# Discussion: 3D Quantum Particle-in-a-Box

Here we will solve the three dimensional particle in a box. To describe the system, we imagine a box with zero potential enclosed in dimensions $$0<x<a$$, $$0<y<b$$, $$0<z<c$$. Outside the box is the region where the particle’s wavefunction does not exist. Hence, we let the potential outside the box be infinite (which means impossible to exist in physics).

$\begin{cases} 0\quad \quad \text{inside the box} \\ \infty \quad \text{outside the box} \end{cases}$

Boundary Conditions $\psi(0,y,z)=\psi(a,y,z)=0$ $\psi(x,0,z)=\psi(x,b,z)=0$ $\psi(x,y,c)=\psi(x,y,c)=0$

As always, wavefunctions are found by solving the Schrödinger equation.

$E \Psi (x,y,z) = -\frac{\hbar}{2m}\left[\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}\right]\Psi (x,y,z) + V\Psi (x,y,z)$

We use the separation of variables to solve this partial differential equation (PDE). Consider the ansatz: $\Psi(x,y,z) = X(x)Y(y)Z(z)$

We also partition the total energy into its rightful components: $E={E}_{x} + {E}_{y} + {E}_{z}.$

Now we decompose the PDE into three ordinary differential equations (ODE):

$\frac{{d}^{2}X}{d{x}^{2}} + {k}_{x}^{2}X=0$

where $${k}_{x}=\sqrt{\frac{2m{E}_{x}}{{\hbar}^{2}}} =\frac{n \pi}{a}$$ is the wavenumber, where $$n$$ is the mode.

For the $$y$$ and $$z$$ variables, we replace the above equation with $$Y$$ and $$Z$$. Likewise,

${k}_{y}=\sqrt{\frac{2m{E}_{y}}{{\hbar}^{2}}} =\frac{n \pi}{b}$ ${k}_{z}=\sqrt{\frac{2m{E}_{z}}{{\hbar}^{2}}} =\frac{n \pi}{c}.$

Solving the individual ODEs we find that

$X(x) = {A}_{x}sin({k}_{x}x)+ {B}_{x}cos({k}_{x}x)$ $Y(y) = {A}_{y}sin({k}_{y}y)+ {B}_{y}cos({k}_{y}y)$ $Z(z) = {A}_{z}sin({k}_{z}z)+ {B}_{z}cos({k}_{z}z).$

By the boundary conditions:

$X(0)=X(a)=0$ $Y(0)=Y(b)=0$ $Z(0)=Z(c)=0$

hence

${B}_{x}=0$ ${B}_{y}=0$ ${B}_{z}=0.$

We normalize the wavefunctions by performing the integral

$\int _{ 0 }^{ a }{ {\left| { { A }_{ x } } \right|} ^{2} } { sin }^{ 2 }\left( { k }_{ x }x \right) dx=1$

and applying the boundary condition $$X(0)=X(a)=0$$.

Thus $${A}_{x}=\sqrt{\frac{2}{a}}$$.

Combine the solutions of all components to the ansatz yields

$\Psi (x,y,z) = \sqrt {\frac{8}{abc}}sin\left(\frac{n \pi}{a}x\right)sin\left(\frac{n \pi}{b}y\right)sin\left(\frac{n \pi}{c}z\right).$

Visit my set Lectures on Quantum Mechanics for more notes.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
3 years, 9 months ago

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thanksssssssssssssssssssssssssssssssssssssssss

- 3 years, 9 months ago

You're welcome!

- 3 years, 9 months ago

ohhh. that makes sense. lol

- 3 years, 7 months ago

wow..wow....quantum mechanics..hummm...at first i thought it would be some KTOG problem....really inspired....nice note...please keep on posting

- 3 years, 8 months ago

What's a KTOG problem?

- 3 years, 8 months ago

Kinetic Theory Of Gases......

- 3 years, 8 months ago

Elegant way to bring back those good ol' equations. Nice!

- 3 years, 9 months ago

Thank you!

- 3 years, 9 months ago