Question: I understand that Hamiltonian has to be hermitian. For a two level system, why does the general form of an Hamiltonian of a qubit have all real entries :
$$
\hat{H} = \frac{1}{2}\left( \begin{array}{cc} \epsilon & \delta \\ \delta & -\epsilon \end{array} \right)
$$
where $\epsilon$ as the offset between the potential energy and $\delta$ as the matrix element for tunneling between them.

Or equivalently in the following form why is there an absence of $\sigma\_y$ term:
$$H=\frac{1}{2} \delta\sigma\_x + \frac{1}{2}\epsilon\sigma\_z \,.$$

Answer: 
Perfectly incompressible bodies are actually prohibited by stability of thermodynamic equilibrium (entropy maximization). Section 21 of Landau & Lifshitz Vol 5: *Statistical Physics* has a nice discussion and derivation of the general thermodynamic inequality $C\_V > 0$.

So the equation of state $T = T(V)$ is pathological, and lets you derive all sorts of silly results:

Assume that $T = T(V)$, and imagine going from $(P\_1,V\_1) \to (P\_1,V\_2)$ in two different ways. On one hand, you can just go from $V\_1 \to V\_2$ at fixed $P = P\_1$. The change in internal energy is $\Delta U = Q\_1 + P\_1 (V\_1 - V\_2)$ by the first law. On the other hand, you can also go along the path $(P\_1, V\_1) \to (0,V\_1) \to (0,V\_2) \to (P\_1,V\_2)$. The first and third legs must have $\Delta U = 0$ because nothing depends on $P$. The second leg will have $\Delta U = Q\_0$. If internal energy is a state variable, then these should be equal:
$$
Q\_1 + P\_1(V\_1 - V\_2) = Q\_0
$$
Here $Q\_1$ is the amount of heat needed to change the temperature of the body from $T(V\_1)$ to $T(V\_2)$ at $P = P\_1$, and $Q\_0$ is the amount of heat needed to change the temperature of the body from $T(V\_1)$ to $T(V\_2)$ at $P = 0$. The equation above says that these must differ. But this contradicts the original assumption that the equation of state $T = T(V)$ is independent of $P$!
This is what I think: A true thermodynamic variable must be a function of two others. Therefore $T=T(V)$ is not a thermodynamic variable. However we have encountered such a pathology before: for ideal gas we have $U=U(T)$. In your very first equation you are piling both of these pathologies together to get $\delta Q=dU+p~dV=C\_vdT+p~dV$. I suspect nothing good will come out of this! In this equation $T$ and $V$ are *not* independent variables. But in writing down $\delta Q=p~dV$ you have assumed that they are independent. More correctly: $\delta Q=\left( C\_v\frac{dT}{dV}+p \right)dV$.

If you are not convinced, the same equation may be derived more formally, by avoiding writing in terms of $\delta Q$. Say the fundamental equation of the system is given by $S=f(U,V)$ (see *Thermodynamics* by Callen). But since $U=U(T(V))$, $S$ really depends on only one variable, $V$. Now:
\begin{align}
dS & =\frac{\partial f}{\partial U}dU+\frac{\partial f}{\partial V}dV \\
& = \left[ \frac{\partial f}{\partial U}\frac{dU}{dT}\frac{dT}{dV}+\frac{\partial f}{\partial V} \right] dV \\
& = \left[ \frac{C\_v}{T}\frac{dT}{dV}+\frac{p}{T} \right] dV
\end{align}
which is the same equation as before (when multiplied by $T$ throughout).

Anyway, you will be happy to know that when Clausius inequality is applied, the absurd conclusion $p\_1\leq p\_2$ results nevertheless. Therefore thermodynamics seems to prohibit a body with a fundamental equation that depends on a single extensive thermodynamic variable such as $V$. Two independent extensive thermodynamic variables are absolutely required (such as $U,V$).

P.S. I am not sure why you keep calling the body in your question, characterized by $T=T(V)$, an incompressible body which actually requires $\frac{\partial V}{\partial p}=0$.