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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
2 years, 5 months ago

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thanksssssssssssssssssssssssssssssssssssssssss Himanshu Tuteja · 2 years, 5 months ago

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@Himanshu Tuteja You're welcome! Steven Zheng · 2 years, 5 months ago

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ohhh. that makes sense. lol Greg Nye · 2 years, 4 months ago

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wow..wow....quantum mechanics..hummm...at first i thought it would be some KTOG problem....really inspired....nice note...please keep on posting Manish Bhargao · 2 years, 4 months ago

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@Manish Bhargao What's a KTOG problem? Steven Zheng · 2 years, 4 months ago

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@Steven Zheng Kinetic Theory Of Gases...... Manish Bhargao · 2 years, 4 months ago

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Elegant way to bring back those good ol' equations. Nice! Kyouhei James · 2 years, 5 months ago

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@Kyouhei James Thank you! Steven Zheng · 2 years, 5 months ago

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