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The general one particle state in a simple infinite well of size $L$ is a superposition of all the Hamiltonian eigen-states: $$\tag{1} \psi(x, t) = \sqrt{\frac{2}{L}} \sum_{n = 1}^{\infty} c_n \, e^{-\, i E_n t/\hbar} \, \sin (n \pi x / L), $$ where the energy levels are $$\tag{2} E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}, $$ and $c_n$ are arbitrary constant coefficients. Using Mathematica, I'm interested in plotting the evolution of a nice wave packet inside the infinite well that moves semi-classically, so I'm looking for coefficients $c_n$ that gives a wave packet that moves in a nice way. Is there such a set of $c_n$?

Of course, I could use a simple superposition of the first two eigen-states: $c_1 = c_2 = 1/\sqrt{2}$ for example, which produces a flip-flop wave motion inside the well, but I'm looking for something less trivial. Any idea about coefficients $c_n$ that produce a nice semi-classical motion inside the infinite square well? I'm looking for a traveling wave that bounces on the two walls (like a particle in a box), instead of a trivial flip-flopping.

Comment: I guess that a localized wave packet would disperse and spread around during its motion, and the reflections on the walls would produce an apparent random sloshing of the wave on the whole box length. So it may be impossible to get what I'm looking, for on a relatively long period of time.

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  • $\begingroup$ Not a full answer to your question, but in the context of a quantum harmonic oscillator, the states that behave like what you are looking for are the so-called "coherent states". This is the wikipedia page on them: en.wikipedia.org/wiki/Coherent_state $\endgroup$
    – ProfM
    Jan 8 at 18:51
  • $\begingroup$ @ProfM there are no coherent states in a square well. Although essentially you want the same kind of position-space wavefunction: a wavepacket. Only problem is that it will spread out (as the question says in the last paragraph) just like a wavepacket in free space does. $\endgroup$
    – AXensen
    Jan 8 at 23:12
  • $\begingroup$ @AXensen agreed, and should probably have explicitly stated this. Just wanted to point OP to a nice/simple example of a different potential in which this happens. $\endgroup$
    – ProfM
    Jan 10 at 6:56
  • $\begingroup$ @ProfM, I'm well aware of the oscillator case, it is a well known subject (coherent states AKA Glauber states, or eigen states of the annihilation operator). Apparently, there's nothing similar for the potential well. $\endgroup$
    – Cham
    Jan 10 at 14:23

2 Answers 2

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Adding to the answer by LPZ:

It can be difficult to find parameters that produce a nice-looking simulation if you just use trial-and-error. But you can derive an important constraint, as follows:

Suppose your initial state is a Gaussian wavepacket, centered at the middle of the well with momentum $p$: $$ \Psi(x,0) \propto e^{ipx} \: e^{-(x - L/2)^2/{2\sigma^2}}.$$

i) The initial packet width $\sigma$ should be several times smaller than $L$, so that the wavefunction doesn't span the whole well. That is, $\sigma = L/f_{\sigma}$, with $f_{\sigma}$ at least 3 or 4.

ii) The wavepacket should not spread too much in the time it takes for the peak to bounce back and forth a few times in the well. The time for $f_b$ bounces is about $T = \frac{f_b L}{p/m}$. During this time, the fractional spread in the wavefunction is about $\frac{1}{\sigma}\sqrt{\frac{T}{m}}$ (from the formula for the time-evolution of a free Gaussian wavepacket). Call this value 1/$f_s$. You'd want $f_s$ to be at least 2 or 3.

Combining these constraints gives $p \ge f_b f_{\sigma}^2 f_s^2/L$, or about $70/L.$ Thus the initial momentum $p$ needs to be quite large. And this means you'll need a fairly large number of basis states (also of order 70) in your expansion for $\Psi$.

Your final result will look something like this.

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One natural candidate in the same vein of a coherent state is to take a gaussian wave packet with a caveat. I describe two equivalent ways of constructing it in a previous answer of mine to Ansatz for wavefunction infinite square well with linear perturbation $\alpha\cdot x$. Essentially, you choose $c_n$ to be gaussian or equivalently, using the method of images you take reflected gaussian wavepackets to satisfy the boundary conditions.

For the time evolution, you can use the usual eigenbasis expansion and add the time dependent phase with the calculated spectrum. I think that for your application, the second expression is more transparent as the images can be interpreted as the reflections. While in principle the reflections are exact, due to the spreading of the wave packets, these reflections will become harder and harder to observe due to the increasingly overlapping image wave functions.

The method is entirely general, and you can start from any trial wavefunction in $\mathbb R$. Either discretize its momentum wave function along the energy eigenbasis. Alternatively, antiperiodize it using the method of images and apply the time evolution of the entire line on the each image.

Mathematically, say you start with a spatial wave function on $\mathbb R$ and its associate momentum wave function related by Fourier transform: $$ \begin{align} \psi &= \int \frac{dk}{2\pi}e^{ikx}\phi(k) & \phi &= \int \frac{dk}{2\pi}e^{-ikx}\psi(k) \end{align} $$ Then you can equivalently project them onto wavefunctions of the finite well: $$ \begin{align} \psi(x) &\to \psi_L(x) = \sum_{n\in\mathbb Z} \psi(x-2nL)-\psi(-x-2nL) \\ \phi(k) &\to \phi_L(k) = \sum_{n\in\mathbb Z^*} (\phi(\pi n/L)-\phi(-\pi n/L))2L\delta(k-\pi n/L) \end{align} $$ with $\psi_L, \phi_L$ are still Fourier transforms of one another.

Hope this helps.

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