Wednesday, March 30, 2011

Particles in Three Dimensional box

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 $V(x,y,z)$ 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 $p_x$, $p_y$ and $p_z$ 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 $x$, $y$ and $z$ directions. Because the energy is a simple sum of energies for the $x$, $y$ and $z$ 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 $n_x$, $n_y$ and $n_z$:
\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 $A$ 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 $n_x$, $n_y$ and $n_z$ 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 $f(x,y,z)$ is the complete set of points $x$, $y$, and $z$ for which $f(x,y,z) = C$, where $C$ is a chosen constant. Hence the name isosurface - the value of the function $f(x,y,z)$ 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 $C$ 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
$\displaystyle P(x\in[a,b]\;{\rm and}\;y\in[c,d]\;{\rm and}\; z\in[f,g])$ $\textstyle =$ $\displaystyle \int_a^b dx\;\int_c^d dy\;\int_f^g dz\; \vert\psi_{n_x n_y n_z}(x,y,z)\vert^2$
$\textstyle =$ $\displaystyle \left[\int_a^b \psi_{n_x}^2(x)dx\right] \left[\int_c^d \psi_{n_y}^2(y)dy\right] \left[\int_f^g \psi_{n_z}^2(z)dz\right]$
Posted by chemistry at 7:56 PM
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