Wave vector $\vec{k}$ vs position vector $\vec{x}$ My question is about the $k$-vectors in first Brillouin zone. If I am not misunderstood, the relation k = 2π/(Na) tells that when k goes to zero, we are very very far away from the reference atom and when k = 2π/a, we are one lattice constant (a) away in real space from reference atom. Now, for nearest neighbour tight binding approximation, the energy at k = 0 is (E-2t) and that at k = π/a is (E+2t). With respect to k = 0 eigenvalue, how do i make the connection in real space picture. Can we say that when we are infinitely far away from our reference atom, the energy is E-2t? If we cannot, then how should we interpret it? Or if we can then what is the justification with respect to real space? 
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If I am not misunderstood, the relation k = 2π/(Na) tells that when k goes to zero, we are very very far away from the reference atom and when k = 2π/a, we are one lattice constant (a) away in real space from reference atom.

I think you're a little misunderstood. The momentum space vector $\bf{k}$ has little to do with where an object is physically positioned. Rather, it tells you the direction and spatial frequency of a plane wave. So $|\mathbf{k}|=\frac{2\pi}{a}$ is the spatial wavevector of a plane wave which makes one full oscillation every interatomic distance. Likewise, when $|\mathbf{k}|\ll\frac{\pi}{a}$, you should visualize a plane wave which has a very long wavelength, on the order of several hundred interatomic distances or more.

With respect to k = 0 eigenvalue, how do i make the connection in real space picture. Can we say that when we are infinitely far away from our reference atom, the energy is E-2t?

Not quite. The intuition of the tight binding model is that when individual atoms are brought together, the wavefunctions one would obtain for the crystal by a brute-force computation would look very similar to what you'd see if you just placed a single-atom wavefunction at every lattice point (giving a function which I'll denote $\psi(\mathbf{x})$) and then multiplied the whole assembly by a spatial plane wave, $e^{i\mathbf{k\cdot x}}$.
There is also a simple intuition as to why the $\mathbf{k}=0$ assembly has a lower energy than the $|\mathbf{k}|=\pi/a$ assembly. Recall from freshman physics that things with short wavelength (electromagnetic, de Broglie, quantum plane waves, etc.) have higher energy than things with long wavelength. Try to visualize $\psi(\mathbf{x})e^{i\mathbf{0\cdot x}}$. Now try to visualize $\psi(\mathbf{x})e^{i\mathbf{k\cdot x}}$ where $|\mathbf{k}|>0$. Which has a shorter wavelength? The extra factor of $e^{i\mathbf{k\cdot x}}$ adds oscillations, which you'd intuitively expect to boost the energy.
So to summarize, the crystal has a low-energy wavefunction which looks very similar to $\psi(\mathbf{x})e^{i\mathbf{0\cdot x}}$ and which has energy $E-2t$, it has a high-energy highly-oscillatory wavefunction which looks very similar to $\psi(\mathbf{x})e^{i\mathbf{k\cdot x}}$ where $|\mathbf{k}|=\pi /a$ and which has energy $E+2t$, and it has a bunch of modes in between. And just to reiterate, it doesn't really have anything to do with how far you are from any particular reference atom.
Additionally, have you heard of "bonding" and "anti-bonding" orbitals? Note that when $|\mathbf{k}|=\pi /a$, the wavefunction flips sign at each lattice point, whereas when $|\mathbf{k}|=0$, the wavefunction is the same sign at each lattice point. Thus the zero-momentum state can be considered a low-energy "bonding" state, and the states near the edges of the Brillouin zone can be considered high-energy "anti-bonding" states.
