Numerical Solution of SE using FFT

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Our goal is to numerically exactly solve the 1D-Schrödinger equation with an arbitrary potential \( V(x) \).

As well known from the courses on quantum mechanics Schrödinger equation reads:

$$\hat{H}|\psi\rangle=E|\psi\rangle$$

Here, \( \hat H \) is a Hamiltonian consisting of a kinetic energy and potential energy operators:

$$\hat{H}=\frac{\hat{p}^{2}}{2m}+V(\hat{x})$$

so that the eigenproblem can be written as:

$$\left[\frac{\hat{p}^{2}}{2m}+V(\hat{x})\right]|\psi\rangle=E|\psi\rangle$$

or explicitly,

$$\frac{\hat{p}^{2}}{2m}|\psi\rangle+V(\hat{x})|\psi\rangle=E|\psi\rangle$$

So far, we do not use any particular basis but represent the state \( |\psi \rangle \) as an arbitrary vector in the Hilbert space. Now we will project the eigenproblem onto a continous position eigenstate basis \( |x\rangle \) giving rise to the position representation of the Schrödinger equation:

$$\langle x|\frac{\hat{p}^{2}}{2m}|\psi\rangle+\langle x|V(\hat{x})|\psi\rangle=E\langle x|\psi\rangle$$

In the first part of the left hand side we now have a kinetic energy operator acting on a state \( \psi \). Since the kinetic energy operator is a square of the momentum operator, the most natural way to calculate this expression is to turn into the momentum representation by using the completeness of the momentum operator eigenfunctions:

$$\mathbb{1} = \int_{-\infty}^{+\infty}\text{d}p |p\rangle \langle p| $$

By inserting the completeness relation between the kinetic energy operator and the state \(|\psi\rangle \) we obtain:

$$\frac{\hat{p}^{2}}{2m}|\psi\rangle=\int_{-\infty}^{+\infty}\text{d}p\frac{\hat{p}^{2}}{2m}|p\rangle\langle p|\psi\rangle,$$

which leads to:

$$\int_{-\infty}^{+\infty}\text{d}p\langle x|\frac{\hat{p}^{2}}{2m}|p\rangle \langle p|\psi\rangle+\langle x|V(\hat{x})|\psi\rangle=E\langle x|\psi\rangle.$$

Now we can exploit the fact that the momentum eigenstate is also an eigenstate of the kinetic energy operator,

$$ \frac{\hat{p}^{2}}{2m}|p\rangle=\frac{p^{2}}{2m}|p\rangle, $$

which leads to: $$\int_{-\infty}^{+\infty}\text{d}p\frac{p^{2}}{2m}\langle x|p\rangle \fcolorbox{red}{#f2d5d3}{$\langle p|\psi\rangle$}+\langle x|\fcolorbox{green}{#d3f2d9}{$V(\hat{x})|\psi\rangle$}=E\langle x|\psi\rangle.$$ The first highlighted expression is the momentum representation of the state \( |\psi\rangle \), which can be obtained from the position representation by inserting the completeness relation for the position eigenstates:

$$\fcolorbox{red}{#f2d5d3}{$\langle p|\psi\rangle$}=\int_{-\infty}^{+\infty}\text{d}x'\langle p|x'\rangle\langle x'|\psi\rangle.$$

The second highlighted expression can be further expanded by inserting the completeness relation for the position eigenstates between the potential operator and the state \( |\psi\rangle\):

$$\fcolorbox{green}{#d3f2d9}{$V(\hat{x})|\psi\rangle$}=\int_{-\infty}^{+\infty}\text{d}x'V(\hat{x})|x'\rangle\langle x'|\psi\rangle$$

By introducing these expressions into the Schrödinger equation and interchanging the order of integration we obtain:

$$\int_{-\infty}^{+\infty}\text{d}x'\int_{-\infty}^{+\infty}\text{d}p\frac{p^{2}}{2m}\langle x|p\rangle\langle p|x'\rangle\langle x'|\psi\rangle+\int_{-\infty}^{+\infty}\text{d}x'\langle x|V(\hat{x})|x'\rangle\langle x'|\psi\rangle=E\langle x|\psi\rangle$$

or

$$\int_{-\infty}^{+\infty}\text{d}x'\left[\int_{-\infty}^{+\infty}\text{d}p\frac{p^{2}}{2m}\langle x|p\rangle\langle p|x'\rangle+\langle x|V(\hat{x})|x'\rangle\right]\langle x'|\psi\rangle=E\langle x|\psi\rangle$$

We now introduce the following shorthand notation: $$\psi(x):=\langle x|\psi\rangle$$

$$\psi(x'):=\langle x'|\psi\rangle$$

$$H(x,x'):=\int_{-\infty}^{+\infty}\text{d}p\frac{p^{2}}{2m}\langle x|p\rangle\langle p|x'\rangle+\langle x|V(\hat{x})|x'\rangle$$

$$\int_{-\infty}^{+\infty}\text{d}x'H(x,x')\psi(x')=E\psi(x)$$

With this, we have converted the Schrödinger equation into an eigenvalue problem for a continous ''matrix'' \( H(x,x') \)! In the next step, we will discretize the continous position variables \(x\) and \(x'\) and turn the infinite dimensional eigenvalue problem for a continous matrix \(H(x,x')\) into a usual discrete eigenvalue problem for a finite dimensional matrix!