Global representation of $\Delta$ and Fast Fourier Transform

A really global consideration can only be achieved by either involving all available interpolations points $x_{j}$ in the semilocal approach or, which is the same, calculate the derivative by multiplying the wave vectors in the Fourier transformed representation of the wavefunction. This means that the momentum operator is local in momentum space:

$$\psi (k) = \frac{1}{\sqrt{2\pi }}\int\limits _{-\infty }^{\infty }\psi (x)e^{-ikx}dx=FT[\psi (x)],$$

$$\psi (x) = \frac{1}{\sqrt{2\pi }}\int\limits _{-\infty }^{\infty }\psi (k)e^{ikx}dk=FT^{-1}[\psi (k)],$$

$$\frac{d\psi (x)}{dx} = \frac{1}{\sqrt{2\pi }}\int\limits _{-\infty }^{\infty }\psi (k)(ik)e^{ikx}dk=FT^{-1}[(ik\psi (k)],$$

$$\frac{d^{2}\psi (x)}{dx^{2}} = \frac{1}{\sqrt{2\pi }}\int\limits _{-\infty }^{\infty }\psi (k)(-k^{2})e^{ikx}dk=FT^{-1}[(-k^{2})\psi (k)].$$

The Fourier representation yields the exact energy spectrum

$$T_{FT}(k)=\frac{\hbar ^{2}k^{2}}{2m},$$

while the finite difference approach gives

$$T_{FD}(k)=\frac{\hbar ^{2}}{2m}\left[ \frac{2\sin (k\Delta x/2)}{\Delta x}\right]$$

which converges to the exact spectrum for very small $\Delta x$. So for an application of the finite difference scheme one has to choose a finer mesh for the representation of the wavefunction.