Solving Schrodinger's equation: Free particles and uncertainty¶

We'll be exploring the evolution of a Gaussian packet, corresponding to the ground state of Hydrogen atom in a harmonic oscillator with frequency one Terahertz. We start by defining some constants.

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hbar = 1.054571628251774e-27;
omega = 1.e12;
protonMass = 1.672621637e-24;
m = protonMass;
a0 = sqrt(...); # RMS width of Gaussian

We define $\psi$ on a lattice of Np points x, from -L/2 to L/2.

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L = ...;
Np = ...;
dx = ...;
x = linspace(-L/2,L/2-dx,Np)

Now we define the initial wavefunction $\psi(0)$ on our grid.

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psi0 = (...)**(1./4.) * exp(-...)

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plot(x,abs(psi0)**2)
xlim(-2*a0,2*a0);

If $\psi(k)$ is expressed in Fourier space, applying the kinetic energy is pointwise multiplication. $p = -i \hbar k$ , so $p^2/2 m = -\hbar^2 k^2 / 2 m$ . We convert from real space into Fourier space using a FFT (Fast Fourier Transform), which returns $\tilde\psi(k)$ at Np points separated by $dk = 2 \pi/L$ :

$k = [0, dk, 2 dk, \dots, (Np/2) dk, -(Np/2 - 1) dk, \dots, -2 dk, -dk]$ Notice that

$k$ grows until halfway down the list, and then jumps to negative values and shrinks. We define

$k$ and

$k2=k^2$ as arrays

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dk = ...;
k = zeros(Np);
for j in range(0,...):
    k[j] = ...*dk 
for j in range(...,Np):
    k[j] = (j-Np) * dk 
k2 = k**2

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plot(k2)

We now define the time evolution operator $\tilde U_{kin}(t)$ in Fourier space. Note that the square root of minus one is 1j in Python.

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def UkinTilde(t):
    return exp(-1j * ...)

and we define $\psi(t)$ using Fourier transforms.

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from scipy import fft, ifft
def psi(t):
    return ifft(...*fft(psi0))

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P = ...
plot(x, psi(P/4).real, x, psi(P/4).imag)
figure()
plot(x, ..., x, ...)
figure()
plot(x, ..., x, ...)

We can understand this spread as a consequence of the uncertainty principle. We can calculate the wavepacket width $\sqrt{\langle x^2(t)\rangle}$ by the discrete approximation to the integral $\langle x^2(t)\rangle \sim \sum x^2 |\psi(x,t)|^2 dx$

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def width(t):
    return sqrt(sum(... * abs(...)**2 * ...))

Our initial wavepacket has a RMS width $\Delta x = a0$

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width(0), a0

We know that a minimum uncertainty wavepacket like ours has $\Delta x \Delta p = \hbar/2$ , so we expect the packet width to grow like $v t$ with $v$ given by the momentum uncertainty. We plot the comparison...

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deltaP = ...;
v = ...;
times = linspace(0,2*P,100);
widths = [width(t) for t in times];
plot(times, widths, times, ...)