3D Quantum Particle in a Box

## **Problem**

Imagine a box with zero potential enclosed in dimensions

(

0
<
x
<
a

)

,

(

0
<
y
<
b

)

,

(

0
<
z
<
c

)

{\displaystyle \left(0 < x < a \right), \left(0 < y < b \right), \left(0 < z < c \right) }

. Outside the box is the region where the particle’s wavefunction does not exist. Hence, the potential outside the box be infinite. Obtain the wavefunction of the particle in the box where the potential function is

# V

{

0

inside the box

∞

outside the box

{\displaystyle V = \begin{cases} 0\quad \quad \text{inside the box} \\ \infty \quad \text{outside the box} \end{cases} }

subject to the boundary conditions

{

Ψ
(
0
,
y
,
z
# )
Ψ
(
a
,
y
,
z
# )
0

Ψ
(
x
,
0
,
z
# )
Ψ
(
x
,
b
,
z
# )
0

Ψ
(
x
,
y
,
0
# )
Ψ
(
x
,
y
,
c
# )
0

.

{\displaystyle
\begin{cases}
\Psi(0,y,z) = \Psi(a,y,z) = 0 \\[5 pt]
\Psi(x,0,z) = \Psi(x,b,z) = 0 \\[5 pt]
\Psi(x,y,0) = \Psi(x,y,c) = 0 \\[5 pt]
\end{cases}.
}

Obtain the time-independent wavefunction of the particle in the box.

## **Solution**

In quantum mechanics, wavefunctions are found by solving the Schrödinger equation.

E
Ψ
(
x
,
y
,
z
# )
−

ℏ
2

2
m

[

∂

2

∂

x

2

∂

2

∂

y

2

∂

2

∂

y

2

]

Ψ
(
x
,
y
,
z
)
V
Ψ
(
x
,
y
,
z
)

{\displaystyle
E \Psi (x,y,z) = -\frac{\hbar^2}{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)
}

Use the separation of variables method to solve this partial differential equation (PDE). Consider the ansatz

Ψ
(
x
,
y
,
z
# )
X
(
x
)
Y
(
y
)
Z
(
z
)

{\displaystyle
\Psi(x,y,z) = X(x)Y(y)Z(z)
}

with the partition of the total energy into its rightful components

# E

E

x

E

y

E

z

.

{\displaystyle
E = {E}\_{x} + {E}\_{y} + {E}\_{z}.
}

Decompose the PDE into three ordinary differential equations (ODE):

d

2

X

d

x

2

k

x

2

X

=
0

[
10
p
t
]

d

2

Y

d

y

2

k

y

2

Y

=
0

[
10
p
t
]

d

2

Z

d

z

2

k

z

2

Z

=
0

{\displaystyle
\begin{align}
\frac{{d}^{2}X}{d{x}^{2}} + {k}\_{x}^{2}X &= 0 \\ [10 pt]
\frac{{d}^{2}Y}{d{y}^{2}} + {k}\_{y}^{2}Y &= 0 \\ [10 pt]
\frac{{d}^{2}Z}{d{z}^{2}} + {k}\_{z}^{2}Z &= 0
\end{align}
}

where

k

x

=

2
m

E

x

ℏ

2

=

n
π

a

{\displaystyle {k}\_{x}=\sqrt{\frac{2m{E}\_{x}}{{\hbar}^{2}}} =\frac{n \pi}{a} }

 is the wavenumber, and

n

{\displaystyle n}

 is the mode. Likewise,

k

y

=

2
m

E

y

ℏ

2

=

n
π

b

{\displaystyle
{k}\_{y}=\sqrt{\frac{2m{E}\_{y}}{{\hbar}^{2}}} =\frac{n \pi}{b}
}

k

z

=

2
m

E

z

ℏ

2

=

n
π

c

{\displaystyle
{k}\_{z}=\sqrt{\frac{2m{E}\_{z}}{{\hbar}^{2}}} =\frac{n \pi}{c}
}

The new boundary conditions for the decomposed ODEs is:

{

X
(
0
# )
X
(
a
# )
0

Y
(
0
# )
Y
(
b
# )
0

Z
(
0
# )
Z
(
c
# )
0

.

{\displaystyle
\begin{cases}
X(0) = X(a) = 0 \\[5 pt]
Y(0) = Y(b) = 0 \\[5 pt]
Z(0) = Z(c) = 0 \\[5 pt]
\end{cases}.
}

Solve the individual ODEs:

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
)

{\displaystyle
\begin{align}
X(x) &= {A}\_{x}\sin({k}\_{x}x) + {B}\_{x}\cos({k}\_{x}x) \\[10 pt]
Y(y) &= {A}\_{y}\sin({k}\_{y}y) + {B}\_{y}\cos({k}\_{y}y) \\[10 pt]
Z(z) &= {A}\_{z}\sin({k}\_{z}z) + {B}\_{z}\cos({k}\_{z}z)

\end{align}
}

Apply the boundary conditions for

X
,
Y
,
Z

{\displaystyle X, Y, Z}

 to the solutions to arrive at the conclusion

B

x

=

B

y

=

B

z

=

{\displaystyle
{B}\_{x} = {B}\_{y} = {B}\_{z} = 0.
}

Normalize the wavefunctions by performing the integral

∫

0

a

A

x

2

sin

2

(

k

x

x

)

d
# x
1

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

and apply the boundary condition

X
(
0
# )
X
(
a
# )
0

{\displaystyle X(0) = X(a) = 0 }

. Thus,

A

x

=

2
a

{\displaystyle {A}\_{x} = \sqrt{\frac{2}{a}} }

. Repeat the normalization procedure for the

y

{\displaystyle y}

 and

z

{\displaystyle z}

 components to get

A

y

=

2
b

{\displaystyle {A}\_{y} = \sqrt{\frac{2}{b}} }

 and

A

z

=

2
c

{\displaystyle {A}\_{z} = \sqrt{\frac{2}{c}} }

. Lastly, combine the solutions of all components of the ansatz to assemble the wavefunction

Ψ
(
x
,
y
,
z
# )

8

a
b
c

sin
⁡

(

n
π
x

a

)

sin
⁡

(

n
π
y

b

)

sin
⁡

(

n
π
z

c

)

.

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