The solution of infinite well, positioned from $x=0$ $x=l$, is $$ \Psi_n(x,t)= \sqrt{\frac{2}{l}}\sin\left(\frac{n\pi}{l}x\right)e^{iE_nt} $$ But the most general solution of this problem is : $$ \Psi(x,t)= \displaystyle\sum_{n=1}^{\infty} C_n\sqrt{\frac{2}{l}}\sin\left(\frac{n\pi}{l}x\right)e^{iE_nt} $$ and $$ \Psi(x,0)=\displaystyle\sum_{n=1}^{\infty}C_n\Psi_n(x,0) $$ summing up linearly all possible states give me the general solution. But what does it mean physically? Does the summation mean all possible states constitute a wave packet in the well?

Actually I can go further and using Fourier transform show that ( shortly I will say $\Psi(x)$ for $\Psi(x,0)$ ) $$ \int\Psi(x)=\displaystyle\sum_{n=1}^{\infty} \int C_n\Psi_n(x) $$ $$ \int \Psi_m(x)^*\Psi(x)dx = \displaystyle\sum_{n=1}^{\infty} C_n\int \Psi_m(x)^*\Psi_n(x)dx $$ $$ \int \Psi_m(x)^*\Psi_n(x)dx =\delta_{mn} $$ $$ \int \Psi_m(x)^*\Psi(x)dx = C_m $$ This equation makes me not much sense. If you integrate an energy eigenstate with the general state, you get the probability of this eigenstate, if you take the square of constant $C_m$. Why?


There are three answers posted but so far nobody has posted the obvious physical interpretation. The energy eigenstates all have the particle spread out within the box, stationary in time. If you want the particle to bounce back and forth between the walls of the box, then you do this by combining eigenstates. The simplest case is just to mix the ground with the first excited state. If you look carefully at the resulting function, you should see that it bounces back and forth between the left and right hand sides of the box.

In this example, the wave function still isn't very precisely localized at any instant, but if you want to do better you just add more eigenstates.

  • $\begingroup$ If I sum all the possible states with $n=1,2,3...$ I should get sth like this !pls ignore the boundaries and this supports my wave packet assumption indeed. $\endgroup$ – aQuestion Jan 13 '15 at 20:20
  • $\begingroup$ You're assuming you add them all in phase at the middle of the box and with equal amplitudes. Yes, you get a wave packet at that instant but I don't think it holds together...I think it explosed all over the box. I'm not sure it's all that easy to get coherent packets that stick together, like the ones you can create in the harmonic oscillator potential. $\endgroup$ – Marty Green Jan 13 '15 at 22:41
  • $\begingroup$ I assume all amplitudes the same. You are right, actually it might seems different than that with different amplitudes but I am not sure now. But however, isn't it logical to fix the amplitudes i.e if you are stuck in a well from $x=0$ to $x=l=2$ ? $\endgroup$ – aQuestion Jan 14 '15 at 11:10
  • 1
    $\begingroup$ The amplitudes don't remain fixed in time. You need to think about the simple example I gave where you mix just the the ground function and the first excited state. Since the electron is bouncing back and forth, it is radiating. So it loses energy. The amplitude "drains" from the excited state to the ground state until eventually it stops moving. It's the exact same thing that happens to a hydrogen atom in a superposition of the ground (1s) state and first excited (2p) state. $\endgroup$ – Marty Green Jan 14 '15 at 12:25
  • $\begingroup$ And don't forget that all the states you mix together have a time component...they are multiplied by exp(jwt), where w depends on the energy level. So the pattern is changing with time as the relative phases move with respect to one another. $\endgroup$ – Marty Green Jan 14 '15 at 12:26

The physical meaning of the superposition

$\Psi(x, 0)=\displaystyle\sum_{n=1}^{\infty}C_n\Psi_n(x, 0), \ \ \ \displaystyle\sum_{n=1}^{\infty}|C_n|^2 = 1$,

appears when you measure energies. At a measurement of energy of the particle in the well, $|C_n|^2$ is the probability to find the energy $E_n$.

For clarity, I suggest a slight change in the notation of the eigenstates,

$\Psi(x, 0)=\displaystyle\sum_{n=1}^{\infty}C_n\Psi(x,0; E_n)$.

Assume that you have some procedure to measure the energy of the particle in the well, i.e. you "entangle" the particle in the well with another particle $M$ that can get out from the well, s.t. the state of both particles be

$\Psi(x,y,0)=\displaystyle\sum_{n=1}^{\infty}C_n\Psi(x,0; E_n) \Phi(y; V_n)$,

where $V_n$ is a property of the particle $M$ that can be detected by an apparatus, i.e. the apparatus can report one of the values $V_1, V_2, V_3, . . . $. If the apparatus reported the value $V_n$, then we infer that the particle in the well had the energy $E_n$.

Then, by repeating the measurement many times, each time with another particle in the well and another particle $M$, the apparatus will indicate $V_n$, i.e. the energy $E_n$, with the probability that the state $\Psi(x,0; E_n)$ appears in $\Psi(x, 0)$, and this is $|C_n|^2$.

About your last question, $C_m$ is the amplitude of probability to find the state $\Psi(x,0; E_n)$ appears in $\Psi(x, 0)$, and the probability is the absolute square of the amplitude. This is the formalism of the QM.


Each of the stationary states $\Psi_n$ is a solution to the Schrödinger equation with definite energy $E_n$, and since the Schrödinger equation is linear, a linear combination $\sum_n C_n \Psi_n$ is also a solution.

What does this mean physically? That the particle is not in a state of definite energy (unless the initial condition $\Psi\left(x,0\right)$ happens to be one of the $\Psi_n$).

The initial condition $\Psi\left(x,0\right)$ is what determines $C_n$, as you showed. If there is any confusion about the process of calculating $C_n$, it is analogous to the following problem.

Let's say I tell you that a vector ${\bf V}$ is a linear combination of some Cartesian basis vectors $\hat{\bf e}_n$, analogous to the stationary states $\Psi_n$. That is, ${\bf V} = \sum_n C_n \hat{\bf e}_n$. To determine $C_n$, you use the orthonormality of the basis vectors $\hat{\bf e}_n \cdot \hat{\bf e}_m = \delta_{nm}$, analogous to $\int dx \Psi^*_m \Psi_n = \delta_{nm}$: $$ {\bf V} \cdot \hat{\bf e}_n = \left(\sum_m C_m \hat{\bf e}_m\right) \cdot \hat{\bf e}_n = \sum_m C_m \delta_{mn} = C_n $$ If I tell you ${\bf V} \cdot \hat{\bf e}_n$, analogous to specifying $\Psi\left(x,0\right)$, then you can calculate the $C_n$s.

Why does $\lvert C_n\rvert^2$ then give the probability of getting $E_n$ when the energy is measured? This article might help.


There are alot of physical meaning which are not in quantum mechanics but analogous to it. For example in microwave theory where waveguide are used the wave is trapped in a box. Analogously an infinite well for electron and the electromagnetic wave in waveguide no longer becomes transverse electromagnetic but it is transverse magnetic where range frequencies and wavelengths only pass depending on the dimensions of the waveguide. Where there are a lot of infinite modes carrying ranges of frequencies depending on the dimension of the waveguide.


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