A java applet to study wave packet dynamics

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Wave packet and wave packet dynamics

How does one represent a freely moving particle in quantum mechanics?

Intuitively, one may consider a travelling plane wave represented by a wave-function, which in one dimension is the solution of the Schrodinger time dependent equation with $V=0$ :

\psi(x,t) = Ae^{i(kx-\omega t)}

where $k$ and $\omega$ can be found out from the relations $p=\hbar k$ and $E=\hbar \omega$. There is a problem though. ${\vert\psi(x,t)\vert}^{2}$ then turns out to be $\vert A\vert^{2}$, which is constant. This would mean that there is equal probability of the particle to be present anywhere along the $x$-axis, which is unacceptable. Moreover, a wave represented by the above wave function travels in the $+x$ direction with the phase velocity $v_{p}=\omega/k$. Now, since
E = \frac{p^{2}}{2m} \;\;\;\;
\hbar\omega = \frac{\hbar^{2...
\omega = \frac{\hbar k^{2}}{2m} = constant\; k^{2} \;\;\;\;

(the so called dispersion relation)
v_{p} = \frac{\omega}{k} = \frac{p}{2m}

which implies that the phase velocity is one-haf of the classical particle velocity.

Thus the above plane wave wave-function does not have any physical significance and as such cannot truely represent a freely moving particle.

Instead, one may take the linear combination of different plane wave wave-functions (with different wave numbers or $k$ values):
\psi(x,t) = \int_{-inf}^{+inf} A(k)e^{i(kx-\omega t)} dk

$A(k)$ is the amplitude of the plane-wave having wave number $k$ or momentum $p=\hbar k$. This superposition, at each instant of time, leads to constructive interference in a small region along $x$-axis and destructive interference outside the region leading to $\psi(x,t)$ being localized in $x$ and being spread out in $k$ or $p$. The above wave-function represents what is known as a wave-packet, which in turn truely represents a classically free particle in quantum mechanics. With the change of time, the centre of the wave packet moves with a group velocity:
v_{g} = \frac{d\omega}{dk} \;\;\;\;
= \frac{d}{dk} \frac{\h...
...} \;\;\;\;
= \frac{\hbar k}{m} \;\;\;\;
= \frac{p}{m} \;\;\;\;

which agrees perfectly with the classical velocity of the free particle.

In order to obtain $A(k)$ we find $\psi(x,0)$:
\psi(x,0) = \int_{-inf}^{+inf} A(k)e^{i kx} dk

$\psi(x,0)$ is the inverse Fourier transform (IFT) of the function $A(k)$. Therefore $A(k))$ = Fourier transform (FT) of $\psi(x,0)$.
A(k) = \int_{-inf}^{+inf} \psi(x,0) e^ {i kx} dx

We may start with a given $\psi(x,0)$, find $A(k)$ from it and then use it to find $\psi(x,t)$ which would then give us the complete temporal evolution of the wave-packet.

To illustrate this technique we take $\psi(x,0)$ as follows:
\psi(x,0) = const \;\; e^{-(x-x_{0})^{2}/(4\sigma_{0}^{2})} e^{i k_{0} x}

The probability density then turns out to be:
\vert\psi(x,0)\vert^{2} = const \;\; e^{-(x-x_{0})^{2}/(2\sigma_{0}^{2})}

which represents a Gaussian distribution. $x_{0}=<x>=$mean of the distribution at $t=0$ and $\sigma_{0}=\sqrt{<(x-<x>)^{2}>}=\sqrt{<x^{2}>-<x>^{2}}=$standard deviation (square root of the variance) of the distribution at $t=0$. $x_{0}$ also equals the initial position of the center of the wave packet and hence the position of the free particle. $k_{0}$ corresponds to the initial momentum of the wave-packet. Then we find $A(k)$ and then substitute it in the expression for $\psi(x,t)$, perform the relevant integration to find $\psi(x,t)$ and hence $\vert\psi(x,t)\vert^{2}$. The important observations are as follows: $\vert\psi(x,t)\vert^{2}$ turns out to be a Gaussian distribution too:
\vert\psi(x,t)\vert^{2} = (const/\sigma_{t}^{2})\;\;e^{-(x-x_{0}-p_{0}t)^{2}/(2 \sigma_{t}^{2})}

\sigma_{t} = \sigma_{0}\;\;\sqrt{1+t^{2}/(4\sigma_{0}^{4})}

Thus the wave-packet localized around $x=x_{0}$ at $t=0$ spreads out (broadens or disperses) as it moves with time with group velocity $v_{g}=p_{0}/m=p_{0}$. The position of its centre changes with time from $x_{0}$ to $x_{0}+p_{0}t$. For $t=0:\;\;\Delta x=\sigma_{0}/\sqrt{2}$ For $t=0:\;\;\Delta p=\hbar /(\sqrt{2}\sigma_{0})$ Thus for $t=0:\;\; \Delta p\;\;\Delta x=\hbar/2$ i.e. Heisenberg's uncertaity relation is satisfied with our choice of initial wave-packet. Moreover the uncertainty product is minimum. For $t>0:\;\;\Delta x=\sigma_{t}/\sqrt{2}$ For $t>0:\;\;\Delta p=\hbar /(\sqrt{2}\sigma_{0})$ i.e. independent of $t$. Thus for $t>0:\;\; \Delta p\;\;\Delta x>\hbar/2$. (by making the substitutions)

Hence, Heisenberg's uncertaity relation is again satisfied for $t > 0$. However, the uncertainty product is no more the minimum. $\hbar = 1$ and $m = 1$ throughout if not explicitly mentioned above (atomic units employed).

One may note that plane electromagnetic or light waves and also light pulses (which are wave packets) travelling through vacuum have a linear dispersion relation in $\omega=c*k$ ( $c=constant=3\times 10^{8}\;m/s$). Hence such waves travel without any dispersion. For such plane light waves phase velocity $=v_{p}=c=\omega/k$ and for light pulse wave packets group-velocity is given by

However, since dispersion relation of particle wave-packets is non-linear, particle wave packets disperse or broaden with time.

The interactive java applet has been created by me to facilitate the study of the different aspects of the dynamics of a free particle wave packet as well as that of a wave packet tunnelling through a rectangular potential barrier or rectangular potential well. A Gaussian wave packet as described above has been used to launch the simulation.

Instead of following any analytical approach, I have used a numerical approach (The Crank Nicolson implicit method) to solve the relevant time dependent Schrodinger equation.

Input different values in the boxes housed at the top of the applet and press enter to start the simulation. For best results with the applet, keep yourself confined within the following range of values:

0.25 <= x0factor <= 0.75

(x0factor corresponds to the center of the initial wave packet)

- 1.0 < k0factor < 0.0 (for 0.5 <= x0factor <= 0.75)

0.0 < k0factor < 1.0 (for 0.25 <= x0factor <= 0.5)

(k0factor corresponds to the velocity of the initial wave packet)

0.0 < sigma0factor < 0.5

(sigma0factor corresponds to the width of the initial wave packet)

0.0 < barrierwidthfactor < 1.0

(barrierwidthfactor corresponds to the width of the potential barrier)

-1.0 < barrierheightfactor < 1.0

(barrierheightfactor corresponds to the height of the potential barrier; negative values of barrierheightfactor correspond to the case of a rectangular potential well while a zero value corresponds to the motion of a free particle).

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