(Note: This post focuses on a single simple example, however I'm asking about the error in general in my logic).

Consider the infinite potential well "particle in a box" system described by

$$V(x)=\begin{cases}0&\text{if }0<x<L\\\infty&\text{otherwise}\end{cases}.$$

It's fairly easy to find the wavefunctions $\psi_n(x)=\langle E_n\vert\psi\rangle$ by solving the time independent Schroedinger equation:


Now, since $\mathcal{\hat H}$ is Hermitian we know there is a complete set of eigenstates $\vert E_n\rangle$ such that, for any initial state $\vert\psi,0\rangle$ we can write

$$\vert\psi,0\rangle = \sum_k a_k\vert E_k\rangle$$

The problem of evolving the state $\vert\psi,0\rangle$ in time is easily reduced to

$$\vert\psi,t\rangle = \sum_k a_k e^{-iE_n t/\hbar}\vert E_k\rangle$$

But the wavefunction of this state is given by

$$\Psi(x,t) =\sum_ka_ke^{-iE_n t/\hbar}\psi_n(x) = \sum_ka_k\sqrt{\frac 2 L}e^{-iE_n t/\hbar}\sin\left(\frac{n\pi}{L}x\right)$$

and taking $\vert\vert^2$s to obtain the probability distribution yields a time-independent function. Hence the time evolution of the probability this system is apparently trivial for any initial state, but I have heard from multiple sources and a demonstration applet that even for a superposition of two stationary states the particle oscillates throughout the box. What have I done wrong here?

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    $\begingroup$ $\lvert x + y \rvert \neq \lvert x \rvert + \lvert y \rvert$ $\endgroup$ – ACuriousMind Nov 14 '14 at 22:33
  • $\begingroup$ @ACuriousMind Unbelievable... I get so caught up in equations that I forget little mathematical tidbits like that. If you post that as an answer I'll accept. $\endgroup$ – theage Nov 14 '14 at 22:38

Are you asking why taking the squared modulus of a superposition of (eigen)states turns out to be considerably more complicated than the squared modulus of a single eigenstates? It this is so, I'd say because eigenstates of different energies evolve differently, and when you do the superposition and consider the square modulus you have to take into account all the interference terms like $$|a+b|^2 = |a|^2 + |b|^2 + 2 \Re (ab^*) $$ which are usually highly non trivial.

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  • $\begingroup$ Yes, this is right - in this analysis I assumed $\vert x+y\vert = \vert x\vert + \vert y\vert$. Thanks. $\endgroup$ – theage Nov 14 '14 at 22:46
  • $\begingroup$ @theage: You, probably, thought of orthogonality of different $\psi_n$, but for that one should integrate. The integral is indeed trivial. $\endgroup$ – Vladimir Kalitvianski Nov 14 '14 at 23:02

Edit after seem Acuriousmind and glance's contributions.... they have the answer sorted out above.

Interesting experimental example of this is Zewails work - for example this paper, which is not behind a 'pay wall' where evolution on femtosecond timescale of a molecular vibrational 'wavepacket' make up of 'superposition of many states was observed'.

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  • $\begingroup$ I think you can... isn't that guaranteed by the TISE? $\endgroup$ – theage Nov 14 '14 at 22:38
  • $\begingroup$ @theage - Ok with the TISE (time indep Schrod. Eq I guess) you get lots of nice stationary states and with a long laser pulse (for example) you could excite just one of these states - but then you would not have a very good time zero for your system - you would not know exactly when it was excited during the laser pulse - does that make sense? $\endgroup$ – tom Nov 14 '14 at 22:44
  • $\begingroup$ One way or the other, unless I'm extremely far from the mark a small time translation isn't the difference between trivial and nontrivial evolution. I made a mathematical error, but thanks for the insights anyway. $\endgroup$ – theage Nov 14 '14 at 22:50
  • $\begingroup$ @theage - I now see the point of what acuriousmind and glance have put - so I edit my answer... $\endgroup$ – tom Nov 14 '14 at 22:51

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