Particle in the Box

The particle in the box can be represented by the following diagram:R1(22)

The goal of this section is to show that a particle with energy $E$ and mass $m$ in the box-potential V(x) defined as

$$V(x)= \begin{cases} 0, & \text{when} \hspace{.3cm} 0 \leq x \leq a, \\ \infty, & \hspace{.3cm} \text{otherwise}, \end{cases}$$

has stationary states and a discrete absorption spectrum (i.e., the particle absorbs only certain discrete values of energy called quanta). To that end, we first solve the equation $\hat{H}\tilde{\phi}(x) = E\tilde{\phi}(x)$, and then we obtain the stationary states $\psi(x,t) = \tilde{\phi}(x)$exp$(- \frac{i}{\hbar}Et)$.

Since $\tilde{\phi}(x)$ has to be continuous, single valued and square integrable (see Postulate 1), $\tilde{\phi}(0)$ and $\tilde{\phi}(a)$ must satisfy the appropriate boundary conditions both inside and outside the box. The boundary conditions inside the box lead to:

$$-\frac{\hbar^2}{2m} \frac{\partial}{\partial{x^2}} \Phi (x) = E \Phi (x), \qquad \Rightarrow \qquad \Phi (x) = A ({K \hspace{.10cm} x}).$$ (6)

Functions $\Phi(x)$ determine the stationary states inside the box. The boundary conditions outside the box are,

$$-\frac{\hbar^2}{2m} \frac{\partial}{\partial{x^2}} \Phi(x) + \infty \Phi(x) = E \Phi(x), \qquad \Rightarrow \qquad \Phi(x)=0,$$

and determine the energy associated with $\Phi(x)$ inside the box as follows. From Eq. (6), we obtain: $\frac{\hbar^2}{2m} A K^2 = E A, \hspace{.3cm}$ and, $\hspace{.3cm} \Phi(a) = A$Sin$(K\hspace{.10cm} a) = 0,$

$\qquad \Rightarrow \qquad Ka= n \pi,$   with$\hspace{.3cm} n = 1, 2, ... \qquad \Rightarrow \qquad$

Note that the number of nodes of $\Phi$ (i.e., the number of coordinates where $\Phi(x)=0)$, is equal to $n-1$ for a given energy, and the energy levels are,

$$E=\frac{\hbar^2}{2m} \frac{n^2\pi^2}{a^2},$$   with$$\hspace{.10cm} \ n=1, \ 2, \ ...$$

e.g.,

$$E(n=1)=\frac{\hbar^2}{2m} \frac{\pi^2}{a^2},$$

$$E(n=2)=\frac{\hbar^2}{2m} \frac{4\pi^2}{a^2}, \ ...$$

Conclusion: The energy of the particle in the box is quantized! (i.e., the absorption spectrum of the particle in the box is not continuous but discrete).

For learning how to use Mathematica to solve the time independent Schrodinger equation for the particle in the box click here.
 

Exercise 6: (i) Using the particle in the box model for an electron in a quantum dot (e.g., a nanometer size silicon material) explain why larger dots emit in the red end of the spectrum, and smaller dots emit blue or ultraviolet as visualized in the following photograph,

 

(ii) Consider the molecule hexatriene $CH_2=CH-CH=CH-CH=CH_2$ and assume that the 6 $\pi$ electrons move freely along the molecule. Approximate the energy levels using the particle in the box model. The length of the box is the sum of bond lengths with C-C = 1.54 Å, C=C = 1.35 Å, and an extra 1.54 Å, due to the ends of the molecule. Assume that only 2 electrons can occupy each electronic state and compute:

(A) The energy of the highest occupied energy level.

(B) The energy of the lowest unoccupied energy level.

(C) The energy difference between the highest and the lowest energy levels, and compare such energy difference with the energy of the peak in the absorption spectrum at $\lambda_{MAX}$=268nm.

(D) Predict whether the peak of the absorption spectrum for $CH_2=CH-(CH=CH)_n-CH=CH_2$ would be red- or blue-shifted relative to the absorption spectrum of hexatriene.