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<a0<x<a, 0<y<b0<y<b, 0<z<c0<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).

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

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

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

EΨ(x,y,z)=2m[2x2+2y2+2y2]Ψ(x,y,z)+VΨ(x,y,z)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: Ψ(x,y,z)=X(x)Y(y)Z(z)\Psi(x,y,z) = X(x)Y(y)Z(z)

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

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

d2Xdx2+kx2X=0\frac{{d}^{2}X}{d{x}^{2}} + {k}_{x}^{2}X=0

where kx=2mEx2=nπa{k}_{x}=\sqrt{\frac{2m{E}_{x}}{{\hbar}^{2}}} =\frac{n \pi}{a} is the wavenumber, where nn is the mode.

For the yy and zz variables, we replace the above equation with YY and ZZ. Likewise,

ky=2mEy2=nπb{k}_{y}=\sqrt{\frac{2m{E}_{y}}{{\hbar}^{2}}} =\frac{n \pi}{b} kz=2mEz2=nπc.{k}_{z}=\sqrt{\frac{2m{E}_{z}}{{\hbar}^{2}}} =\frac{n \pi}{c}.

Solving the individual ODEs we find that

X(x)=Axsin(kxx)+Bxcos(kxx)X(x) = {A}_{x}sin({k}_{x}x)+ {B}_{x}cos({k}_{x}x) Y(y)=Aysin(kyy)+Bycos(kyy)Y(y) = {A}_{y}sin({k}_{y}y)+ {B}_{y}cos({k}_{y}y) Z(z)=Azsin(kzz)+Bzcos(kzz).Z(z) = {A}_{z}sin({k}_{z}z)+ {B}_{z}cos({k}_{z}z).

By the boundary conditions:

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

hence

Bx=0{B}_{x}=0 By=0{B}_{y}=0 Bz=0.{B}_{z}=0.

We normalize the wavefunctions by performing the integral

0aAx2sin2(kxx)dx=1\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)=0X(0)=X(a)=0.

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

Combine the solutions of all components to the ansatz yields

Ψ(x,y,z)=8abcsin(nπax)sin(nπby)sin(nπcz).\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
4 years, 10 months ago

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thanksssssssssssssssssssssssssssssssssssssssss

Himanshu Tuteja - 4 years, 10 months ago

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You're welcome!

Steven Zheng - 4 years, 10 months ago

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

Kyouhei James - 4 years, 10 months ago

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Thank you!

Steven Zheng - 4 years, 10 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 - 4 years, 8 months ago

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What's a KTOG problem?

Steven Zheng - 4 years, 8 months ago

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Kinetic Theory Of Gases......

manish bhargao - 4 years, 8 months ago

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ohhh. that makes sense. lol

greg nye - 4 years, 8 months ago

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