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

[CAF123]

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?

 

[CAF123]

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!

 

[vanhees71]

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
$$|\pi^+ \rangle \propto |1/2,1/2 \rangle = |u \bar{d} \rangle,$$
$$|\pi^- \rangle \propto |-1/2,-1/2 \rangle=|d \bar{u} \rangle,$$
$$|\pi^0 \rangle \propto |1/2,-1/2 \rangle + |-1/2,1/2 \rangle= |u \bar{u} \rangle+|d \bar{d} \rangle.$$

 

[CAF123]

Hi vanhees71,

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 $\bar{u}$.

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!

 

[vanhees71]

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.

 

[CAF123]

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.

 

[vanhees71]

This looks right now.

 

[CAF123]

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!

 

[vanhees71]

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!

 

[CAF123]

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!