# Particle in a 3-dimensional box

### 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:

• $$\dfrac{d^{2}\psi}{dx^{2}} = YZ\dfrac{d^{2}X}{dx^{2}} \Rightarrow \dfrac{1}{X}\dfrac{d^{2}X}{dx^{2}}$$
• $$\dfrac{d^{2}\psi}{dy^{2}} = XZ\dfrac{d^{2}Y}{dy^{2}} \Rightarrow \dfrac{1}{Y}\dfrac{d^{2}Y}{dy^{2}}$$
• $$\dfrac{d^{2}\psi}{dz^{2}} = XY\dfrac{d^{2}Z}{dz^{2}} \Rightarrow \dfrac{1}{Z}\dfrac{d^{2}Z}{dz^{2}}$$

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

$$E$$ 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 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.

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

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 level, add Equation $$\ref{3.9.9}$$ and Equation $$\ref{3.9.6}$$.

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