Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Feb 17, 2013 #1
    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.
     
    Last edited: Feb 17, 2013
  2. jcsd
  3. Feb 17, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    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.
     
  4. Feb 17, 2013 #3

    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?
     
  5. Feb 17, 2013 #4

    Bill_K

    User Avatar
    Science Advisor

    Yep, that's correct.
     
  6. Feb 18, 2013 #5
    Thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Notation confusion; |\pi N; I, I_3> states
  1. (pi+)(n)-> (p)(gamma) (Replies: 5)

Loading...