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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestthanksssssssssssssssssssssssssssssssssssssssss

Log in to reply

You're welcome!

Log in to reply

ohhh. that makes sense. lol

Log in to reply

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

Log in to reply

What's a KTOG problem?

Log in to reply

Kinetic Theory Of Gases......

Log in to reply

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

Log in to reply

Thank you!

Log in to reply