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What is the analogue of the unitarity constraint of chiral perturbation theory in superchiral perturbation theory?

+ 3 like - 0 dislike
173 views

I am trying to reconstruct something and I would appreciate someone helping me filling the gaps. To motivate my question let's first consider chiral perturbation theory with up and down quarks. This is the lowest order Lagrangian

$$\begin{equation*}
\mathcal{L}_2=\frac{f^2}{4}\langle{}\partial_{\mu}U^{\dagger}\partial^{\mu}U\rangle
\end{equation*}
$$

where $f$ is some parameter with mass dimensions and $U$ is defined via

$$U=e^{it^a\phi_a/f}$$

where $t^a$ are the Pauli matrices and $\phi_a$ are complex scalar fields. For some reason I don't know, people take this matrix to be unitary. So my first question is, why does $U$ have to be unitary?

In any case, if we demand $U$ to be unitary

$$UU^{\dagger}=1$$
$$e^{it^a\phi_a/f}e^{-it^a\phi_a^*/f}=1$$
and introducing $e^{-it^a\phi_a/f}$ on both sides

$$e^{-it^a\phi_a^*/f}=e^{-it^a\phi_a/f}$$
$$t^a\phi_a^*=t^a\phi_a$$
$$\begin{pmatrix}
\pi_3&\pi_1-i\pi_2\\
\pi_1+i\pi_2&-\pi_3
\end{pmatrix}=
\begin{pmatrix}
\pi_3^*&\pi_1^*-i\pi_2^*\\
\pi_1^*+i\pi_2^*&-\pi_3^*
\end{pmatrix}$$
now, redefining $\pi_3\equiv\pi_0$, $\pi_1+i\pi_2\equiv\sqrt{2}\pi_-$ and $\pi_1-i\pi_2\equiv\sqrt{2}\pi_+$ we see that the unitarity of $U$ imposes that $\pi_0$ is a real field after all and $\pi_-^*=\pi_+$

This up till now is nothing but the usual pion theory. I want to supersymmetrize this theory and this is where most of my doubts arise.

I have been told that I can embed the Lagrangian considered so far in

$$\mathcal{L}=f^2\int{}d^4\theta\langle\mathcal{U}^{\dagger}\mathcal{U}\rangle$$

where
$$\mathcal{U}=e^{it^a\Phi_a/f}$$
where $t^a$ are still the Pauli matrices and $\Phi_a$ are chiral superfields. Now my second question. In the nonsupersymmetric theory we imposed a unitarity constraint in the matrix $U$. I have been told that there is an analogous constraint on $t^a\Phi_a$ but I don't know which it is, let alone where it comes from. So which constraint must I apply and why? My third question would be about how I can relate the pion fields of the nonsupersymetric theory with the superfields I have just introduced, but of course I cannot properly state this question without first applying the constraint on $t^a\Phi_a$ (If the post gets anwsered this will go in a follow up question). 

asked Jul 13, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]

1 Answer

+ 3 like - 0 dislike

Since no answers appeared yet, I will try to make a contribution. My answer concerns only the first of your questions, i.e. why is $U$ unitary.

To my understanding, the unitarity of $U$ is a consequence of the nature of the chiral symmetry breaking. The assumption is, that it is the bilinear combination of quarks that form a condensate breaking the symmetry

$$(\bar{q}_{R})_j(q_L)^i=v^3 \delta^{i}_j$$

Here $i,j$ are flavour indices while spinor and color indices are summed over to form a Lorentz and a color scalar.  Let for the sake of definiteness only consider $u$ and $d$ quarks. Under the $SU(2)_L\times SU(2)_R$ the quarks transform as follows

$$q_L^i\to L^i_{i'}q_L^{i'},\quad q_R^i\to R^j_{j'}q_R^{j'}$$

The vacuum condensate then transforms as

$$(\bar{q}_{R})_j(q_L)^i\to (R^\dagger)^{j'}_{j} v^3\delta^{i'}_{j'}L^{i}_{i'}=v^3(LR^\dagger)^i_j$$

For the vectorial subgroup of $SU(2)_L\times SU(2)_R$ characterized by $L=R$ the vacuum condensate stays invariant.  On the other hand, for $L\neq R^\dagger$ the vaccum condensate changes by a unitary matrix   $\mathcal{U}=LR^\dagger$. Value   $\mathcal{U}=1$ corresponds to the absence of the Goldstone bosons (pions). On general grounds  we know that a chiral transformation with  $L\neq R^\dagger$ will produce pions (non-linear realization of symmetry). Hence, the quark condensate with some unitary matrix  $\mathcal{U}$ in place of the unit matrix describes a state with pions rather then the vacuum.   By letting  $\mathcal{U}$ vary with the space-time point   $\mathcal{U}=\mathcal{U}(x)$ we account for a dynamical pion field. Thus we identify  $\mathcal{U}(x)=U(x)$, the same $U(x)$ that appears in your Lagrangian. It is unitary by construction and under the chiral transformations it changes as

$$U(x)\to LU(x)R^\dagger$$

This explanation is not crystal-clear to me. Some conclusions do not seem to follow inevitably. Moreover, it can be plain wrong. Hopefully, someone more qualified will correct me.

answered Jul 15, 2015 by Weather Report (210 points) [ no revision ]

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