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I Raising and lowering operators for a composite isospin SU(2)

  1. Apr 20, 2016 #1

    CAF123

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    Consider pion states composed of ##q \bar q## pairs where ##q \in \left\{u,d \right\}## transforms under an ##SU(2)## isospin flavour symmetry. These bound states transform in the tensor product ##R_1 \otimes R_2## of two representations ##(R_1, R_2)## of ##SU(2)##. Take ##R_2## as the fundamental representation of isospin with generators ##I^i = \sigma^i/2## and ##R_1## is the conjugate fundamental with generators ##-(\sigma^{i})^*/2##. If the third component of isospin is ##I_{\pm}^{R_1 \otimes R_2} = \frac{1}{2} \left( \sigma_1^{R_1 \otimes R_2} \pm i \sigma_2^{R_1 \otimes R_2}\right)## I can try and form a representation of this operator using the standard Pauli matrices. Take ##|\pi^+ \rangle = |u\rangle |\bar d \rangle \equiv |u \rangle \otimes | \bar d \rangle \equiv |u \bar d \rangle##



    Then ##I_{+}^{R_1 \otimes R_2} |u \bar d \rangle = \frac{1}{2} \left( \sigma_1^{R_1 \otimes R_2} \pm i \sigma_2^{R_1 \otimes R_2}\right)|u \bar d \rangle = \frac{1}{2}\left( \sigma_1^{R_1} |\bar d\rangle \otimes \text{Id} |u \rangle + \text{Id} |\bar d \rangle \otimes \sigma_1^{R_2} |u \rangle \pm i(1 \leftrightarrow 2)\right)##



    1)My first question is if I take ##|u \rangle \rightarrow (1,0), |\bar d \rangle = (0,1)## then I have $$I_+^{R_1 \otimes R_2} |\pi^+ \rangle = I_+^{R_1 \otimes R_2} |u \bar d \rangle $$ This is equal to, $$ \frac{1}{2} \left( \sigma_1^{R_1} | \bar d \rangle \otimes \text{Id}_{2\times 2} |u \rangle + \text{Id}_{2 \times 2} |\bar d \rangle \otimes \sigma_1^{R_2}|u \rangle + i(1 \rightarrow 2) \right)$$ Inputting relevant matrices for sigma_i I get the zero vector as I should since pi^+ is the state of greatest weight and I am applying the raising operator. I just wondered would I get this result independent of the choice of what I take for ##| u \rangle ## and ##|d \bar \rangle## basis vectors? As long as the choice for u and d transforming in fundamental and ubar and dbar transforming in conjugate fundamental are linearly independent?



    2) Alternatively, I could just construct the representations for ##I_{\pm}^{R_1 \otimes R_2}## and I would end up with ##4 \times 4## matrices. But what would the ##4 \times 1## objects that these operators act on represent? Would a generic vector be something like ##(u, d, \bar u, \bar d)## so for example I would write ##u = (1,0,0,0)## and ##\bar d = (0,0,0,1)## for example?
     
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  3. Apr 25, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. May 7, 2016 #3

    CAF123

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    I would still be interested in obtaining a response so I thought some more about my questions-

    1) I think the choice of |u> = (1,0) and |dbar> = (0,1) is fixed if the doublet ##\begin{pmatrix} u \\ d \end{pmatrix}## transforms under an SU(2) isospin?

    2) My 4x4 matrix for $$I_+^{R_1 \otimes R_2} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 \end{pmatrix}$$ So the state of greatest weight is one in which, in the composite system, can be written in terms of the generic vector ##( u, d, \bar u, \bar d) = (x,y,z,w)##. So want to find such a vector such that $$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ This means y=0, x=w but z arbitrary. If I set z=0, and x=w=1 I get the state (1,0,0,1) which is a u dbar in the notation defined above (indeed the state of greatest weight). Is that correct? And, if so, what is stopping me from writing a generic vector in the space as ##(u, d, \bar d, \bar u)## for example thereby concluding another state of greatest weight?

    Thanks!
     
  5. May 9, 2016 #4

    vanhees71

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    I'm not sure, what you are really asking, but I try to give an answer.

    The quarks in the SU(2) (flavor!) model have isospin ##T=1/2## and the ##u## quark ##t_3=+1/2## and the ##d## quark ##t=-1/2##. For the usual Pauli matrices thus you associate ##|u \rangle=|+1/2 \rangle## with ##\binom{1}{0}## and ##|d \rangle=|-1/2 \rangle## with ##\binom{0}{1}## by choice of the eigenvectors of ##\hat{t}_3=\hat{\sigma}_3/2##.

    The antiquarks follow the same isospin representation, because for SU(2) the conjugate complex of the fundamental representation is equivalent to the fundamental representation itself. Only, of curse you have ##t_3=-1/2## for ##\bar{u}## and ##t_3=+1/2## for ##\bar{d}##.

    Now you consider meson states, consisting of one quark and one antiquark. This makes ##2 \times 2## flavor states. The product basis is of course ##q_1 \bar{q}_2 \rangle=|q_1 \rangle \otimes |\bar{q}_2 \rangle##. Now, if you consider exact isospin symmetry of the strong interactions (which is not exactly true, because the masses of ##u## and ##d## quarks are different, but that's a small difference of a few MeV compared to the typical hadronic scale of ##1 \;\text{GeV}##; so it's still a good approximate symmetry, as is chiral symmetry by the way).

    Thus it makes sense to group the mesons in irreducible isospin representations. This works with the usual Clebsch-Gordan coefficients, and these tell you that the four product states are grouped into the antisymmetric (leading to isoscalar ##T=0## states) and symmetric combination (leading to isovectof ##T=1## states). The lowest-mass multiplets of the here considered scalar particles are the ##f_0## (or ##\sigma##) meson, which is a very broad resonance state with masses around 450 MeV (previously the mass was considered higher, and usually the particle is labeled as ##f_0(600)##) and the pions. So you have
    $$|\sigma \rangle \propto |1/2,-1/2 \rangle-|-1/2,1/2 \rangle =|u \bar{u} \rangle - |d \bar{b} \rangle$$
    and
    $$|\pi^+ \rangle \propto |1/2,1/2 \rangle = |u \bar{d} \rangle,$$
    $$|\pi^- \rangle \propto |-1/2,-1/2 \rangle=|d \bar{u} \rangle,$$
    $$|\pi^0 \rangle \propto |1/2,-1/2 \rangle + |-1/2,1/2 \rangle= |u \bar{u} \rangle+|d \bar{d} \rangle.$$
     
    Last edited: May 9, 2016
  6. May 9, 2016 #5

    CAF123

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    Hi vanhees71,
    Is there a typo in that last line? So, in the notation introduced in the OP I would write ##R_2 = \begin{pmatrix} u \\ d \end{pmatrix}## but how to write ##R_1##? As you just said, I've seen that SU(2) is pseudoreal so the conjgate fundamental transforms like ## \sim \epsilon \bar 2## which would mean $$R_1 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} \bar u \\ \bar d \end{pmatrix} = \begin{pmatrix} \bar d \\ -\bar u \end{pmatrix} $$ So dbar and ubar have, respectively, +1/2 and -1/2 third component of isospin associated with them. Is that correct? So in the ## \bar 2## rep , we assign the eigenvector (0,1) for u and (1,0) for d?

    I got the matrix representation for the raising operator for isospin in the composite system as $$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 0 \\ - 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 \end{pmatrix}$$ This will act on the 'vector' $$\begin{pmatrix} R_1 \\ R_2 \end{pmatrix} = \begin{pmatrix} -\bar d \\ \bar u \\ u \\ d \end{pmatrix}$$ Is that right? If so, I must have my matrix wrong then because acting with it with the state (1,0,1,0) I don't get zero. ((1,0,1,0) comes about because the state of greatest weight is u dbar = pi^+))

    Thanks!
     
  7. May 9, 2016 #6

    vanhees71

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    I've corrected the typo, and your mapping is also correct. I still don't understand what you mean by the ##4 \times 4## matrix. The SU(2) isospin matrices act on the direct product of the spinors as they should act on the direct product, i.e.,
    $$|q_1 \rangle \otimes |\bar{q}_2 \rangle \rightarrow U |q_1 \rangle \otimes U |\bar{q}_2 \rangle,$$
    and from this you can read off how your ##4 \times 4##-matrix should operate.
     
  8. May 9, 2016 #7

    CAF123

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    I think I can write $$I_+^{R_1 \otimes R_2} (|q \rangle \otimes | \bar q \rangle ) = I_+^{R_2} |q \rangle \otimes I_+^{R_1}|q \rangle$$ and I can represent my 4x4 matrix like so $$I_+^{R_1 \otimes R_2} = \begin{pmatrix} I_{+}^{R_1} & 0 \\ 0 & I_+^{R_2} \end{pmatrix}$$ acting on ## \begin{pmatrix} R_1 \\ R_2 \end{pmatrix}##. I find the state of greatest weight in R_1 and R_2 are |d bar> and |u> respectively, so the state of greatest weight in the composite system is then |u dbar>.

    Ok so I have the answer now. I guess this means the 4x4 matrix I wrote down in my previous posts is incorrect then (for one it is not diagonal). Do you agree with this? Thanks.
     
  9. May 9, 2016 #8

    vanhees71

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    This looks right now.
     
  10. May 9, 2016 #9

    CAF123

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    Thanks! Ok the only other thing I want to check is why I am getting the error in my 4x4 matrix. Here are my steps: $$I_+^{R_1 \otimes R_2} = \frac{1}{2} (\sigma_1^{R_1 \otimes R_2} + i \sigma_2^{R_1 \otimes R_2}),$$ where $$\sigma_i^{R_1 \otimes R_2} = \sigma_i^{R_1} \otimes \text{Id}_{R_2} + \text{Id}_{R_1} \otimes \sigma_i^{R_2}$$ Now, ##\sigma_1^{R_1} = -\sigma_1^* = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}## and ##\sigma_1^{R_2} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}##Therefore $$\sigma_1^{R_1 \otimes R_2} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & 1 & 0 \end{pmatrix}$$

    Similarly for $$\sigma_2^{R_1 \otimes R_2} = i \begin{pmatrix} 0 & -1 & -1 & 0 \\ 1 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \end{pmatrix}$$ Adding them together gives me the correct block diagonal terms but in the bottom left I have non zero entries. I can't really see where I went wrong - I guess it's something to do with writing the Pauli matrices in different representations but I think conjugate generators ##T_a^{conj} ## are always given by ##-T_a^*##, as I applied to the Pauli matrices in the above.

    Can you spot anything that looks wrong? I tried to write out most of my steps.
    Thanks!
     
  11. May 10, 2016 #10

    vanhees71

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    I don't understand, why you insist on this clumsy notation in terms of ##4 \times 4## matrices. To work with the tensor products is much more straight forward!
     
  12. May 10, 2016 #11

    CAF123

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    I insist because I want to make sure everything hangs together and also because I am instructed to obtain a 4x4 matrix representation for the raising operator in the composite system. I think I have nearly done this - it is just the lower left block that comes out wrongly yet I cannot see where I am making my mistake. Any ideas?

    Thanks!
     
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