Why does the bulk of a superconducting material expel magnetic field but not that of a perfect conductor?

It is well-known that when a superconductor (SC) is cooled below the transition temperature $$T_c$$, the magnetic field passing through the bulk of the SC is completely expelled. In Zee's book on Quantum Field Theory in a Nutshell, he explains why a superconductor abhors magnetic field in the bulk along the following lines

A hallmark of superconductivity is the Meissner effect, in which an external magnetic field $$\textbf{B}$$ permeating the material is expelled from it as the temperature drops below $$T_c$$. This indicates that a constant magnetic field inside the material is not favored energetically. The effective laws of electromagnetism in the material must somehow change at $$T_c$$. Normally, a constant magnetic field would cost an energy of the order$$\sim\textbf{B}^2V$$, where V is the volume of the material. Suppose that the energy density is changed from the standard $$\textbf{B}^2$$ to $$\textbf{A}^2$$ (where as usual $$\nabla\times\textbf{A}=\textbf{B}$$). For a constant magnetic field $$\textbf{B}, \textbf{A}$$ grows as the distance and hence the total energy would grow faster than V. After the material goes superconducting, we have to pay an unacceptably large amount of extra energy to maintain the constant magnetic field and so it is more favorable to expel the magnetic field.

Statement 1 "Suppose that the energy density is changed from the standard $$\textbf{B}^2$$ to $$\textbf{A}^2$$..."

Question 1: What does this mean? Energy density depends on $$\textbf{B}^2$$ and not on $$\textbf{A}^2$$.

Statement 2 "After the material goes superconducting, we have to pay an unacceptably large amount of extra energy..."

Question 2 Why?

I'm assuming that we're talking about BCS superconductivity.

Question 2 Why?

Because the energy has now the form mentioned in Q1

Statement 1 "Suppose that the energy density is changed from the standard $$B^2$$ to $$A^2$$..."

Question 1: What does this mean? Energy density depends on $$B^2$$ and not on $$A^2$$.

I'll be short,if you'd like some more detail you can ask in the comments.

Consider just a static $$B$$ field in the coulomb gauge; the field by itself has energy given by the integral of $$\frac{1}{2}B^2 = \frac{1}{2}(\nabla \times A)^2= -\frac{1}{2}A\Delta A$$ ;

We then have to take into account that in a superconductor the field is coupled to fermions. They will have an action (grand canonical ensamble): $$S=\int d\tau d\vec x \,\, \psi^*_\sigma \left( \partial_\tau -\mu +\frac{(\hat P +eA)^2}{2m} \right) \psi_\sigma + g \psi^*_\downarrow \psi^*_\uparrow \psi_\uparrow \psi_\downarrow$$

One could show that the spontaneous breaking of the $$U(1)$$ local symmetry generates a mass term for the gauge field $$A$$ ; putting it together with the term we inherited from the free fields with obtain an action for $$A$$ that has the form: $$\int dx \, \frac{1}{2}A( \tilde m^2 - \Delta) A (*)$$.

The "mass" squared $$\tilde m^2$$ is not a constant and depends on the parameters of the system. It is proportional to the so-called "superfluid density" ($$\tilde m^2 =\frac{n_s}{m_e}$$), which is nonzero only below the critical temperature.

This gets you the desired form of the (free) energy. Notice that the Euler-Lagrange equations we obtain from (*) are no other than the London equations for the $$B$$ field:

$$(\frac{n_s}{m} -\Delta )A= 0 \rightarrow (\frac{n_s}{m} -\Delta )B= 0$$

and they tell us that the magnetic field is exponentially suppressed going from the surface to the bulk of the superconductor.