Infinite Square Well

## **Problem**

Consider the potential that is 0 within the region

0
<
x
<
L

{\displaystyle 0 < x < L }

 and infinite elsewhere. This potential is called the **infinite square well**.

**Part 1:** Solve the time-independent Schrödinger equation for this potential with boundary conditions:

Ψ
(
0
# )
Ψ
(
L
# )
0

{\displaystyle \Psi(0) = \Psi(L) = 0 }

. Make sure to normalize the wavefunction

Ψ
(
x
)

{\displaystyle \Psi(x)}

.

**Part 2:** Now that you have found the solution to the particle in a box, find the set of permissible eigenenergies (energy levels). Remember that the wavenumber of a matter-wave is given by

# k

n
π

L

{\displaystyle k = \frac{n \pi}{L}}

 where

# n
1
,
2
,
3
,
.
.
.

{\displaystyle n=1,2,3,...}

. This will help you simplify

E

n

{\displaystyle {E}\_{n} }

.

**Part 3:** Determine the time-evolution of the stationary states in the form

Ψ
(
x
,
t
# )
Ψ
(
x
,
0
)

e

−
i

E
n

t

/

ℏ

{\displaystyle \Psi(x,t) = \Psi(x,0) {e}^{-iE\_n t/\hbar} }

.

## **Solution**

**Part 1**

First, note that the particle cannot be outside the infinite square well because there is no way to give it infinite energy. So the wavefunction must vanish there. This means there is zero probability of finding the particle in those regions.

Inside the well, the time-independent Schrödinger equation is

−

ℏ

2

2
m

d

2

Ψ

d

x

2

=
E
Ψ
(
x
)

{\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\Psi }{dx^{2}}}=E\Psi (x)}

subject to the boundary conditions

Ψ
(
0
,
t
# )
Ψ
(
L
,
t
# )
0

{\displaystyle \Psi (0,t)=\Psi (L,t)=0}

. Let

# k

2
m
E

ℏ

2

{\displaystyle k={\sqrt {\frac {2mE}{\hbar ^{2}}}}}

, the ordinary differential equation (ODE) simplifies to

d

2

Ψ

d

x

2

=
−

k

2

Ψ

{\displaystyle {\frac {d^{2}\Psi }{dx^{2}}}=-k^{2}\Psi }

The solution to this problem is now trivial (for people who have learned ODEs):

Ψ
(
x
# )
A
sin
⁡
(
k
x
)
B
cos
⁡
(
k
x
)
.

{\displaystyle \Psi (x)=A\sin(kx)+B\cos(kx).}

Applying the boundary condition, we get

Ψ
(
0
)

=
A
sin
⁡
(
0
)
B
cos
⁡
(
0
)

Ψ
(
L
)

=
A
sin
⁡
(
k
L
)
B
cos
⁡
(
k
L
)
.

{\displaystyle {\begin{aligned}\Psi (0)&=A\sin(0)+B\cos(0)\\[5pt]\Psi (L)&=A\sin(kL)+B\cos(kL).\end{aligned}}}

The first boundary condition implies that

# B
0

{\displaystyle B = 0}

; the second boundary condition gives

sin
⁡
(
k
L
# )
0

{\displaystyle \sin(kL)=0}

. Therefore,

Ψ
(
x
# )
A
sin
⁡

(

n
π
x

L

)

{\displaystyle \Psi (x)=A\sin \left({\frac {n\pi x}{L}}\right)}

where

k
# L
n
π

{\displaystyle kL=n\pi }

, where

# n
1
,
2
,
3
,
.
.
.

{\displaystyle n=1,2,3,...}

 (the positive integers). Note that we omit

# n
0

{\displaystyle n=0}

 because the ground state is the minimum non-zero energy (more on this in Part 3).

Normalize the wavefunction by calculating the integral

∫

0

L

A

2

sin

2

(

n
π
x

L

)

d
# x
1

{\displaystyle \int \_{0}^{L}{{\left|A\right|}^{2}}{\sin }^{2}\left({\frac {n\pi x}{L}}\right)dx=1}

and by applying the boundary condition

Ψ
(
0
# )
Ψ
(
L
# )
0

{\displaystyle \Psi(0) = \Psi(L) = 0 }

. Subsequently,

1

A

2

=

∫

0

L

1
2

[

1
−
cos
⁡

(

2
π
n
x

L

)

]

d
x

{\displaystyle {\frac {1}{A^{2}}}=\int \_{0}^{L}{\frac {1}{2}}\left[1-\cos \left({\frac {2\pi nx}{L}}\right)\right]dx}

1

A

2

=

1
2

[

x
−

L

2
π
n

sin
⁡

(

2
π
n
x

L

)

]

0

L

{\displaystyle {\frac {1}{A^{2}}}={\frac {1}{2}}\left[x-{\frac {L}{2\pi n}}\sin \left({\frac {2\pi nx}{L}}\right)\right]\_{0}^{L}}

1

A

2

=

2
L

{\displaystyle {\frac {1}{A^{2}}}={\frac {2}{L}}}

# A

2
L

.

{\displaystyle A={\sqrt {\frac {2}{L}}}.}

Therefore, the solution is

Ψ
(
x
# )

2
L

sin
⁡

(

n
π
x

L

)

.

{\displaystyle \Psi (x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right).}

**Part 2**

Since

# k

2
m
E

ℏ

2

{\displaystyle k={\sqrt {\frac {2mE}{\hbar ^{2}}}}}

, and

k
# L
n
π

{\displaystyle kL=n\pi }

, where

# n
1
,
2
,
3
,
.
.
.

{\displaystyle n=1,2,3,...}

, we find that the permissible eigenenergies are

E

n

=

π

2

ℏ

2

n

2

2
m

L

2

.

{\displaystyle E\_{n}={\frac {\pi ^{2}\hbar ^{2}n^{2}}{2mL^{2}}}.}

**Part 3**

Since no forces act upon the particle, the part of the wavefunction inside the well oscillates through space and time as a free particle. The time-evolution solution follows the property

Ψ
(
x
,
t
# )
Ψ
(
x
,
0
)

e

−
i

E
n

t

/

ℏ

{\displaystyle \Psi(x,t) = \Psi(x,0) {e}^{-iE\_n t/\hbar} }

hence,

Ψ
(
x
,
t
# )

2
L

sin
⁡

(

n
π
x

L

)

e

−
i

E

n

t

/

ℏ

.

{\displaystyle \Psi (x,t)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)e^{-iE\_{n}t/\hbar }.}