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The particle in a box problem is a common question that people are taught in order to get some practice using Schrödinger’s equation. For this kind of problem one usually solves the equation for energy eigenvalues

$$ \hat{H} \psi_n (x) = E_n \psi_n (x)\;, $$ where you get some $\psi_n (x)$ with their respective quantized values of energy $E_n$.

My question is, what is the real state of the particle? I guess the $\psi_n (x)$ are the space dependent part of some stationary state. So, intuition gives me the answer: $$ \Psi (x,t) = \sum_{n=1}^{\infty}{\alpha_n \psi_n (x) e^{-i E_n t / \hbar}}. $$

However, if this is true, what are the values for each $\alpha_n$? The solutions for $\psi_n (x)$ let $n$ to run in the set $\{n \in \mathbb{N} \; \vert \; n > 0\}$, so it looks something annoying to think of infinite states with same probability for all of them, and those probabilities restricted to sum one.

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  • $\begingroup$ The values of $\alpha_n$ are determined by the initial wavefunction. Are you asking what the initial wavefunction is? Because that's something you decide for yourself. It's an initial condition for the problem. $\endgroup$ – J. Murray Jul 30 '18 at 16:18
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The "real" state of the particle completely depends on the initial conditions of the wavefunction. And while you are asking about the particle in a box, this answer can be applied to pretty much any intro QM problem. Since you have not gone into anything dealing with the particle in a box specifically, I will stay on the more general side as well.

The formula you have given $\Psi(x,t)=\sum \alpha _n \psi _n(x)e^{-iE_nt/\hbar}$ is the general solution to this problem, where $\psi _n(x)$ are the eigenfunctions of the Hamiltionian $\hat H$. Without any further information this is all you can really say.

If we know the initial wavefunction, then we can express this wavefunction in the eigenbasis $$\psi(x,t=0)=\psi_0=\sum \beta_n \psi_n(x)$$

where $$\beta_n=\int \psi_0^*\space \psi_n(x)dx$$

it looks something annoying to think of infinite states with same probability for all of them, and those probabilities restricted to sum one...

The probability of measuring our particle in state $n$ is given by $|\beta_n|^2$ assuming everything is normalized. This does not mean that all of these probabilities are equal (i.e., it is not true that $\beta_1=\beta_2=\beta_3=...$). Also, the restriction that these all sum to being equal to $1$ is needed so that what we mean by probability actually makes sense. We can have infinite sums of unequal terms whose sum approaches $1$. I would hardly call it annoying. It is extremely useful, and I also think pretty cool, that we can describe a host of functions in the same way: an infinite sum of eigenfunctions.

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  • $\begingroup$ and when the box is infinitely wide, we call this a Fourier transform. $\endgroup$ – JEB Jul 30 '18 at 21:17
  • $\begingroup$ @JEB Which part? $\endgroup$ – Aaron Stevens Jul 30 '18 at 21:50
  • $\begingroup$ That you can expand the initial state into a superposition of eigenstates. When there is no box, that is a just an FT. $\endgroup$ – JEB Jul 30 '18 at 22:20
  • $\begingroup$ @JEB I see what you are saying, but if the length goes to infinity don't all of the eigenfunctions go to 0 due to the 1/L dependencies? How does this work out mathematically? $\endgroup$ – Aaron Stevens Jul 30 '18 at 23:53
  • $\begingroup$ @JEB I see what you are saying, but if the length goes to infinity don't all of the eigenfunctions go to 0 due to the 1/L dependencies? How does this work out mathematically? $\endgroup$ – Aaron Stevens Jul 30 '18 at 23:53
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The values of the $\alpha_n$ depend on the initial conditions for a particular wavefunction. If you prepare the $n=1$ energy eigenstate $\psi_1$ (or measure the energy of a state and find it to be $E_1$), then $\alpha_1=1$ and all others are zero, for all time (this is because the $\psi_n$ form a diagonal complete basis for the Hamiltonian, and are therefore stationary states).

If you prepare some other arbitrary state with wavefunction $\phi(x)$, then the $\alpha_n$ are determined by the overlap between $\phi$ and $\psi_n$, since the energy eigenstates form a basis:

$$|\phi\rangle=\sum_{n=1}^\infty\langle\psi_n|\phi\rangle|\psi_n\rangle\equiv\sum_{n=1}^\infty\alpha_n|\psi_n\rangle$$

and therefore:

$$\alpha_n=\langle\psi_n|\phi\rangle=\int\langle\psi_n|x\rangle\langle x|\phi\rangle \;dx\equiv\int\psi_n^*(x)\phi(x)\;dx$$

which works because the position states $\{|x\rangle\}$ form a continuous complete basis.

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