Particle in a Box

Figure 9.2: First three modes for wave function of a particle in a box.

We now imagine how a particle confined to a region $0 \le x \le a$ on the $x$ axis must behave. As with the displacement of a guitar string, the wave function must be zero at $x = 0$ and $a$, i. e., at the ends of the guitar string. A single complex exponential plane wave cannot satisfy this condition, since $\vert \exp [i(kx - \omega t)]\vert^2 = 1$ everywhere. However, a superposition of leftward and rightward traveling waves creates a standing wave, as illustrated in figure 9.2:

$$\psi = \exp [i(kx - \omega t)] - \exp [i(-kx - \omega t)] = 2i \exp (-i \omega t) \sin (kx) .$$ (10.16)

Notice that the time dependence is still a complex exponential, which means that $\vert \psi \vert^2$ is independent of time. This insures that the probability of finding the particle somewhere in the box remains constant with time.

Because we took a difference rather than a sum of plane waves, the condition $\psi = 0$ is already satisfied at $x = 0$. To satisfy it at $x = a$, we must have $ka = n \pi$, where $n = 1,2,3, \ldots$. Thus, the absolute value of the wavenumber must take on the discrete values

$$k_n = \frac{n \pi}{a} , ~~~~~ n = 1,2,3, \ldots .$$ (10.17)

(The wavenumbers of the two plane waves take on plus and minus this absolute value.) This implies that the absolute value of the particle momentum is $\Pi_n = \hbar k_n = n \pi \hbar / a$, which in turn means that the energy of the particle must be

$$E_n = ( \Pi_n^2 c^2 + m^2 c^4 )^{1/2} = (n^2 \pi^2 \hbar^2 c^2 / a^2 + m^2 c^4 )^{1/2} ,$$ (10.18)

where $m$ is the particle mass. In the non-relativistic limit this becomes

$$E_n = \frac{\Pi_n^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2ma^2} ~~~ (\mbox{non-relativistic})$$ (10.19)

where we have dropped the rest energy $mc^2$ since it is just a constant offset. In the ultra-relativistic case where we can ignore the particle mass, we find

$$E_n = \vert \Pi_n \vert c = \frac{n \pi \hbar c}{a} ~~~ \mbox{(zero mass)} .$$ (10.20)

In both limits the energy takes on only a certain set of possible values. This is called energy quantization and the integer $n$ is called the energy quantum number. In the non-relativistic limit the energy is proportional to $n^2$, while in the ultra-relativistic case the energy is proportional to $n$.

Figure: Allowed energy levels for the non-relativistic particle in a box. The constant $E_0 = \pi^2 \hbar^2 /(2ma^2 )$. See text for the meanings of symbols.

We can graphically represent the allowed energy levels for the particle in a box by an energy level diagram. Such a diagram is shown in figure 9.3 for the non-relativistic case.

One aspect of this problem deserves a closer look. Equation (9.16) shows that the wave function for this problem is a superposition of two plane waves corresponding to momenta $\Pi_1 = + \hbar k$ and $\Pi_2 = - \hbar k$ and is therefore a kind of wave packet. Thus, the wave function is not invariant under displacement and does not correspond to a definite value of the momentum -- the momentum's absolute value is definite, but its sign is not. Following Feynman's prescription, equation (9.16) tells us that the amplitude for the particle in the box to have momentum $+ \hbar k$ is $\exp [i(kx - \omega t)]$, while the amplitude for it to have momentum $- \hbar k$ is $- \exp [i(-kx - \omega t)]$. The absolute square of the sum of these amplitudes gives us the relative probability of finding the particle at position $x$:

$$P(x) = \vert 2i \exp (-i \omega t) \sin (kx) \vert^2 = 4 \sin^2 (kx) .$$ (10.21)

Which of the two possible values of the momentum the particle takes on is unknowable, just as it is impossible in principle to know which slit a particle passes through in two slit interference. If an experiment is done to measure the momentum, then the wave function is irreversibly changed, just as the interference pattern in the two slit problem is destroyed if the slit through which the particle passes is unambiguously determined.