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Classically, a particle will be able to exhibit sideways motion as it bounces from one end of the box to the other. Can we form a Gaussian like wavepacket from the stationery state solutions? If so, when these solutions evolve over time, do they cause the position distribution to exhibit 'bouncing'? And what does the momentum space distribution look like in this case? The momentum distribution for a single eigenstate is an even function, but surely with a bouncing particle it should be localised?

How does Q.M correspond to classical predictions in the limit of large L?

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  • $\begingroup$ "no matter what the length of the box is" is not correct. Quantum mechanics is used for small dimensions,, commensurate to h_bar. A box of dimensions larger than nanometers cannot be modeled by a square well, there are the innumerable potentials of those atoms on the sides Even in nanometer dimensions much more complicated potentials are needed to fit the data. $\endgroup$
    – anna v
    Mar 19, 2021 at 11:40
  • $\begingroup$ physics.stackexchange.com/questions/91221/… suggestion is that this is impossible. Why $\endgroup$
    – user86425
    Mar 19, 2021 at 11:49
  • $\begingroup$ I have stated in my comment the range of values that need QM solution. Real boxes cannot be larger than nanometer to be able to truly model the data. Mathematically one can play games. $\endgroup$
    – anna v
    Mar 19, 2021 at 14:27

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The classical limit of the particle in a box would indeed be a wave packet with some range $\Delta n$ of values. This page discusses this limit in a bit of detail, and provides a handy applet that allows you to play around with various wavepackets and see their evolution in the position and momentum representations.

To summarize in case of link rot: a superposition of states between $n_0-\Delta n$ and $n_0+\Delta n$, with $\Delta n \ll n_0$, gives a wavepacket of the type you want. The wavefunction will then be $$ \Psi(x,t) = \sum_{n=n_0-\Delta n}^{n_0+\Delta n} C_n \sin \left( \frac{n \pi x}{a} \right) e^{-i E_n t/\hbar} $$ The coefficients $C_n$ for the $n$th eigenstate are somewhat arbitrary; we choose them to be $$ C_n \propto \cos^2 \left[\frac{(n- n_0) \pi}{2 n_0 + 1} \right] e^{-i (n-n_0)\pi/2} $$ (Simply replacing the cosine above with a constant will yield similar results but the initial wavepacket won't be as smooth.) This leads to a "smooth" wavepacket that "bounces around" inside the box for some time. However, it also disperses with time (i.e., $\Delta x$ increases), as we would expect from a free particle. The wave packet also interferes with itself substantially when it bounces off of the "walls" of the box.

enter image description here

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  • $\begingroup$ why are we allowed to control phase in the applet. When solutions are only sine waves? $\endgroup$
    – user86425
    Mar 19, 2021 at 17:08
  • $\begingroup$ @ggmate: The phase is the relative phase between the various sine waves. If you look at the expression above, you'll see that successive sine waves in the sum differ by a phase of $\pi/2$. $\endgroup$ Mar 19, 2021 at 17:12
  • $\begingroup$ This applet is very satisfactory. But I'm struggling to make intuitive sense of it. Surely the energy eigenstates are dirac deltas in the momentum space. Then why is the graph in the momentum space smoothed off? $\endgroup$
    – user86425
    Mar 19, 2021 at 17:23
  • $\begingroup$ @ggmate: I think it's just an artifact of the graphics used, which smoothly interpolate between $p$ values. $\endgroup$ Mar 19, 2021 at 19:20
  • $\begingroup$ but the dirac deltas should be spaced out for various waves. i don't get it? $\endgroup$
    – user86425
    Mar 20, 2021 at 8:28
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The ubiquitous 'particle in an infinite square well' examples are using the time-independent Schroedinger equation to look for stationary states. Because those are very simple to find. Unsurprisingly, they find stationary states, and have no time dependence. That's not really a feature of Quantum Mechanics, just of the simple illustrative example.

It is possible to model non-stationary states where the particle starts with some initial position and momentum functions and observe how they change over time, with the modal position moving around the box.

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  • $\begingroup$ yes but these state would fail to produce a valid wavefunction in the energy space? $\endgroup$
    – user86425
    Mar 19, 2021 at 11:50
  • $\begingroup$ not eigenvectors of energy. but a gaussian wavepacket is not a linear combination of them either $\endgroup$
    – user86425
    Mar 19, 2021 at 16:49

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