What is paramagnetic current-current correlation? I know what paramagnetism is. But first I want to know about the paramagnetic current and then the above-mentioned correlation?
Actually, I am working on a paper on superconductivity where I have seen the term:
Scalapino, White, and Zhang.
 A: Typically the current operator is a sum of to parts: a so-called "paramagnetic" term and a "diamagnetic" one. 
This is just notation, and bears no real meaning, since the true, physical current operator is the sum of them both.
For example, consider an electron gas in an external electromagnetic field:
$H=\int d^3 x \frac{1}{2m} \psi^\dagger (\mathbf x) (-i \nabla -e/c \mathbf A(\mathbf x,t))^2 \psi (\mathbf x) + e \phi(\mathbf x,t) \psi^\dagger (\mathbf x) \psi (\mathbf x)$
Then the current operator is 
$\mathbf J(\mathbf x)= \frac{e}{2m} \psi^\dagger (\mathbf x) (-i \nabla -e/c \mathbf A(\mathbf x,t)) \psi (\mathbf x) +h.c$ (where h.c is hermitian conjugate)
The term in the current operator which is proportional to A is the diamagnetic term, the other is the paramagnetic one.
As you can see, each term on its own is not gauge invariant (there is an explicit dependence on A in the diamagnetic one!). Only the sum is.
The paramagnetic current correlation function is just the correlation function of the paramagnetic parts of the current operator. The correlation function often appears when things are calculated to first order in A.
A: The correct derivation for this is that
$$ j^A_\mu = \frac{\delta S}{\delta A_\mu}$$
where the action is indeed for this minimal coupling Hamiltonian. You can 'prove' this is the correct Hamiltonian to use either by showing that the Lagrangian that gives the Lorentz force is the Legendre transform of this Hamiltonian and then "corresponding" to a quantum-mechanical Hamiltonian, or by showing that this is the only Hamiltonian that gives a gauge-invariant Schrodinger equation.
We have
$$
S[\bar{\psi},\psi, A] = \sum_\sigma \int d x \quad \bar{\psi}_{\sigma}(\partial_{\tau}+\phi+\frac{\hbar}{2 m}(-i \nabla-\frac{q}{c}\mathbf{A})^{2}-\mu+V_{0})\psi_{\sigma}
$$
where we can expand the square and drop any $A$ independent terms to give
$$
  \sum_\sigma \int d x \quad \bar{\psi}_{\sigma}\frac{i \hbar q  }{2 m c}(\nabla\cdot \mathbf{A}+\nabla\cdot \mathbf{A})\psi_{\sigma} + \frac{q^2}{2m} A^2 \bar{\psi}_{\sigma}\psi_{\sigma}
$$
Integration by parts and discarding the boundary terms gives
$$
  \sum_\sigma \int d x \quad \mathbf{A} \cdot \frac{i \hbar q  }{2 m c}((\nabla\bar{\psi}_{\sigma})\psi_{\sigma} - \bar{\psi}_{\sigma}(\nabla\psi_{\sigma})) + \frac{q^2}{2m} A^2 \bar{\psi}_{\sigma}\psi_{\sigma}
$$
and finally performing the functional derivative gives
$$\vec{j}^A_{\sigma} = \vec{j}^P_{\sigma}  +  \vec{j}^D_{\sigma} $$
where now you can see the paramagnetic current is
$$
\vec{j}^D_{\sigma}  = \frac{i \hbar q  }{2 m c}((\nabla\bar{\psi}_{\sigma})\psi_{\sigma} - \bar{\psi}_{\sigma}(\nabla\psi_{\sigma}))
$$
and the diamagnetic term is
$$
 \vec{j}^D_{\sigma} = \frac{q^2}{m} A \bar{\psi}_{\sigma}\psi_{\sigma}
$$
jjjj's answer seems to be ambiguous about this functional derivative, so it's ambiguous where the derivative operator acts and the factor of 2 in the diamagnetic current.
For current-current correlator, you have to look at the linear response function as a second derivative of the path integral partition function, or derive the Kubo formula another way.
