Wave functions and eigenenergies

We first study the case where the motion of the particle is restricted within the interval $[x_1,x_2]=[0,L], L>0$ between the infinitely high walls of the potential

$$V(x)=\left\{ \begin{array}{cc} \infty, & -\infty<x\le 0 \\ 0, & 0<x\le L \\ \infty & L<x< \infty \end{array} \right.$$ (94)

Outside the interval $[0,L]$ the particle can not exist and its wave function must be zero, i.e.

$$\psi(x)=\left\{ \begin{array}{cc} 0, & -\infty<x\le 0 \\ a e^{ikx}+be^{-ikx}, & 0<x\le L \\ 0, & L<x< \infty \end{array} \right.$$ (95)

1. We demand that the wave function vanishes at $x=0$ and $x=L$ so that it is continuous a these points. Clearly, this makes physically sense because at $x=0,L$ the potential is infinitely high and the probability density $\vert\psi(x)\vert^2$ to find the particle there should be zero. We obtain

$$\psi(0) = 0 \leadsto 0= a+b \leadsto \psi(x)= c \sin(kx),\quad 0\le x\le L,\quad c = const.$$

$$\psi(L) = 0 \leadsto \sin(kL)=0.$$ (96)

The first condition tells us that the wave function must be a sine-function. The second condition is more interesting: it sets a condition for the possible values $k_n$ that $k$ can have,

$$kL=n\pi \leadsto k\equiv k_n=\frac{n\pi}{L},\quad n=1,2,3,...$$ (97)

The second boundary condition at $x=L$ restricts the possible values of the energy $E$, because $k:=\sqrt{ (2m/{\hbar^2})\left(E-V\right)}= \sqrt{ (2m/{\hbar^2})E}$. Therefore, the energy can only acquire values

$$E_n= \frac{\hbar^2 k_n^2}{2m}= \frac{n^2 \hbar^2 \pi^2}{2mL^2},\quad n=1,2,3,...$$ (98)

This is the first case where we encounter a quantisation of energy. The reason for the quantization here is obvious: the wave functions $\psi(x)$ have to `fit' into the well, similar to classical waves in a resonator which only allows waves with certain wave lengths . The allowed wave vectors $k_n$ then are related to the energy by the de Broglie relation $p=\hbar k \leadsto p_n=\hbar k_n$, and the energy within the well is just the kinetic energy $E=p^2/2m$ (since the potential is zero there) whence (2.21) follows.

2. The potential well gives only rise to discrete values of the energy. One says that the spectrum of energies is discrete. If we did not have the confinement potential, the wave functions would just be plane waves $e^{\pm ikx}$ with arbitrary values $k$ and therefore arbitrary, continues values for the energies $E=\hbar^2k^2/2m$. In such a case the the spectrum is called a continuous spectrum.

3. In order to interpret the absolute square wave of the wave functions $\psi_n(x)= c \sin(k_nx)$ as a probability density, we have to demand

$$1 = \int_0^Ldx \vert\psi_n(x)\vert^2 = \int_0^Ldx \vert c\vert^2 \sin^2(n\pi x/L)$$

$$= \frac{1}{2}\int_0^Ldx \vert c\vert^2 [1- \cos(n 2\pi x/L)] = \frac{\vert c\vert^2L}{2}$$

$$\vert c\vert^2 = \frac{2}{L} \leadsto c = \sqrt{\frac{2}{L}}e^{i\phi} \leadsto \psi_n(x) = \sqrt{\frac{2}{L}}\sin (n\pi x/L )e^{i\varphi},$$ (99)

where $\varphi \in R$ is a (real) phase factor. This normalization condition determines the wave functions $\psi_n(x)$ uniquely only up to a phase factor: if $\Psi$ is a normalized solution of the Schrödinger equation, so is $\Psi e^{i\varphi}$, i.e. the same wave function multiplied with a constant overall phase factor. Usually, we do not distinguish between such wave functions since they describe the same state of the particle, and one says that the state is only determined `up to a phase' which is irrelevant when calculating, for example, the probability density $\vert\Psi\vert^2$ or expectation values.

This is different, however, for superpositions of two different wave functions, where the relative phase difference is important and leads, for example, to interference.