# 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 \(\PageIndex{1}\).

\(n_x^2+n_y^2+n_z^2\) |
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

- 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