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

Question

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). 

Answer 1

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.