# Full solution of the wavefunction for the particle in a box problem

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.

• 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. Jul 30, 2018 at 16:18

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.

• and when the box is infinitely wide, we call this a Fourier transform.
– JEB
Jul 30, 2018 at 21:17
• @JEB Which part? Jul 30, 2018 at 21:50
• That you can expand the initial state into a superposition of eigenstates. When there is no box, that is a just an FT.
– JEB
Jul 30, 2018 at 22:20
• @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? Jul 30, 2018 at 23:53
• @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? Jul 30, 2018 at 23:53

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.