7B: Particles in Three-Dimensional Boxes
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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 wavefunctions in three dimensions. The potential energy \(V(x,y,z)\) for the cubic box is defined to be 0 if \(x\epsilon [0,L]\), \(y\epsilon [0,L]\) and \(z\epsilon [0,L]\) and infinite otherwise. This means that the wavefunction \(\psi (x,y,z)\) 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
\[\dfrac{p_{x}^{2}}{2m}+\dfrac{p_{y}^{2}}{2m}+\dfrac{p_{z}^{2}}{2m}=E\]
where \(p_x\), \(p_y\) and \(p_z\) are the three components of the particle's momentum vector \(p\). Thus, we can write the energy as
\[E=\varepsilon_x +\varepsilon_y +\varepsilon_z\]
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 wavefunction will be a product of wavefunction forms for the one-dimensional box, and in order to satisfy the first three of the boundary conditions, we can take the \(\sin\) functions:
\[\psi (x,y,z)=A\sin\left ( \sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}x \right ) \sin \left ( \sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}y \right ) \sin \left ( \sqrt{\dfrac{2m\varepsilon_z}{\hbar^2}}z \right )\]
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 \(n_z\):
\[\varepsilon_{n_x}=\dfrac{\hbar^2 \pi^2}{2mL^2}n_{x}^{2}\;\;\;\; \varepsilon_{n_y}=\dfrac{\hbar^2 \pi^2}{2mL^2}n_{y}^{2}\;\;\;\; \varepsilon_{n_z}=\dfrac{\hbar^2 \pi^2}{2mL^2}n_{z}^{2}\]
so that the allowed values of the total energy are
\[E_{n_x n_y n_z}=\dfrac{\hbar^2 \pi^2}{2mL^2}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})\]
and the wavefunctions become
\[\psi_{n_x n_y n_z}(x,y,z)=A\sin\left ( \dfrac{n_x \pi x}{L} \right ) \sin \left ( \dfrac{n_y \pi y}{L} \right ) \sin \left ( \dfrac{n_z \pi z}{L} \right )\]
The constant \(A\) is determined from the normalization condition
\[\begin{align*}\int_{0}^{L}\int_{0}^{L}\int_{0}^{L}|\psi_{n_x n_y n_z}(x,y,z)|^2 dx\,dy\,dz &= 1\\ A^2 \int_{0}^{L}\sin^2 \left ( \dfrac{n_x \pi x}{L} \right ) dx\int_{0}^{L}\sin^2\left ( \dfrac{n_y \pi y}{L} \right ) dy\int_{0}^{L}\sin^2 \left ( \dfrac{n_z \pi z}{L} \right )dz &= 1\\ A^2\dfrac{L}{2}\cdot \dfrac{L}{2}\cdot \dfrac{L}{2} &= 1\\ A &= \left ( \dfrac{2}{L} \right )^{3/2}\end{align*}\]
Thus, the wavefunctions are
\[\psi_{n_x n_y n_z}(x,y,z)=\left ( \dfrac{2}{L} \right )^{3/2}\sin\left ( \dfrac{n_x \pi x}{L} \right ) \sin\left ( \dfrac{n_y \pi y}{L} \right ) \sin \left ( \dfrac{n_z \pi z}{L} \right )\]
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
\[E_{111}=\dfrac{3(\bar{h})^2 \pi^2}{2mL^2}\]
Visualizing the wavefunctions 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 wavefunctions, where the sign can be positive or negative, it is useful to base the value of \(C\) not on the wavefunction value but rather on the probability density \(p_{n_x n_y n_z}(x,y,z)=|\psi_{n_x n_y n_z}(x,y,z)|^2\). The figure below shows two isosurfaces of the wavefunction \(\psi_{123}(x,y,z)\). The first occurs at a probability density value of \(0.64\) and the other occurs at \(0.04\). These value then imply that the wavefunction 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, \(\psi_{123}(x,y,z)=\pm 0.2\).
![\includegraphics[scale=0.7]{threeD_box.eps}](http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_7/img119.png) |
As in the two-dimensional case, the fact that the wavefunction \(\psi_{n_x n_y n_z}(x,y,z)\) is a product
\[\psi_{n_x n_y n_z}(x,y,z)=\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z)\]
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 \(x\epsilon [a,b]\), \(y\epsilon [c,d]\) and \(z\epsilon [f,g]\) is
\[\begin{align*}P(x\epsilon [a,b] \ and \ y\epsilon [c,d] \ and \ z\epsilon [f,g]) &= \int_{a}^{b}dx \int_{c}^{d} dy \int_{f}^{g} dz|\psi_{n_x n_y n_z}(x,y,z)|^2\\ &= \left [ \int_{a}^{b}\psi_{n_x}^{2}(x)dx \right ] \left [ \int_{c}^{d}\psi_{n_y}^{2}(y)dy \right ]\left [ \int_{f}^{g}\psi_{n_z}^{2}(z)dz \right ]\end{align*}\]