# 3.9: A Particle in a Three-Dimensional Box

- Page ID
- 13401

Learning Objectives

- To see the particle in 1-D box can easily extrapolate to boxes of higher dimensions.
- Introduction to nodal surfaces (e.g., nodal planes)

The quantum particle in the 1D box problem can be expanded to consider a particle within a higher dimensions as demonstrated elsewhere for a quantum particle in a 2D box. Here we continue the expansion into a particle trapped in a 3D box with three lengths \(L_x\), \(L_y\), and \(L_z\). As with the other systems, there is NO FORCE (i.e., no potential) acting on the particles *inside *the box (Figure \(\PageIndex{1}\)).

The potential for the particle inside the box

\[V(\vec{r}) = 0\]

- \(0 \leq x \leq L_x\)
- \(0 \leq y \leq L_y\)
- \(0 \leq z \leq L_z\)
- \(L_x < x < 0\)
- \(L_y < y < 0\)
- \(L_z < z < 0\)

\(\vec{r}\) is the vector with all three components along the three axes of the 3-D box: \(\vec{r} = L_x\hat{x} + L_y\hat{y} + L_z\hat{z}\). When the potential energy is infinite, then the wavefunction equals zero. When the potential energy is zero, then the wavefunction obeys 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.g., 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: \(xyz\):

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

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

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

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

\(E\) 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,

\[\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} \varepsilon_{x}X = 0\]

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

\(\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} \varepsilon_{x}X = 0 \label{3.9.5a}\)

\(\dfrac{d^{2}Y}{dy^{2}} + \dfrac{2m}{\hbar^{2}} \varepsilon_{y}Y = 0 \label{3.9.5b}\)

\(\dfrac{d^{2}Z}{dz^{2}} + \dfrac{2m}{\hbar^{2}} \varepsilon_{z}Z = 0 \label{3.9.5c}\)

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

\[\varepsilon_{x}+ \varepsilon_{y} + \varepsilon_{z} = E \label{3.9.6}\]

Do these equations look familiar? They should because we have now reduced the 3D box into three particle in a 1D box problems!

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

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:

\[\psi(x) =

\begin{cases}

\sqrt{\dfrac{2}{L_x}}\sin{\dfrac{n \pi x}{L_x}} & \mbox{if } 0 \leq x \leq L \\

0 & \mbox{if } {L < x < 0}

\end{cases}\]

Normalization wavefunction equation for each variable

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

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

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

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:

\[\varepsilon_{x} = \dfrac{n_{x}^{2}h^{2}}{8mL_x^{2}} \label{3.9.9}\]

with \(n_{x} = 1...\infty\)

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.

\[\psi(r) = \sqrt{\dfrac{8}{V}}\sin \left( \dfrac{n_{x} \pi x}{L_x} \right) \sin \left(\dfrac{n_{y} \pi y}{L_y}\right) \sin \left(\dfrac{ n_{z} \pi z}{L_z} \right) \label{3D wave}\]

with

\[V = \underbrace{L_x \times L_y \times L_z}_{\text{volume of box}}\]

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

\[E_{n_x,n_y,n_z} = \dfrac{h^{2}}{8m}\left(\dfrac{n_{x}^{2}}{L_x^{2}} + \dfrac{n_{y}^{2}}{L_y^{2}} + \dfrac{n_{z}^{2}}{L_z^{2}}\right) \label{3.9.10}\]

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

## Degeneracy in a 3D Cube

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 \(n_z=1\). This energy (\(E_{1,1,1}\)) is hence

\[E_{1,1,1} = \dfrac{3 h^{2}}{8mL^2}\]

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 (Equation \ref{3D wave}). The particle having a particular value of energy in the excited state *MAY *has several different stationary states or wavefunctions. 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 wavefunction has same energy:

\[E_{2,1,1} =E_{1,2,1} = E_{1,1,2} = \dfrac{6 h^{2}}{8mL^2}\]

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 \(\PageIndex{1}\).

\(n_x^2+n_y^2+n_z^2\) | Combinations of Degeneracy (\(n_x\), \(n_y\), \(n_z\)) |
Total Energy (\(E_{n_x,n_y,n_z}\)) |
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) | \(\dfrac{11 h^{2}}{8mL^2}\) | 3 | |||

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) | \(\dfrac{14 h^{2}}{8mL^2}\) | 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 |

Example \(\PageIndex{1}\): Accidental Degeneracies

When is there degeneracy in a 3-D box when none of the sides are of equal length (i.e., \(L_x \neq L_y \neq L_z\))?

**Solution**

From simple inspection of Equation \ref{3.9.10} or Table \(\PageIndex{1}\), it is clear that degeneracy originates from different combinations of \(n_x^2/L_x^2\), \(n_y^2/L_y^2\) and \(n_z^2/L_z^2\) that give the same value. These will occur at common multiples of at least two of these quantities (the Least Common Multiple is one example). For example

if

\[\dfrac{n_x^2}{L_x^2} = \dfrac{n_y^2}{L_y^2} \nonumber\]

there will be a degeneracy. Also degeneracies will exist if

\[\dfrac{n_y^2}{L_y^2} = \dfrac{n_z^2}{L_z^2} \nonumber\]

or if

\[\dfrac{n_x^2}{L_x^2} = \dfrac{n_z^2}{L_z^2} \nonumber\]

and especially if

\[\dfrac{n_x^2}{L_x^2} = \dfrac{n_y^2}{L_y^2} = \dfrac{n_z^2}{L_z^2} \nonumber.\]

There are two general kinds of degeneracies in quantum mechanics: degeneracies due to a symmetry (i.e., \(L_x=L_y\)) and accidental degeneracies like those above.

Exercise \(\PageIndex{1}\)

The 6^{th} energy level of a particle in a 3D Cube box is 6-fold degenerate.

- What is the energy of the 7th energy level?
- What is the degeneracy of the 7th energy level?

**Answer a**-
\(\dfrac{17 h^{2}}{8mL^2}\)

**Answer b**-
three-fold (i.e., there are three wavefunctions that share the same energy.

## References

- Atkins, Peter. Physical Chemistry 5th Ed. USA. 1994.
- Fitts, Donald. Principles of Quantum Mechanics. United Kingdom, Cambridge. University Press. 1999
- McQuarrie. Donald. Physical Chemistry A Molecular Approach. Sausalito, CA. University Science Books. 1997.
- Riggs. N. V. Quantum Chemistry. Toronto, Ontario. The Macmillan Company. 1969
- C. A. Hollingsworth, Accidental Degeneracies of the Particle in a Box, J. Chem. Educ., 1990, 67 (12), p 999