# 3.9: A Particle in a Three-Dimensional Box

Skills to Develop

• To see the particle in 1-D box can easily extrapolate to boxes of higher dimensions.

The 1D particle in the box problem can be expanded to consider a particle within a 3D box for three lengths $$a$$, $$b$$, and $$c$$. When there is NO FORCE (i.e., no potential) acting on the particles inside the box.

Figure $$\PageIndex{1}$$: A particle in a 3-D box scheme with equal lengths on all three dimensions. Image used with permission (CC BY-SA-ND 3.0; www.a-levelphysicstutor.com)

The potential for the particle inside the box

$V(\vec{r}) = 0$

• $$0 \leq x \leq a$$
• $$0 \leq y \leq b$$
• $$0 \leq z \leq c$$
• $$a < x < 0$$
• $$b < y < 0$$
• $$c < z < 0$$

$$\vec{r}$$ is the vector with all three components along the three axes of the 3-D box: $$\vec{r} = a\hat{x} + b\hat{y} + c\hat{z}$$. When the potential energy is infinite, then the wave function equals zero. When the potential energy is zero, then the wave function obeys the Schrödinger equation.

### The Time-Independent Schrodinger Equation

$-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi(r) + V(r)\psi(r) = E\psi(r) \label{3.9.1}$

Since we are dealing with a 3-dimensional figure, we need to add the 3 different axes into the Schrondinger equation:

$-\dfrac{\hbar^{2}}{2m}\left(\dfrac{d^{2}\psi(r)}{dx^{2}} + \dfrac{d^{2}\psi(r)}{dy^{2}} + \dfrac{d^{2}\psi(r)}{dz^{2}}\right) = E\psi(r) \label{3.9.2}$

The easiest way in solving this partial differential equation is by having the wavefunction equal to a product of individual function for each independent variable (e.., the Separation of Variables technique):

$\Psi{(x,y,z)} = X(x)Y(y)Z(z) \label{3.9.3}$

Now each function has its own variable:

• $$X(x)$$ is a function for variable $$x$$ only
• $$Y(y)$$ function of variable $$y$$ only
• $$Z(z)$$ function of variable $$z$$ only

Now substitute Equation $$\ref{3.9.3}$$ into Equation $$\ref{3.9.2}$$ and divide it by the product, x, y, z:

$\dfrac{d^{2}\psi}{dx^{2}} = YZ\dfrac{d^{2}X}{dx^{2}} \Rightarrow \dfrac{1}{X}\dfrac{d^{2}X}{dx^{2}}$

$\dfrac{d^{2}\psi}{dy^{2}} = XZ\dfrac{d^{2}Y}{dy^{2}} \Rightarrow \dfrac{1}{Y}\dfrac{d^{2}Y}{dy^{2}}$

$\dfrac{d^{2}\psi}{dz^{2}} = XY\dfrac{d^{2}Z}{dz^{2}} \Rightarrow \dfrac{1}{Z}\dfrac{d^{2}Z}{dz^{2}}$

$\left(-\dfrac{\hbar^{2}}{2mX} \dfrac{d^{2}X}{dx^{2}}\right) + \left(-\dfrac{\hbar^{2}}{2mY} \dfrac{d^{2}Y}{dy^{2}}\right) + \left(-\dfrac{\hbar^{2}}{2mZ} \dfrac{d^{2}Z}{dz^{2}}\right) = E \label{3.9.4}$

$$E$$ is an energy constant, and is the sum of x, y, and z. For this to work, each term must equal its own constant. For example,

$\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} E_{x}X = 0$

Now separate each term in Equation $$\ref{3.9.4}$$ to equal zero:

$$\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} E_{x}X = 0 \label{3.9.5a}$$

$$\dfrac{d^{2}Y}{dy^{2}} + \dfrac{2m}{\hbar^{2}} E_{y}Y = 0 \label{3.9.5b}$$

$$\dfrac{d^{2}Z}{dz^{2}} + \dfrac{2m}{\hbar^{2}} E_{z}Z = 0 \label{3.9.5c}$$

Now we can add all the energies together to get the total energy:

$E_{x}+ E_{y} + E_{z} = E \label{3.9.6}$

Do these equations look familiar? They should because we have now reduced the 3D box into a particle in a box problem!

$\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} E_{x}X = 0 \approx \dfrac{d^{2}\psi}{dx^{2}} = -\dfrac{4\pi^{2}}{\lambda^{2}}\psi \label{3.9.7}$

Now the equations are very similar to a 1-D box and the boundary conditions are identical, i.e.,

$n = 1, 2,..\infty$

Use the normalization wavefunction equation for each variable:

Normalization wavefunction Equation

$\psi(x) = \sqrt{2/L}\sin{{n \pi x}/L}$

$Limit: 0 \leq x \leq L$

$\psi(x) = 0$

$Limit: L < x < 0$

Normalization wavefunction equation for each variable

$X(x) = \sqrt{\dfrac{2}{a}} \sin \left( \dfrac{n_{x}\pi x}{a} \right) \label{3.9.8a}$

$Y(y) = \sqrt{\dfrac{2}{b}} \sin \left(\dfrac{n_{y}\pi y}{b} \right) \label{3.9.8b}$

$Z(z) = \sqrt{\dfrac{2}{c}} \sin \left( \dfrac{n_{z}\pi z}{c} \right) \label{3.9.8c}$

The limits of the three quantum numbers

• $$n_{x} = 1, 2, 3 ...\infty$$
• $$n_{y} = 1, 2, 3 ...\infty$$
• $$n_{z} = 1, 2, 3 ...\infty$$

For each constant use the de Broglie Energy equation:

$E_{x} = \dfrac{n_{x}^{2}h^{2}}{8ma^{2}} \label{3.9.9}$

$$n_{x} = 1...\infty$$

Do the same for variables $$n_y$$ and $$n_z$$. Combine Equation $$\ref{3.9.3}$$ with Equations $$\ref{3.9.8a}$$-$$\ref{3.9.8c}$$ to find the wavefunctions inside a 3D box.

$\psi(r) = \sqrt{\dfrac{8}{V}}\sin \left( \dfrac{n_{x} \pi x}{a} \right) \sin \left(\dfrac{n_{y} \pi y}{b}\right) \sin \left(\dfrac{ n_{z} \pi z}{c} \right)$

$V = a \times b \times c = volume \; of \; box$

To find the Total Energy, add Equation $$\ref{3.9.9}$$ and Equation $$\ref{3.9.6}$$.

$E_{n_x,n_y,x_z} = \dfrac{h^{2}}{8m}\left(\dfrac{n_{x}^{2}}{a^{2}} + \dfrac{n_{y}^{2}}{b^{2}} + \dfrac{n_{z}^{2}}{c^{2}}\right) \label{3.9.10}$

Notice the similarity between the energies a particle in a 3D box (Equation $$\ref{3.9.10}$$) and a 1D box.

### Degeneracy

The energy of the particle in a 3-D cube (i.e., $$a=L$$, $$b=L$$, and $$c= L$$) in the ground state is given by Equation $$\ref{3.9.10}$$ with $$n_x=1$$, $$n_y=1$$, and $$n_y=1$$. This energy ($$E_{111}$$) is hence

$E_{111} = \dfrac{3 h^{2}}{8mL^2}$

The ground state has only one wavefunction and no other state has this specific energy; the ground state and the energy level are said to be non-degenerate. However, in the 3-D cubical box potential the energy of a state depends upon the sum of the squares of the quantum numbers. The particle having a particular value of energy in the excited state MAY has several different stationary states or wave functions. If so, these states and energy eigenvalues are said to be degenerate.

For the first excited state, three combinations of the quantum numbers $$(n_x,\, n_y, \, n_z )$$ are $$(2,\,1,\,1),\, (1,2,1),\, (1,1,2)$$. The sum of squares of the quantum numbers in each combination is same (equal to 6). Each waveunction has same energy:

$E_{211} =E_{121} = E_{12} = \dfrac{6 h^{2}}{8mL^2}$

Corresponding to these combinations three different wavefunctions and three different states are possible. Hence, the first excited state is said to be three-fold or triply degenerate. The number of independent wavefunctions for the stationary states of an energy level is called as the degree of degeneracy of the energy level. The value of energy levels with the corresponding combinations and sum of squares of the quantum numbers

$n^2 \,= \, n_x^2+n_y^2+n_z^2$

as well as the degree of degeneracy are depicted in Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Degeneracy properties of the particle in a 3-D cube with $$a=L$$, $$b=L$$, and $$c= L$$.
$$n_x^2+n_y^2+n_z^2$$ Combinations of Degeneracy ($$n_x$$, $$n_y$$, $$n_z$$) Total Energy ($$E_{n_x,n_y,x_z}$$) Degree of Degeneracy
3 (1,1,1)           $$\dfrac{3 h^{2}}{8mL^2}$$ 1
6 (2,1,1) (1,2,1) (1,1,2)       $$\dfrac{6 h^{2}}{8mL^2}$$ 3
9 (2,2,1) (1,2,2) (2,1,2)       $$\dfrac{9 h^{2}}{8mL^2}$$ 3
11 (3,1,1) (1,3,1) (1,1,3)       $$\dfrac{11 h^{2}}{8mL^2}$$ 1
12 (2,2,2)           $$\dfrac{12 h^{2}}{8mL^2}$$ 1
14 (3,2,1) (3,1,2) (2,3,1) (2,1,3) (1,3,2) (1,2,3) $$\dfrac{13 h^{2}}{8mL^2}$$ 6
17 (2,2,3) (3,2,2) (2,3,2)       $$\dfrac{17 h^{2}}{8mL^2}$$ 3
18 (1,1,4) (1,4,1) (4,1,1)       $$\dfrac{18 h^{2}}{8mL^2}$$ 3
19 (1,3,3) (3,1,3) (3,3,1)       $$\dfrac{19 h^{2}}{8mL^2}$$ 3
21 (1,2,4) (1,4,2) (2,1,4) (2,4,1) (4,1,2) (4,2,1) $$\dfrac{21 h^{2}}{8mL^2}$$ 6

### References

1. Atkins, Peter. Physical Chemistry 5th Ed. USA. 1994.
2. Fitts, Donald. Principles of Quantum Mechanics. United Kingdom, Cambridge. University Press. 1999
3. McQuarrie. Donald. Physical Chemistry A Molecular Approach. Sausalito, CA. University Science Books. 1997.
4. Riggs. N. V. Quantum Chemistry. Toronto, Ontario. The Macmillan Company. 1969