_{IJK}\int \psi ^{a\dagger}_J \Pi^a_{\phi_K} - \frac{2i}{e^2} \epsilon_{IJK}\int \psi^{a}_J (D \phi_K)^a\label{ext9}\\
&\hskip.2in -\int \omega ^{a\dagger} \Pi^a_{\phi_I} + \frac{2i}{e^2}\int \omega ^{a  }( D \phi )^a - \frac{im_k}{e^2}\epsilon_{IJK}\int \psi^{a\dagger}_J\phi^a_K - \frac{im_k}{e^2}\int \omega ^{a\dagger} \phi^a_I \nonumber\\
&\hskip.2in -  \frac{i}{2e^2}\int f^{abc}\epsilon_{ABI} \phi ^b_A\phi^c_B\omega ^{a\dagger} - \frac{i}{e^2} \int f^{abc} \phi ^b_K \phi^c_I \psi^{a\dagger}_K\nonumber
\end{align}
We can now construct the Hamiltonian from the anticommutator of the
supercharges given above. It is a straightforward, but lengthy, calculation; the result
is ${\mathcal H} = {\mathcal H}_0 + {\mathcal H}_m$, where, in terms of the gauge-covariant variables,
\begin{align}
{\mathcal H}_0 &= \frac{e^2}{2}\int (\bar D ^{-1})^{ab}(x,y) \left(D\frac{\delta}{\delta A}\right)^b_y \frac{\delta}{\delta A^a(x)}\nonumber\\
&\hskip.2in+\frac{1}{2}\int  (\bar D ^{-1})^{ab}(x,y) f^{bmn}\left(\psi^{ m \dagger}_L\psi^n_L + \omega^{ m \dagger}\omega^n + e ^2 \phi^m_L\frac{\delta}{\delta \phi^n_L}\right)_y \frac{\delta}{\delta A^a(x)}\nonumber\\
&\hskip.2in +\frac{1}{e^2}\int \epsilon_{ABC}f^{amn}\phi^m_A\psi^{a\dagger }_B\psi^n_C +\frac{1}{e^2}\int f^{abc}\left(\phi^c_A\psi^b_A\omega^{a\dagger} + \phi^c_A\omega^a\psi^{b\dagger}_A\right)\label{Hamil1}\\
&\hskip.2in+ \frac{1}{2e^2}\int B^aB^a + \frac{2}{e^2}\int (D\phi_L)^a(\bar D \phi_L)^a + \frac{e^2}{2}\int \Pi_{\phi_L}^a  \Pi_{\phi_L}^a\nonumber\\
&\hskip.2in+\frac{1}{e^2}\int (\psi^a_LD\psi^a_L - \psi^{a \dagger}_L\bar{D}\psi^{a\dagger}_L) +\frac{1}{e^2}\int (\omega^aD\omega^a - \omega^{a \dagger}\bar{D}\omega^{a\dagger})
\nonumber\\
&\hskip.2in+\frac{1}{4}f^{amn}f^{apq}\phi^m_M\phi^n_N\phi^p_M\phi^q_N\nonumber
\end{align}
\beq\begin{split}
{\mathcal H}_m &= \frac{e^2(k + (2-n)c_A)}{4\pi}\int(A-a)^a\frac{\delta}{\delta A^a} + \frac{e^2(k + (2-n)c_A)}{4\pi e^2}\int \psi^{a\dagger}_L\psi^a_L\\
&\hskip.2in -\frac{k}{4\pi}\int \omega^{a\dagger}\omega ^a + \frac{1}{2}\left[\frac{k}{4\pi}\right]^2e^2\int \phi^a_I\phi^a_I
\end{split}
\label{Hamil2}
\end{equation}
The second term of ${\mathcal H}$, which is the mass term
${\mathcal H}_m$, we see that the masses of the gauge field and the $SO(3)$ fermions get a shift proportional to $(2-n)c_A$ while the scalars and the fourth fermion do not. Obviously this is the result of the fact that only the $SO(3)$ fermions couple to the electric field in the supercharge, and are hence affected by the singular contributions proportional to $\bar D^{-1}(x,x)$ brought about by the use of the Gauss law and normal ordering, see the discussion following
(\ref{bos-mass}).
 For the theory to be supersymmetric, one must necessarily have equal
 masses for these degrees of freedom; this is obtained
 only when $n=2$, which makes the mass term $\mathcal{H}_m$ vanish when $k = 0$.
We can also verify more explicitly, with another lengthy calculation, that 
\begin{equation}
[q_I, H] = 0 \hspace{.3cm}\Longrightarrow \hspace{.3cm} n=2  \hspace{.2cm}\mbox{for}  \hspace{.2cm} \mathcal{N}\geq 2\label{ext10}
\end{equation}
Thus in the case of extended supersymmetry, the requirement of supersymmetry invariance forces the measure to be exactly what we had presented earlier from anomaly considerations, leading to a gapless spectrum for the gauge theories in the $k = 0$ limit.

For completeness, we also give here
the formulae for ${\mathcal H}_0$ and ${\mathcal H}_m$ in terms of the gauge invariant variables.
\begin{align}
{\mathcal H}_0 &= \frac{e^2c_A}{2\pi}\int \Omega ^{ab}(x,y)\frac{\delta}{\delta J^a(x)}\frac{\delta}{\delta J^b(y)}\nonumber\\
&\hskip.2in +\frac{ic_A}{2\pi}\int \bar{\mathcal{G}}(x,y)f^{abc}\left((H\chi^{b\dagger}_L)\chi^c_L + \chi_\omega ^{b \dagger}(H\chi_\omega)^c + \Phi^b_L\frac{\delta}{\delta \Phi^c_L}\right)(y)\frac{\delta}{\delta J^a(x)}
\nonumber\\
&\hskip.2in -\frac{1}{e^2}\int f^{abc}\epsilon_{ABC}\Phi^a_A(H\chi^\dagger)^b_B\chi^c_C - \frac{1}{e^2}\int f^{abc}(\Phi^a_A\chi^b_A\chi^\dagger _\omega +  \mbox{h.c})\label{invn=40}\\
&\hskip.2in +\frac{2\pi^2}{e^2c_A^2} \int \bar{\partial}J^a \bar{\partial}J^a + \frac{2}{e^2}\int \bar{\partial}\Phi^a_A(\mathcal{D}_J \Phi_L)^a + \frac{e^2}{2}\int \Pi^a_{\Phi_L} \Pi^a_{\Phi_L}\nonumber\\
&\hskip.2in + \frac{1}{e^2}\int (\chi^a_L(\mathcal{D}_J\chi_L)^a - \chi^{a\dagger}_L(\bar{\mathcal{D}}_{\bar{J}}\chi_L)^{a\dagger}) + \frac{1}{e^2}\int(\chi_\omega ^a\partial \chi^a_\omega - \chi_\omega ^{a\dagger}\bar{\partial} \chi^{a\dagger}_\omega)\nonumber\\
&\hskip.2in +\frac{1}{4e^2}\int f^{lpq}f^{kbc}\Phi^p_K\Phi^q_I\Phi^{\dagger b}_K\Phi^{\dagger c}_IH^{lk}
\nonumber\\
{\mathcal H}_m &=  \frac{e^2(k + (2-n)c_A)}{4\pi}\int \left[ J^a\frac{\delta}{\delta J^a} + \frac{1}{e^2} \chi^{a \dagger }_L(H^{-1})^{ab}\chi^b_L\right]\nonumber\\
&\hskip.2in -\frac{k}{4\pi}\int \chi^{a\dagger}_\omega H^{ab}\chi^b_\omega + \frac{1}{2}\left[\frac{k}{4\pi}\right]^2e^2\int \Phi^a_I