Particle in a box

Our first application of the de Broglie hypothesis dealt with probability patterns that correspond to waves interfering and diffracting as they traveling through various obstacles. We now consider an application which corresponds to the first phenomenon which we have studied, normal modes. From the general wave behavior of normal modes we will learn several important general lessons about the behavior of elementary particles.

As a first example, we consider a ``particle in a (one-dimensional) box'' as in Figure 7. Here, a particle of mass $m$ is constrained to move along a frictionless rod oriented along the $x$ axis. A very strong repulsive force is applied at $x=0$ so that the probability ${\mathcal P}(x)$ of finding the particle at points $x<0$ becomes extremely small, ${\mathcal P}(x<0) \rightarrow 0$. The ``box'' is completed by applying a similar repulsive force at $x=L$ so that the probability of finding the particle at locations $x>L$ is also extremely small, ${\mathcal P}(x>L) \rightarrow 0$. In general in quantum mechanics, we cannot make these probabilities go exactly to zero, but we can consider the idealization that the forces are sufficiently strong to make the probabilities as close to zero as we like.

Figure 7: Realization of particle of mass $m$ in a one-dimensional ``box''

Just as there are many possible answers for the intensity of waves moving along a string depending upon how we generate the waves, there are many possible answers for ${\mathcal P}(0<x<L)$, the probability of finding the particle at any of the allowed points between $x=0$ and $x=L$, depending upon how exactly we place the particle into the box, the so called ``state'' of the particle. For the string, although many states of motion are possible, there are a few which are very natural and easy to generate, the normal modes. Similarly, it is quite common to find particles states corresponding to normal modes. Thus, we often find ${\mathcal P}(x) = \vert \underline{A}(x) \vert^2$, where $\underline{A}(x)$ is the complex amplitude for a normal mode of classical wave motion. In quantum mechanics, the complex amplitude has a special name, the wave function and is usually written with the Greek letter `psi' so that $\underline{A}(x) \equiv \Psi(x)$. This ``wave function'' is nothing very mysterious, just the same complex amplitude which we have used throughout the course. The standard way of writing out the probability for finding an electron is thus,

$${\mathcal P}(x) = \vert \Psi(x) \vert ^2.$$ (11)

For the present situation, the conditions ${\mathcal P}(x=0) \rightarrow 0$, ${\mathcal P}(x=L) \rightarrow 0$ mean that the wave function (complex amplitude) $\Psi(x)$ must go to zero at the points $x=0$ and $x=L$, corresponding to fixed boundary conditions. We already found the normal mode solutions for such boundary conditions to be $y(x,t)=A_0 \sin(k_n x) \cos(\omega t)$, where every point under goes simple harmonic motion in phase at the same frequency, and

$$k_n=\frac{\pi}{L} n \mbox{\ \ \ ; $n=1, 2, 3, {\ldots}$}$$ (12)

Rewriting this in the complex representation, we have $y(x,t) = \mbox{Re\,}\left(A_0 \sin(k_n x) e^{-i \omega t}\right)$ so that the complex amplitude for the n$^{th}$ mode is $\underline{A}(x) = A_0 \sin(k_n x)$. Correspondingly, we expect the probability of finding a particle in a box in one of these ``normal mode'' states to be given by Eq. (11) with

$$\Psi_n(x) = A_0 \sin(k_n x),$$ (13)

where Eq. (12) determines $k_n$.

Having found for each normal mode $n$ the probability of finding the particle at different points $x$, we next ask for the momentum of the particle. The de Broglie relation gives the magnitude as $p=\hbar k_n$, but momentum is a vector the de Broglie relation says nothing about the direction or sign of $p$. To address this issue, we note that any standing wave can be written as a sum of a forward and a backward traveling wave. Mathematically, we can rewrite the solution Eq. (13) using Euler's formula to find

$$\Psi_n(x) = A_0 \frac{e^{+i k_n x}-e^{-ik_n x}}{2i},$$ (14)

so that we see that the particle ``state'' consists of both a forward $+i k_n x$ and a backward $-i k_n x$ moving part, indicating that it is possible to find either possibility, $p=\pm \hbar k_n$. As probabilities in quantum mechanics are always given by intensities, to determine the probability for each of these two alternatives, we look at the factors multiplying the two component waves and take their square magnitude,

$${\mathcal P}_\pm \propto \left\vert \pm \frac{A_0}{2i} \right\vert^2 = \frac{\vert A_0\vert^2}{4}.$$ (15)

The two possibilities have equal intensities and thus are equally likely. The changes of finding $p=\pm \hbar k_n$ are 50%:50%.

Finally, we consider the energy of the particle in state $n$. This is tricky because we do not know where we will find the particle or what its momentum will be, and so we must consider all of the possibilities. The total energy is the sum of the potential and kinetic energies. Because the particle is always found in the allowed region $0<x<L$ where there is no force acting, the potential energy is always $U=0$. The kinetic energy is given by $KE=\frac{p^2}{2m}$ and the momentum will be either $\pm \hbar k_n$ so that the kinetic energy will also always have a single value, $KE=\frac{\hbar^2 k_n^2}{2m}$. Thus, although the position and momentum are uncertain, we always find a single, well defined value for the energy of

$$E_n=KE+U=\frac{(\pm \hbar k_n)^2}{2m}+0=\frac{\hbar^2 k_n^2}{2m}=\frac{\hbar^2 \pi^2}{2mL^2} n^2$$

when the particle is in state $n$.

Having a single, well-defined value for the energy turns out to be a general feature of particles in normal mode quantum states, and so these states are typically called ``energy states''. Ultimately, this feature results from the frequency-energy part of the de Broglie hypothesis. It arises from the fundamental property of normal modes that every point undergoes harmonic motion at the same frequency frequency, which the frequency-energy relation translates into a single, well-defined value for the energy.