Diagram By Felix Kling

In the above diagram, the oscillating curve on the left of the potential barrier is a standing wave pattern that results from the interference between

the incident matter wave and
the reflected matter wave that has a smaller amplitude than the incident matter wave.

These two waves, of the same type, travelling in opposite directions interfere with each other thus setting up a standing wave pattern.

Within the potential barrier, the wave function decreases exponentially.

To the right of the potential barrier, the wave of lower amplitude but of the same energy as the incident matter wave emerges, indicating a small probability of detecting an electron in this region.

For a particle that tunnels through a potential barrier, a transmission coefficient T can be defined such that

T = e^-2kd

, where
d = barrier thickness/width,

, in which

m = mass of particle
U = potential energy of the potential barrier
E = initial mechanical energy of the electron
h = Planck's constant.

The transmission coefficient T gives the probability which an approaching electron will be transmitted through the barrier (probability that tunnelling will occur).

Since T is an exponential function, its value is very much dependent on variables such as

the mass m of the particle
the barrier thickness
the energy difference (U-E)

Since part of the incident matter wave will be transmitted and the other part reflected, a reflection coefficient R, which gives the probability which an approaching electron will be reflected, can also be defined such that

P(matter wave is reflected) R = 1 - P(matter wave is transmitted) T

Such that R + T = 1

Particle in a three-dimensional box

Generalization of the results for a two-dimensional square box to a three-dimensional cubic box is straightforward. Since we live in a three-dimensional world, this generalization is an important one, and we need to be able to think about energy levels and wave functions in three dimensions. The potential energy for the cubic box is defined to be 0 if $x\in[0,L]$ , $y\in[0,L]$ and $z\in[0,L]$ and infinite otherwise. This means that the wave function $\psi(x,y,z)$ must satisfy six boundary conditions $\psi(0,y,z) = 0$ , $\psi(x,0,z) = 0$ , $\psi(x,y,0) = 0$ , $\psi(L,y,z) = 0$ , $\psi(x,L,z) = 0$ and $\psi(x,y,L) = 0$ .

We first note that the classical energy is the sum of three terms

$\begin{displaymath} {p_x^2 \over 2m} + {p_y^2 \over 2m} + {p_z^2 \over 2m} = E \end{displaymath}$

where

and

are the three components of the particle's momentum vector ${\bf p}$ . Thus, we can write the energy as

$\begin{displaymath} E = \varepsilon_x + \varepsilon_y + \varepsilon_z \end{displaymath}$

corresponding to the kinetic energy in the

and

directions. Because the energy is a simple sum of energies for the

and

directions, the wave function will be a product of wave function forms for the one-dimensional box, and in order to satisfy the first three of the boundary conditions, we can take the $\sin$ functions:

$\begin{displaymath} \psi(x,y,z) = A\sin\left(\sqrt{2m\varepsilon_x \over \hbar^2... ...\right) \sin\left(\sqrt{2m\varepsilon_z \over \hbar^2}z\right) \end{displaymath}$

As in the two-dimensional case, applying second three boundary conditions yields the allowed values of $\varepsilon_x$ , $\varepsilon_y$ and $\varepsilon_z$ , which now require three integers

and

$\begin{displaymath} \varepsilon_{n_x} = {\hbar^2 \pi^2 \over 2mL^2}n_x^2 \;\;\;\... ...;\;\;\;\; \varepsilon_{n_z} = {\hbar^2 \pi^2 \over 2mL^2}n_z^2 \end{displaymath}$

so that the allowed values of the total energy are

$\begin{displaymath} E_{n_xn_yn_z} = {\hbar^2 \pi^2 \over 2mL^2} \left(n_x^2 + n_y^2 + n_z^2\right) \end{displaymath}$

and the wave functions become

$\begin{displaymath} \psi_{n_xn_yn_z}(x,y,z) = A\sin\left({n_x\pi x \over L}\righ... ...({n_y\pi y \over L}\right) \sin\left({n_z\pi z \over L}\right) \end{displaymath}$

The constant

is determined from the normalization condition

$\displaystyle \int_0^L \int_0^L \int_0^L\vert\psi_{n_xn_yn_z}(x,y,z)\vert^2 dx dy dz$	$\textstyle =$	$\displaystyle 1$

$\displaystyle A^2 \int_0^L \sin^2\left({n_x\pi x \over L}\right)dx \int_0^L \si... ...ft({n_y\pi y \over L}\right)dy \int_0^L \sin^2\left({n_z\pi z \over L}\right)dz$	$\textstyle =$	$\displaystyle 1$

$\displaystyle A^2 {L \over 2}\cdot{L \over 2}\cdot{L \over 2}$	$\textstyle =$	$\displaystyle 1$

$\displaystyle A$	$\textstyle =$	$\displaystyle \left({2 \over L}\right)^{3/2}$

Thus, the wave functions are

$\begin{displaymath} \psi_{n_xn_yn_z}(x,y,z) = \left({2 \over L}\right)^{3/2} \si... ...({n_y\pi y \over L}\right) \sin\left({n_z\pi z \over L}\right) \end{displaymath}$

As with the two-dimensional box, the three integers

and

are restricted to the natural numbers 1,2,3,.... Thus, the lowest energy or ground-state energy is

$\begin{displaymath} E_{111} = {3\hbar^2 \pi^2 \over 2mL^2} \end{displaymath}$

Visualizing the wave functions is tricky because of their high dimensionality. The most common method of visualizing functions of three variables is the use of an isosurface. An isosurface of a function is the complete set of points , , and for which , where is a chosen constant. Hence the name isosurface - the value of the function is the same at all points on the surface.

For wave functions, where the sign can be positive or negative, it is useful to base the value of not on the wave function value but rather on the probability density $p_{n_xn_yn_z}(x,y,z) = \vert\psi_{n_xn_yn_z}(x,y,z)\vert^2$ . The figure below shows two isosurfaces of the wave function $\psi _{123}(x,y,z)$ . The first occurs at a probability density value of 0.64 and the other occurs at 0.04. These value then imply that the wave function can have a fixed positive or negative value along the surface. In one case, $\psi_{123}(x,y,z) = \pm 0.8$ , while in the other, $\psi_{123}(x,y,z) = \pm 0.2$ .

**Figure:** Isosurfaces of $\psi _{123}(x,y,z)$ at probability density values of 0.64 and 0.04. Red is positive and blue is negative.
$\includegraphics[scale=0.7]{threeD_box.eps}$

As in the two-dimensional case, the fact that the wave function $\psi_{n_x n_y n_z}(x,y,z)$ is a product

$\begin{displaymath} \psi_{n_x n_y n_z}(x,y,z) = \psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z) \end{displaymath}$

means that the probabilities can be calculated from products of one-dimensional integrals. The probability that a measurement of a particle's position yields a value $x\in[a,b]$ , $y\in[c,d]$ and $z\in[f,g]$ is

$\displaystyle P(x\in[a,b]\;{\rm and}\;y\in[c,d]\;{\rm and}\; z\in[f,g])$	$\textstyle =$	$\displaystyle \int_a^b dx\;\int_c^d dy\;\int_f^g dz\; \vert\psi_{n_x n_y n_z}(x,y,z)\vert^2$

	$\textstyle =$	$\displaystyle \left[\int_a^b \psi_{n_x}^2(x)dx\right] \left[\int_c^d \psi_{n_y}^2(y)dy\right] \left[\int_f^g \psi_{n_z}^2(z)dz\right]$

Master of Science in Chemistry

Pages

Wednesday, March 30, 2011

Transmission coefficient of a particle from a potential barrier

Particles in Three Dimensional box

Particle in a three-dimensional box