangle ## states
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.
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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